Cram Fast In Time Of Exams

Cramming for exams should be avoided at all costs. You should only cram for an exam as a last resort. It's hard to take in and retain a large amount of information in a short period of time. Some of the tips on studying and preparing for a test may overlap with the cramming techniques below.
Eat some food to give you energy to study but avoid consuming excess sugar which will make you hyper and will make it more difficult to study.
An apple does a better job at keeping you focused and awake than caffeine.
Find a well lit place with no distractions around to study but don't get too comfortable or you may fall asleep.
Keep a positive attitude, it is easier to study when you are relaxed than when you are stressed out.
Since your time is limited, you have to choose what you study. Don't attempt to learn everything, focus on things that will get you the most points on the exam.
Focus on the main ideas and learn key formulas. Skip the details for now and only come back to them if you see that you have time after you have learned the key points.
Write down the key ideas/formulas on a sheet of paper and keep on studying from that sheet, repetition is important.
Highlight the important points in your notes, and text and focus on that.
Read the chapter summaries (they usually do a good job at summarizing the important points). If there're no chapter summaries then skim through the text and write down key ideas.
Study from past tests, review questions, homework & review sheets.
Take at least one five minute break an hour so that you can gather your thoughts and let your brain relax.
If time permits, try to get at least 3 hours of sleep (one sleep cycle) before the exam so that you don't fall asleep when taking your exam. Don't forget to set your alarm!

Chapter - Electric Charges And Fields Class - 12

Question 1.1:
What is the force between two small charged spheres having charges of 2 × 10
× 10−7 C placed 30 cm apart in air?
Answer
Repulsive force of magnitude 6 × 10
Charge on the first sphere, q1
Charge on the second sphere,
Distance between the spheres,
Electrostatic force between the spheres is given by the relation,
Where, ∈0 = Permittivity of free space
Hence, force between the two small charged spheres is 6 × 10
same nature. Hence, force between them will be repulsive.
Question 1.2:
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of
charge − 0.8 μC in air is 0.2 N. (a) What is the distance betwe
What is the force on the second sphere due to the first?
Answer
10−3 N
1 = 2 × 10−7 C
econd q2 = 3 × 10−7 C
r = 30 cm = 0.3 m
10−3 N. The charges are of
between the two spheres? (b)
10−7 C and 3
en
Electrostatic force on the first sphere,
Charge on this sphere, q1 = 0.4 μC = 0.4
Charge on the second sphere,
Electrostatic force between the spheres is given by the relation,
Where, ∈0 = Permittivity of free space
The distance between the two spheres is 0.12 m.
Both the spheres attract each other
second sphere due to the first is 0.2 N.
Question 1.3:
Check that the ratio ke2/G me
and determine the value of this ratio. What does the
Answer
F = 0.2 N
× 10−6 C
q2 = − 0.8 μC = − 0.8 × 10−6 C
with the same force. Therefore, the force on the
emp is dimensionless. Look up a Table of Physical Constants
ratio signify?
The given ratio is .
Where,
G = Gravitational constant
Its unit is N m2 kg−2.
me and mp = Masses of electron and proton.
Their unit is kg.
e = Electric charge.
Its unit is C.
∈0 = Permittivity of free space
Its unit is N m2 C−2.
Hence, the given ratio is dimensionless.
e = 1.6 × 10−19 C
G = 6.67 × 10−11 N m2 kg-2
me= 9.1 × 10−31 kg
mp = 1.66 × 10−27 kg
Hence, the numerical value of the given ratio is
This is the ratio of electric force to the gravitational force between a proton and an
electron, keeping distance between them constant.
Question 1.4:
Explain the meaning of the statement ‘electric charge of a body is quantised’.
Why can one ignore quantisation of electric charge when dealing with macroscopic i.e.,
large scale charges?
Answer
Electric charge of a body is quantized. This means that only integral (1, 2, ….,
of electrons can be transferred from one body to the other. Charges are not transferred in
fraction. Hence, a body possesses total
charge.
In macroscopic or large scale charges, the charges used are huge as compared to the
magnitude of electric charge. Hence, quantization of electric charge is of no use on
macroscopic scale. Therefore, it
continuous.
Question 1.5:
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar
phenomenon is observed with many other pairs of bodies. Explain how this observation
consistent with the law of conservation of charge.
Answer
Rubbing produces charges of equal magnitude but of opposite nature on the two bodies
because charges are created in pairs. This phenomenon of charging is called charging by
friction. The net charge on the system of two rubbed bodies is zero. This is because equal
amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk
cloth, opposite natured charges appear on both the bodies. This phenomenon is in
consistence with the law of conservation of energy. A similar phenomenon is observed
with many other pairs of bodies.
charge only in integral multiples of electric
is ignored and it is considered that electric charge is
et e n) number
is
Question 1.6:
Four point charges qA = 2 μC,
corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at
the centre of the square?
Answer
The given figure shows a square of side 10 cm with four charges placed at its corners. O
is the centre of the square.
Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD =
AO = OC = DO = OB =
A charge of amount 1μC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude
but opposite in direction relative to the force of repulsion between the charges placed at
corner C and centre O. Hence, they will cancel each other. Similarly, force
between charges placed at corner B and centre O is equal in magnitude but opposite in
direction relative to the force of attraction between the charges placed at corner D and
centre O. Hence, they will also cancel each other. Therefore, net f
charges placed at the corner of the square on 1 μC charge at centre O is zero.
qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the
cm
cm
of attraction
force caused by the four
orce
Question 1.7:
An electrostatic field line is a continuous curve. That is, a field line cannot have sudden
breaks. Why not?
Explain why two field lines never cross each other at any point?
Answer
An electrostatic field line is a continuous curve because a charge experiences a
continuous force when traced in an electrostatic field. The field line cannot have sudden
breaks because the charge mo
other.
If two field lines cross each other at a point, then electric field intensity will show two
directions at that point. This is not possible. Hence, two field lines never cross each other.
Question 1.8:
Two point charges qA = 3 μC and
What is the electric field at the midpoint O of the line AB joining the two charges?
If a negative test charge of magnitude 1.5 × 10
force experienced by the test charge?
Answer
The situation is represented in the given figure. O is the mid
Distance between the two charges, AB = 20 cm
moves continuously and does not jump from one point to the
qB = −3 μC are located 20 cm apart in vacuum.
10−9 C is placed at this point, what is the
mid-point of line AB.
ves
∴AO = OB = 10 cm
Net electric field at point O = E
Electric field at point O caused by +3μC charge,
E1 = along OB
Where,
= Permittivity of free space
Magnitude of electric field at point O caused by −3μC charge,
E2 = = along OB
= 5.4 × 106 N/C along OB
Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.
A test charge of amount 1.5 × 10−9 C is placed at mid-point O.
q = 1.5 × 10−9 C
Force experienced by the test charge = F
∴F = qE
= 1.5 × 10−9 × 5.4 × 106
= 8.1 × 10−3 N
The force is directed along line OA. This is because the negative test charge is repelled by
the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 × 10
Question 1.9:
A system has two charges qA
(0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric
dipole moment of the system?
Answer
Both the charges can be located in a coordinate frame of r
figure.
At A, amount of charge, qA = 2.5 × 10
At B, amount of charge, qB =
Total charge of the system,
q = qA + qB
= 2.5 × 107 C − 2.5 × 10−7 C
= 0
Distance between two charges at points A and B,
d = 15 + 15 = 30 cm = 0.3 m
Electric dipole moment of the system is given by,
10−3 N along OA.
A = 2.5 × 10−7 C and qB = −2.5 × 10−7 C located at points A:
reference as shown in the given
10−7C
−2.5 × 10−7 C
eference
p = qA × d = qB × d
= 2.5 × 10−7 × 0.3
= 7.5 × 10−8 C m along positive
Therefore, the electric dipole moment of the system is 7.5 × 10
z−axis.
Question 1.10:
An electric dipole with dipole moment 4 × 10
a uniform electric field of magnitude 5 × 10
acting on the dipole.
Answer
Electric dipole moment, p = 4 × 10
Angle made by p with a uniform electric field,
Electric field, E = 5 × 104 N C
Torque acting on the dipole is given by the relation,
τ = pE sinθ
Therefore, the magnitude of the torque acting
Question 1.11:
A polythene piece rubbed with wool is found to have a negative charge of 3 × 10
Estimate the number of electrons transferred (from which to which?)
z-axis
10−8 C m along positive
10−9 C m is aligned at 30° with the direction of
104 N C−1. Calculate the magnitude of the torque
10−9 C m
θ = 30°
C−1
on the dipole is 10−4 N m.
. 10−7 C.
Is there a transfer of mass from wool to polythene?
Answer
When polythene is rubbed against wool, a number of electrons get transferred from wool
to polythene. Hence, wool becomes positively charged and polythene becomes negatively
charged.
Amount of charge on the polythene piece, q = −3 × 10−7 C
Amount of charge on an electron, e = −1.6 × 10−19 C
Number of electrons transferred from wool to polythene = n
n can be calculated using the relation,
q = ne
= 1.87 × 1012
Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012.
Yes.
There is a transfer of mass taking place. This is because an electron has mass,
me = 9.1 × 10−3 kg
Total mass transferred to polythene from wool,
m = me × n
= 9.1 × 10−31 × 1.85 × 1012
= 1.706 × 10−18 kg
Hence, a negligible amount of mass is transferred from wool to polythene.
Question 1.12:
Two insulated charged copper spheres A and B have their centers separated by a distance
of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 ×
10−7 C? The radii of A and B are negligible compared to the distance of separation.
What is the force of repulsion if each sphere is charged double the above amount, and the
distance between them is halved?
Answer
Charge on sphere A, qA = Charge on sphere B,
Distance between the spheres,
Force of repulsion between the two spheres,
Where,
∈0 = Free space permittivity
= 9 × 109 N m2 C−2
∴
= 1.52 × 10−2 N
Therefore, the force between the two spheres
After doubling the charge, charge on sphere A,
10−7 C = 1.3 × 10−6 C
The distance between the spheres is halved.
qB = 6.5 × 10−7 C
r = 50 cm = 0.5 m
is 1.52 × 10−2 N.
qA = Charge on sphere B, qB
= 2 × 6.5 ×
∴
Force of repulsion between the two spheres,
= 16 × 1.52 × 10−2
= 0.243 N
Therefore, the force between the two spheres is 0.243 N.
Question 1.13:
Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the
same size but uncharged is brought in contact with the first, then brought in contact w
the second, and finally removed from both. What is the new force of repulsion between A
and B?
Answer
Distance between the spheres, A and B,
Initially, the charge on each sphere,
When sphere A is touched with an uncharged sphere C,
transfer to sphere C. Hence, charge on each of the spheres, A and C, is
When sphere C with charge
charges on the system will divide into two equal halves given as,
r = 0.5 m
q = 6.5 × 10−7 C
amount of charge from A will
.
is brought in contact with sphere B with charge
with
q, total
Each sphere will each half. Hence, charge on each of the spheres, C and B, is
Force of repulsion between sphere A having charge
Therefore, the force of attraction bet
Question 1.14:
Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give
the signs of the three charges. Which particle has the highest charge to mass ratio?
Answer
Opposite charges attract each other and same charges repel each other. It can be observed
that particles 1 and 2 both move towards the positively charged plate and repel away from
the negatively charged plate. Hence, these two particles are neg
also be observed that particle 3 moves towards the negatively charged plate and repels
away from the positively charged plate. Hence, particle 3 is positively charged.
and sphere B having charge
between the two spheres is 5.703 × 10−3 N.
negatively charged. It can
.
=
atively
The charge to mass ratio (emf) is directly proportional to the d
deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has
the highest charge to mass ratio.
Question 1.15:
Consider a uniform electric field
through a square of 10 cm on a side whose plane is parallel to the
the flux through the same square if the normal to its plane makes a 60° angle with the
axis?
Answer
Electric field intensity, = 3 × 10
Magnitude of electric field intensity,
Side of the square, s = 10 cm = 0.1 m
Area of the square, A = s2 = 0.01 m
The plane of the square is parallel to the
normal to the plane and electric field,
Flux (Φ) through the plane is given by the relation,
Φ =
= 3 × 103 × 0.01 × cos0°
= 30 N m2/C
Plane makes an angle of 60° with the
Flux, Φ =
= 3 × 103 × 0.01 × cos60°
displacement or amount of
E = 3 × 103 îN/C. (a) What is the flux of this field
yz plane? (b) What is
103 î N/C
= 3 × 103 N/C
m2
y-z plane. Hence, angle between the unit vector
θ = 0°
) x-axis. Hence, θ = 60°
isplacement /xplane.
= 15 N m2/C
Question 1.16:
What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side
20 cm oriented so that its faces are parallel to the coordinate planes?
Answer
All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field
lines entering the cube is equal to the number of field lines piercing out of the cube. As a
result, net flux through the cube is zero.
Question 1.17:
Careful measurement of the electric field at the surface of a black box indicates that the
net outward flux through the surface of the box is 8.0 × 10
charge inside the box? (b) If the net outward flux through the surfac
zero, could you conclude that there were no charges inside the box? Why or Why not?
Answer
Net outward flux through the surface of the box,
For a body containing net charge
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
103 N m2/C. (a) What is the net
surface of the box were
Φ = 8.0 × 103 N m2/C
q, flux is given by the relation,
/e
q = ∈0Φ
= 8.854 × 10−12 × 8.0 × 103
= 7.08 × 10−8
= 0.07 μC
Therefore, the net charge inside the box is 0.07 μC.
No
Net flux piercing out through a body depends on the
net flux is zero, then it can be inferred that net charge inside the body is zero. The body
may have equal amount of positive and negative charges.
Question 1.18:
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10
cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square?
(Hint: Think of the square as one face of a cube with edge 10 cm.)
Answer
The square can be considered as one face of a cube of edge 10 cm with a centre where
charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through
all its six faces.
net charge contained in the body. If
Hence, electric flux through one face of the c
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
q = 10 μC = 10 × 10−6 C
∴
= 1.88 × 105 N m2 C−1
Therefore, electric flux through the square is 1.88 × 10
Question 1.19:
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge.
What is the net electric flux through the surface?
Answer
Net electric flux (ΦNet) through the cubic surface is given by,
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
q = Net charge contained inside the cube = 2.0 μC = 2
cube i.e., through the square,
105 N m2 C−1.
) × 10−6 C
∴
= 2.26 × 105 N m2 C−1
The net electric flux through the surface is 2.26 ×10
Question 1.20:
A point charge causes an electric flux of
Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the
Gaussian surface were doubled, how much flux would pass through the surface? (b) What
is the value of the point charge?
Answer
Electric flux, Φ = −1.0 × 103
Radius of the Gaussian surface,
r = 10.0 cm
Electric flux piercing out through a surface depends on the net charge enclosed inside a
body. It does not depend on the size of the body. If the radius of the Gaussian surface is
doubled, then the flux passing through the surface remains the same i.e.,
Electric flux is given by the relation,
Where,
q = Net charge enclosed by the spherical surface
∈0 = Permittivity of free space = 8.854 × 10
∴
= −1.0 × 103 × 8.854 × 10−12
×105 N m2C−1.
−1.0 × 103 Nm2/C to pass through a spherical
N m2/C
−10
10−12 N−1C2 m−2
/103 N m2/C.
= −8.854 × 10−9 C
= −8.854 nC
Therefore, the value of the point charge is
Question 1.21:
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm
from the centre of the sphere is 1.5 × 10
charge on the sphere?
Answer
Electric field intensity (E) at a distance (
charge q is given by the relation,
Where,
q = Net charge = 1.5 × 103 N/C
d = Distance from the centre = 20 cm = 0.2 m
∈0 = Permittivity of free space
And, = 9 × 109 N m2 C
∴
= 6.67 × 109 C
= 6.67 nC
−8.854 nC.
103 N/C and points radially inward, what is the net
) d) from the centre of a sphere containing net
C−2
)
Therefore, the net charge on the sphere is 6.67 nC.
Question 1.22:
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of
80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving
the surface of the sphere?
Answer
Diameter of the sphere, d = 2.4 m
Radius of the sphere, r = 1.2 m
Surface charge density, = 80.0 μC/m
Total charge on the surface of the sphere,
Q = Charge density × Surface area
=
= 80 × 10−6 × 4 × 3.14 × (1.2)
= 1.447 × 10−3 C
Therefore, the charge on the sphere is 1
Total electric flux ( ) leaving out the surface of a sphere containing net charge
given by the relation,
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
. m2 = 80 × 10−6 C/m2
2
1.447 × 10−3 C.
Q is
Q = 1.447 × 10−3 C
= 1.63 × 108 N C−1 m2
Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10
Question 1.23:
An infinite line charge produces a field of 9 × 10
linear charge density.
Answer
Electric field produced by the infinite line charges at a distance
density λ is given by the relation,
Where,
d = 2 cm = 0.02 m
E = 9 × 104 N/C
∈0 = Permittivity of free space
= 9 × 109 N m2 C−2
= 10 μC/m
104 N/C at a distance of 2 cm. Calculate the
d having linear charge
108 N C−1 m2.
Therefore, the linear charge density is 10 μC/m.
Question 1.24:
Two large, thin metal plates are parallel and close to each other. On their inner faces, the
plates have surface charge densities of opposite signs and of magnitude 17.0 × 10
C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the
second plate, and (c) between the plates?
Answer
The situation is represented in the following figure.
A and B are two parallel plates close to each other. O
I, outer region of plate B is labelled as
labelled as II.
Charge density of plate A, σ = 17.0 × 10
Charge density of plate B, σ =
In the regions, I and III, electric field
the respective plates.
Electric field E in region II is given by the relation,
Where,
∈0 = Permittivity of free space = 8.854 × 10
: Outer region of plate A is labelled as
, III, and the region between the plates, A and B, is
10−22 C/m2
−17.0 × 10−22 C/m2
, E is zero. This is because charge is not enclosed by
10−12 N−1C2 m−2
10−22
uter ,
∴
= 1.92 × 10−10 N/C
Therefore, electric field between the plates is 1.92 × 10
Question 1.25:
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55
× 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm
Estimate the radius of the drop. (
Answer
Excess electrons on an oil drop,
Electric field intensity, E = 2.55 × 10
Density of oil, ρ = 1.26 gm/cm
Acceleration due to gravity, g = 9.81 m s
Charge on an electron, e = 1.6 × 10
Radius of the oil drop = r
Force (F) due to electric field
F = W
Eq = mg
Ene
Where,
q = Net charge on the oil drop =
m = Mass of the oil drop
10−10 N/C.
g = 9.81 m s−2; e = 1.60 × 10−19 C).
n = 12
104 N C−1
cm3 = 1.26 × 103 kg/m3
s−2
10−19 C
) E is equal to the weight of the oil drop (W)
ne
cm−3.
= Volume of the oil drop × Density of oil
= 9.82 × 10−4 mm
Therefore, the radius of the oil drop is 9.82 × 10
Question 1.26:
Which among the curves shown in Fig. 1.35
lines?
(a)
(b)
10−4 mm.
ccaannnnoott ppoossssiibbllyy rreepprreesseenntt eelleeccttrroossttaattiicc ffiieelldd
(c)
(d)
(e)
Answer
The field lines showed in (a) do not represent electrostatic field lines because field lines
must be normal to the surface of the conductor.
The field lines showed in (b) do not represent electrostatic field lines because the field
lines cannot emerge from a negative charge and cannot terminate at a positive charge.
The field lines showed in (c) represent electrostatic field lines. This is because the field
lines emerge from the positive charges and repel each other.
The field lines showed in (d) do not represent electrostatic field lines because the field
lines should not intersect each other.
The field lines showed in (e) do not represent electrostatic field
are not formed in the area between the field lines.
Question 1.27:
In a certain region of space, electric field is along the z
magnitude of electric field is, however, not constant but increases
positive z-direction, at the rate of 10
experienced by a system having a total dipole moment equal to 10
direction?
Answer
Dipole moment of the system,
Rate of increase of electric field per unit length,
Force (F) experienced by the system is given by the relation,
F = qE
= −10−7 × 10−5
= −10−2 N
The force is −10−2 N in the negative z
field. Hence, the angle between electric field and dipole moment is 180°.
Torque (τ) is given by the relation,
lines because closed loops
z-direction throughout. The
uniformly along the
105 NC−1 per metre. What are the force and torque
10−7 Cm in the negative
p = q × dl = −10−7 C m
) z-direction i.e., opposite to the direction of electric
) zdirection
τ = pE sin180°
= 0
Therefore, the torque experienced by the system is
Question 1.28:
A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge
entire charge must appear on the outer surface of the conductor. (b) Another conductor B
with charge q is inserted into the cavity keeping
charge on the outside surface of A is
be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Answer
Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing
the cavity. The electric field intensity
Let q is the charge inside the conductor and
According to Gauss’s law,
Flux,
Here, E = 0
zero.
Q. Show that the
B insulated from A. Show that the total
Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to
E inside the charged conductor is zero.
is the permittivity of free space.
.
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
The outer surface of conductor A has a charge of amount
charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of
amount −q will be induced in the inner surface of conductor A and +
outer surface of conductor A. Therefore, total charge on the outer surface of conductor A
is Q + q.
A sensitive instrument can be shielded from the strong electrostatic field in its
environment by enclosing it fully inside a metallic surface. A closed metallic body acts as
an electrostatic shield.
Question 1.29:
A hollow charged conductor
field in the hole is
and is the surface charge density near the hole.
Answer
Let us consider a conductor with a cavity or a
Let E is the electric field just outside the conductor,
charge density, and is the permittivity of free space.
Charge
According to Gauss’s law,
Q. Another conductor B having
q is induced on the
has a tiny hole cut into its surface. Show that the electric
, where is the unit vector in the outward normal direction,
hole. Electric field inside the cavity is zero.
q is the electric charge,
. is the
Therefore, the electric field just outside the conductor is
superposition of field due to the cavity
conductor . These fields are equal and opposite inside the conductor, and equal in
magnitude and direction outside
Therefore, the field due to the rest of the conductor is
Hence, proved.
Question 1.30:
Obtain the formula for the electric field due to a long thin wire of uniform linear charge
density λ without using Gauss’s law. [
necessary integral.]
Answer
Take a long thin wire XY (as shown in the figure) of uniform linear charge density
Consider a point A at a perpendicular distance
shown in the following figure.
. This field is a
and the field due to the rest of the charged
the conductor.
.
Hint: Use Coulomb’s law directly and evaluate the
l from the mid-point O of the wire, as
.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x
Let q be the charge on this piece.
Electric field due to the piece,
The electric field is resolved into two rectangular components. is the
perpendicular component and is the parallel component.
When the whole wire is considered, the component is cancelled.
Only the perpendicular component affects point A.
Hence, effective electric field at point A due to the element dx is dE1.
On differentiating equation (2), we obtain
From equation (2),
Putting equations (3) and (4) in equation (1), we obtain
The wire is so long that tends from
By integrating equation (5), we obtain the value of field
Therefore, the electric field due to long wire is
Question 1.31:
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter)
are themselves built out of more elementary units called quarks. A proton and a neutron
consist of three quarks each. Two types of quarks, the so called ‘up’ quark (d
of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (
electrons build up ordinary matter. (Quarks of other types have also been found which
give rise to different unusual varieties of matter.) Suggest a poss
a proton and neutron.
Answer
to .
E1 as,
.
, (−1/3) e, together with
possible quark composition of
denoted by u)
, ible
A proton has three quarks. Let there be n up quarks in a proton, each having a charge of
.
Charge due to n up quarks
Number of down quarks in a proton = 3 − n
Each down quark has a charge of .
Charge due to (3 − n) down quarks
Total charge on a proton = + e
Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3 − n = 3 − 2 = 1
Therefore, a proton can be represented as ‘uud’.
A neutron also has three quarks. Let there be n up quarks in a neutron, each having a
charge of .
Charge on a neutron due to n up quarks
Number of down quarks is 3 − n,each having a charge of .
Charge on a neutron due to
Total charge on a neutron = 0
Number of up quarks in a neutron,
Number of down quarks in a neutron = 3
Therefore, a neutron can be represented as ‘udd’.
Question 1.32:
Consider an arbitrary electrostatic field configuration. A small test charge is placed at a
null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test
charge is necessarily unstable.
Verify this result for the simple configu
sign placed a certain distance apart.
Answer
Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and
displaced from its position in any direction, then it experiences a
null point, where the electric field is zero. All the field lines near the null point are
directed inwards towards the null point. There is a net inward flux of electric field
through a closed surface around the null point. Accor
electric field through a surface, which is not enclosing any charge, is zero. Hence, the
equilibrium of the test charge can be stable.
Two charges of same magnitude and same sign are placed at a certain distance. The mid
point of the joining line of the charges is the null point. When a test charged is displaced
along the line, it experiences a restoring force. If it is displaced normal t
down quarks =
n n = 1
− n = 2
configuration of two charges of the same magnitude and
restoring force towards a
According to Gauss’s law, the flux of
to the joining line,
ration ding midpoint
o
then the net force takes it away from the null point. Hence, the charge is unstable because
stability of equilibrium requires restoring force in all directions.
Question 1.33:
A particle of mass m and charge (
initially moving along x-axis with speed
plate is L and an uniform electric field
vertical deflection of the particle at the far edge of the plate is
Compare this motion with motion of a projectile in gravitational field discussed in
Section 4.10 of Class XI Textbook of Physics.
Answer
Charge on a particle of mass
Velocity of the particle = vx
Length of the plates = L
Magnitude of the uniform electric field between the plates =
Mechanical force, F = Mass (
Therefore, acceleration,
Time taken by the particle to cross the field of length
t
In the vertical direction, initial velocity,
According to the third equation of motion, vertical deflection
obtained as,
(−q) enters the region between the two charged plates
vx (like particle 1 in Fig. 1.33). The length of
E is maintained between the plates. Show that the
he qEL2/ (2m ).
m = − q
E
m) × Acceleration (a)
L is given by,
u = 0
s of the particle can be
)
Hence, vertical deflection of the particle at the far edge of the plate is
. This is similar to the motion of horizontal projectiles under gravity.
Question 1.34:
Suppose that the particle in Exercise in 1.33 is an electron projected with velocity
× 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 10
electron strike the upper plate? (|
Answer
Velocity of the particle, vx = 2.0 × 10
Separation of the two plates,
Electric field between the two plates,
Charge on an electron, q = 1.6 × 10
Mass of an electron, me = 9.1 × 10
Let the electron strike the upper plate at the end of plate
Therefore,
102 N/C, where will the
e | =1.6 × 10−19 C, me = 9.1 × 10−31 kg.)
106 m/s
d = 0.5 cm = 0.005 m
E = 9.1 × 102 N/C
10−19 C
10−31 kg
L, when deflection is
vx= 2.0
, s.
Therefore, the electron will strike the other plate after travelling 11.66 cm.

Chapter - Current Electricity Class 12

Question 3.1:
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is
0.4Ω, what is the maximum current that can be drawn from the battery?
Answer
Emf of the battery, E = 12 V
Internal resistance of the battery,
Maximum current drawn from the battery =
According to Ohm’s law,
The maximum current drawn from the given battery is 30 A.
Question 3.2:
A battery of emf 10 V and internal resistance 3 Ω is connecte
in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage
of the battery when the circuit is closed?
Answer
Emf of the battery, E = 10 V
Internal resistance of the battery,
r = 0.4 Ω
I
connected to a resistor. If the current
r = 3 Ω
d
Current in the circuit, I = 0.5 A
Resistance of the resistor = R
The relation for current using Ohm’s law is,
Terminal voltage of the resistor =
According to Ohm’s law,
V = IR
= 0.5 × 17
= 8.5 V
Therefore, the resistance of the resistor is 17 Ω and the
8.5 V.
Question 3.3:
Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of
the combination?
If the combination is connected to a battery of emf 12 V and negligible internal
resistance, obtain the potential drop across each resistor.
Answer
Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of
the combination is given by the algebraic sum of individual resistances.
V
terminal voltage is
Total resistance = 1 + 2 + 3 = 6 Ω
Current flowing through the circuit =
Emf of the battery, E = 12 V
Total resistance of the circuit,
The relation for current using Ohm’s law is,
Potential drop across 1 Ω resistor =
From Ohm’s law, the value of
V1 = 2 × 1= 2 V … (i)
Potential drop across 2 Ω resistor =
Again, from Ohm’s law, the value of
V2 = 2 × 2= 4 V … (ii)
Potential drop across 3 Ω resistor =
Again, from Ohm’s law, the value of
V3 = 2 × 3= 6 V … (iii)
Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V
respectively.
Question 3.4:
Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of
the combination?
If the combination is connected to a battery of emf 20 V and negligible internal
resistance, determine the current through each resistor, and the total current drawn from
the battery.
Answer
I
R = 6 Ω
V1
V1 can be obtained as
V2
V2 can be obtained as
V3
V3 can be obtained as
There are three resistors of resistances,
R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω
They are connected in parallel. Hence, total resistance (R) of the combination is given by,
Therefore, total resistance of the combination is .
Emf of the battery, V = 20 V
Current (I1) flowing through resistor R1 is given by,
Current (I2) flowing through resistor R2 is given by,
Current (I3) flowing through resistor R3 is given by,
Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A
Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the
total current is 19 A.
Question 3.5:
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the
temperature of the element if the resistance is found to be 117 Ω, given that the
temperature coefficient of the material of the
Answer
Room temperature, T = 27°C
Resistance of the heating element at
Let T1 is the increased temperature of the filament.
Resistance of the heating element at
Temperature co-efficient of the material of the
Therefore, at 1027°C, the resistance of the element is 117Ω.
Question 3.6:
resistor is
T, R = 100 Ω
T1, R1 = 117 Ω
filament,
A negligibly small current is passed through a wire of length 15 m and uniform cross
section 6.0 × 10−7 m2, and its resistance is measured to be 5.0 Ω. What is
the material at the temperature of the experiment?
Answer
Length of the wire, l =15 m
Area of cross-section of the wire,
Resistance of the material of the wire,
Resistivity of the material of the wire =
Resistance is related with the resistivity as
Therefore, the resistivity of the material is 2 × 10
Question 3.7:
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C.
Determine the temperature coefficient of resistivity of silver.
Answer
Temperature, T1 = 27.5°C
Resistance of the silver wire at
, the resistivity of
a = 6.0 × 10−7 m2
R = 5.0 Ω
ρ
10−7 Ω m.
T1, R1 = 2.1 Ω
crossthe
Temperature, T2 = 100°C
Resistance of the silver wire at
Temperature coefficient of silver =
It is related with temperature and resistance as
Therefore, the temperature coefficient of silver is 0.0039°C
Question 3.8:
Aheating element using nichrome connected to a 230 V supply draws an initial current of
3.2 A which settles after a few seconds toa steady
temperature of the heating element if the room temperature is 27.0 °C? Temperature
coefficient of resistance of nichrome averaged over the temperature range involved is
1.70 × 10−4 °C −1.
Answer
Supply voltage, V = 230 V
Initial current drawn, I1 = 3.2 A
Initial resistance = R1, which is given by the relation,
Steady state value of the current,
Resistance at the steady state =
T2, R2 = 2.7 Ω
α
C−1.
value of 2.8 A. What is the steady
, I2 = 2.8 A
R2, which is given as
Temperature co-efficient of nichrome,
Initial temperature of nichrome,
Study state temperature reached by nichrome =
T2 can be obtained by the relation for
Therefore, the steady temperature of the heating element is 867.5°C
Question 3.9:
Determine the current in each branch of the network shown in fig 3.30:
Answer
Current flowing through various branches of the circuit is represented in the given figure.
α = 1.70 × 10−4 °C −1
T1= 27.0°C
T2
α,
CCuurrrreenntt fflloowwiinngg tthhrroouugghh vvaarriioouuss bbrraanncchheess ooff tthhee cciirrccuuiitt iiss rreepprreesseenntteedd iinn tthhee ggiivveenn ffiigguurree..
I1 = Current flowing through the outer circuit
I2 = Current flowing through branch AB
I3 = Current flowing through branch AD
I2 − I4 = Current flowing through branch BC
I3 + I4 = Current flowing through branch CD
I4 = Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
10I2 + 5I4 − 5I3 = 0
2I2 + I4 −I3 = 0
I3 = 2I2 + I4 … (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I2 − I4) − 10(I3 + I4) − 5I4 = 0
5I2 + 5I4 − 10I3 − 10I4 − 5I4 = 0
5I2 − 10I3 − 20I4 = 0
I2 = 2I3 + 4I4 … (2)
For the closed circuit ABCFEA, potential is zero i.e.,
−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0
10 = 15I2 + 10I1 − 5I4
3I2 + 2I1 − I4 = 2 … (3)
From equations (1) and (2), we obtain
I3 = 2(2I3 + 4I4) + I4
I3 = 4I3 + 8I4 + I4
3I3 = 9I4
3I4 = + I3 … (4)
Putting equation (4) in equation (1), we obtain
I3 = 2I2 + I4
4I4 = 2I2
I2 = − 2I4 … (5)
It is evident from the given figure that,
I1 = I3 + I2 … (6)
Putting equation (6) in equation (1), we obtain
3I2 +2(I3 + I2) − I4 = 2
5I2 + 2I3 − I4 = 2 … (7)
Putting equations (4) and (5) in equation (7), we obtain
5(−2 I4) + 2(− 3 I4) − I4 = 2
10I4 − 6I4 − I4 = 2
17I4 = − 2
Equation (4) reduces to
I3 = − 3(I4)
Therefore, current in branch
In branch BC =
In branch CD =
In branch AD
In branch BD =
Total current =
Question 3.10:
In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end
when the resistor Y is of 12.5 Ω. Determine the resistance of
between resistors in a Wheatstone or meter bridge made of thick copper strip
Determine the balance point of the bridge above if
What happens if the galvanometer and cell are interchanged at the balance point of the
bridge? Would the galvanometer show any current?
X. Why are the connections
X and Y are interchanged.
A,
. strips?
Answer
A metre bridge with resistors X and Y is represented in the given figure.
Balance point from end A, l1 = 39.5 cm
Resistance of the resistor Y = 12.5 Ω
Condition for the balance is given as,
Therefore, the resistance of resistor X is 8.2 Ω.
The connection between resistors in a Wheatstone or metre bridge is made of thick copper
strips to minimize the resistance, which is not taken into consideration in the bridge
formula.
If X and Y are interchanged, then l1 and 100−l1 get interchanged.
The balance point of the bridge will be 100−l1 from A.
100−l1 = 100 − 39.5 = 60.5 cm
Therefore, the balance point is 60.5 cm from A.
When the galvanometer and cell are interchanged at the balance point of the bridge, the
galvanometer will show no deflection. Hence, no current would flow through the
galvanometer.
Question 3.11:
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V
dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery
during charging? What is the purpose of having a series resistor in the charging circuit?
Answer
Emf of the storage battery, E
Internal resistance of the battery,
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit =
R is connected to the storage battery in series. Hence, it can be
V1 = V − E
V1 = 120 − 8 = 112 V
Current flowing in the circuit =
Voltage across resistor R given by the product,
DC supply voltage = Terminal voltage of battery + Voltage drop across
Terminal voltage of battery = 120
= 8.0 V
r = 0.5 Ω
V1
written as
I, which is given by the relation,
IR = 7 × 15.5 = 108.5 V
R
− 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
Question 3.12:
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm
length of the wire. If the cell is replaced by another cell and the balance point shifts to
63.0 cm, what is the emf of the second cell?
Answer
Emf of the cell, E1 = 1.25 V
Balance point of the potentiometer,
The cell is replaced by another cell of emf
New balance point of the potentiometer,
Therefore, emf of the second cell is 2.25V.
Question 3.13:
The number density of free elect
8.5 × 1028 m−3. How long does an electron take to drift from one end of a wire 3.0 m long
to its other end? The area of cross
current of 3.0 A.
Answer
l1= 35 cm
E2.
l2 = 63 cm
electrons in a copper conductor estimated in Example 3.1 is
. cross-section of the wire is 2.0 × 10−6 m2 and it is carrying a
rons
Number density of free electrons in a copper conductor,
copper wire, l = 3.0 m
Area of cross-section of the wire,
Current carried by the wire, I
I = nAeVd
Where,
e = Electric charge = 1.6 × 10
Vd = Drift velocity
Therefore, the time taken by an electron to drift from one end of the wire to the other is
2.7 × 104 s.
Question 3.14:
The earth’s surface has a negative surface charge
difference of 400 kV between the top of the atmosphere and the surface results (due to the
low conductivity of the lower atmosphere) in a current of only 1800 A over the entire
globe. If there were no mechanism of susta
time (roughly) would be required to neutralise the earth’s surface? (This never happens in
practice because there is a mechanism to replenish electric charges, namely the continual
thunderstorms and lightning in d
m.)
Answer
n = 8.5 × 1028 m−3 Length of the
A = 2.0 × 10−6 m2
= 3.0 A, which is given by the relation,
10−19 C
density of 10−9 C m−2. The potential
sustaining atmospheric electric field, how much
different parts of the globe). (Radius of earth = 6.37 × 10
. ining ifferent 106
Surface charge density of the earth,
Current over the entire globe,
Radius of the earth, r = 6.37 × 10
Surface area of the earth,
A = 4πr2
= 4π × (6.37 × 106)2
= 5.09 × 1014 m2
Charge on the earth surface,
q = σ × A
= 10−9 × 5.09 × 1014
= 5.09 × 105 C
Time taken to neutralize the earth’s surface =
Current,
Therefore, the time taken to neutralize the earth’s surface is 282.77 s.
Question 3.15:
Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω
are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current
drawn from the supply and its terminal voltage?
σ = 10−9 C m−2
I = 1800 A
106 m
t
A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380
Ω. What maximum current can be drawn from the cell? Could the cell drive the starting
motor of a car?
Answer
Number of secondary cells, n
Emf of each secondary cell, E
Internal resistance of each cell,
series resistor is connected to the combination of cells.
Resistance of the resistor, R = 8.5 Ω
Current drawn from the supply =
Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is
11.87 A.
After a long use, emf of the secondary cell,
Internal resistance of the cell,
Hence, maximum current
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is
required to start the motor of a car, the cell cannot be used to start a motor.
= 6
= 2.0 V
r = 0.015 Ω
I, which is given by the relation,
E = 1.9 V
r = 380 Ω
Question 3.16:
Two wires of equal length, one of aluminium and the other of copper have the same
resistance. Which of the two wires is lighter? Hence explain why aluminium wires are
preferred for overhead power cables. (ρAl = 2.63 × 10−8 Ω m, ρCu = 1.72 × 10−8 Ω m,
Relative density of Al = 2.7, of Cu = 8.9.)
Answer
Resistivity of aluminium, ρAl = 2.63 × 10−8 Ω m
Relative density of aluminium, d1 = 2.7
Let l1 be the length of aluminium wire and m1 be its mass.
Resistance of the aluminium wire = R1
Area of cross-section of the aluminium wire = A1
Resistivity of copper, ρCu = 1.72 × 10−8 Ω m
Relative density of copper, d2 = 8.9
Let l2 be the length of copper wire and m2 be its mass.
Resistance of the copper wire = R2
Area of cross-section of the copper wire = A2
The two relations can be written as
It is given that,
And,
Mass of the aluminium wire,
m1 = Volume × Density
= A1l1 × d1 = A1 l1d1 … (3)
Mass of the copper wire,
m2 = Volume × Density
= A2l2 × d2 = A2 l2d2 … (4)
Dividing equation (3) by equation (4), we obtain
It can be inferred from this ratio that
copper.
Since aluminium is lighter, it is preferred for overhead power cables over copper.
Question 3.17:
What conclusion can you draw from the following observations on a resistor made of
alloy manganin?
m1 is less than m2. Hence, aluminium is lighter than
Current
A
Voltage
V
0.2 3.94
0.4 7.87
0.6 11.8
0.8 15.7
1.0 19.7
2.0 39.4
Answer
It can be inferred from the given table that the ratio of voltage with current is a constant,
which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys
Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the re
the conductor. Hence, the resistance of manganin is 19.7 Ω.
Question 3.18:
Answer the following questions:
A steady current flows in a metallic conductor of non
these quantities is constant along the
drift speed?
Is Ohm’s law universally applicable for all conducting elements?
If not, give examples of elements which do not obey Ohm’s law.
A low voltage supply from which one needs high currents must h
resistance. Why?
A high tension (HT) supply of, say, 6 kV must have a very large internal resistance.
Why?
Answer
Current
A
Voltage
V
3.0 59.2
4.0 78.8
5.0 98.6
6.0 118.5
7.0 138.2
8.0 158.0
non-uniform cross- section. Which of
conductor: current, current density, electric field,
have very low internal
resistance of
ave
When a steady current flows in a metallic conductor of non
current flowing through the conductor is constant. Current density, electric field, and drift
speed are inversely proportional to the area of cross
constant.
No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode
semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.
According to Ohm’s law, the relation for the potential is
Voltage (V) is directly proportional to current (
R is the internal resistance of the source.
If V is low, then R must be very low, so that high current can be drawn from the source.
In order to prohibit the current from exceeding the safety limit, a high tension supply
must have a very large internal resistance. If the internal resistance is not large, then the
current drawn can exceed the safety limits in case of a short circuit.
Question 3.19:
Choose the correct alternative:
Alloys of metals usually have (greater/less) resistivity than that of their constituent
metals.
Alloys usually have much (lower/higher) temperature coefficients of resistance than pure
metals.
The resistivity of the alloy manganin is nearly independent of/increases rapidly with
increase of temperature.
The resistivity of a typical insulator (e.g., amber) is greater than that
of the order of (1022/103).
Answer
non-uniform cross-section, the
cross-section. Therefore, they are not
V = IR
) I).
ent of a metal by a factor
Alloys of metals usually have greater resistivity than that of their constituent metals.
Alloys usually have lower temperature coefficients of resistance than pure metals.
The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
The resistivity of a typical insulator is greater than that of a metal by a factor of the order
of 1022.
Question 3.20:
Given n resistors each of resistance
maximum (ii) minimum effective resistance? What is the ratio of the maximum to
minimum resistance?
Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent
resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
Determine the equivalent resistance of networks shown in Fig. 3.31.
Answer
Total number of resistors = n
Resistance of each resistor =
When n resistors are connected in series, effective resistance
the product nR.
R, how will you combine them to get the (i)
R
R1is the maximum, given by
, is
Hence, maximum resistance of the combination, R1 = nR
When n resistors are connected in parallel, the effective resistance (R2) is the minimum,
given by the ratio .
Hence, minimum resistance of the combination, R2 =
The ratio of the maximum to the minimum resistance is,
The resistance of the given resistors is,
R1 = 1 Ω, R2 = 2 Ω, R3 = 3 Ω2
Equivalent resistance,
Consider the following combination of the resistors.
Equivalent resistance of the circuit is given by,
Equivalent resistance,
Consider the following combination of the resistors.
Equivalent resistance of the circuit is given by,
Equivalent resistance, R’ = 6 Ω
Consider the series combination of the resistors, as shown in the given circuit.
Equivalent resistance of the circuit is given by the sum,
R’ = 1 + 2 + 3 = 6 Ω
Equivalent resistance,
Consider the series combination of the resistors, as shown in the given circuit.
Equivalent resistance of the circuit is given by,
(a) It can be observed from the given circuit that in the first small loop, two resistors of
resistance 1 Ω each are connected in series.
Hence, their equivalent resistance = (1+1) = 2 Ω
It can also be observed that two resistors of resistance 2 Ω each are connected in series.
Hence, their equivalent resistance = (2 + 2) = 4 Ω.
Therefore, the circuit can be redrawn as
It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all the four
loops. Hence, equivalent resistance (R’) of each loop is given by,
The circuit reduces to,
All the four resistors are connected in series.
Hence, equivalent resistance of the given circuit is
It can be observed from the given circuit that five resistors of resistance
connected in series.
Hence, equivalent resistance of the circuit =
= 5 R
Question 3.21:
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the
infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.
Answer
The resistance of each resistor connected in the given circuit,
Equivalent resistance of the given circuit =
The network is infinite. Hence, equivalent resistance is given by the relation,
R each are
R + R + R + R + R
R = 1 Ω
R’
Negative value of R’ cannot be accepted.
Internal resistance of the circuit,
Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm’s Law, current drawn from the source is given by the ratio,
3.72 A
Question 3.22:
Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω
maintaining a potential drop across the resistor wire AB. A standard cell which maintains
a constant emf of 1.02 V (for very moderate currents up
point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard
cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the
balance point. The standard cell is then re
balance point found similarly, turns out to be at 82.3 cm length of the wire.
Hence, equivalent resistance,
r = 0.5 Ω
to a few mA) gives a balance
replaced by a cell of unknown emf ε
=
and the
What is the value ε ?
What purpose does the high resistance of 600 kΩ have?
Is the balance point affected by this high resistance?
Is the balance point affected by the internal resistance of the driver cell?
Would the method work in the above situation if the driver cell of the
potentiometer had an emf of 1.0 V instead of 2.0 V?
(f ) Would the circuit work well for determining an extremely small emf, say of the order
of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify
the circuit?
Answer
Constant emf of the given standard cell, E1 = 1.02 V
Balance point on the wire, l1 = 67.3 cm
A cell of unknown emf, ε,replaced the standard cell. Therefore, new balance point on the
wire, l = 82.3 cm
The relation connecting emf and balance point is,
The value of unknown emfis 1.247 V.
The purpose of using the high resistance of 600 kΩ is to reduce the current through the
galvanometer when the movable contact is far from the balance point.
The balance point is not affected by the presence of high resistance.
The point is not affected by the internal resistance of the driver cell.
The method would not work if the driver cell of the potentiometer had an emf of 1.0 V
instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less
than the emf of the other cell, then there would be no balance point on the wire.
The circuit would not work well for determining an extremely small emf. As the circuit
would be unstable, the balance point would be close to end A. Hence, there would be a
large percentage of error.
The given circuit can be modified if a series resistance
The potential drop across AB is slightly greater than the emf measured. The percentage
error would be small.
Question 3.23:
Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance
point with a standard resistor
unknown resistance X is 68.5 cm. Determine the value of
failed to find a balance point with the given cell of emf
Answer
Resistance of the standard resistor,
Balance point for this resistance,
Current in the potentiometer wire =
Hence, potential drop across
Resistance of the unknown resistor =
thod he is connected with the wire AB.
R = 10.0 Ω is found to be 58.3 cm, while that with the
X. What might you do if y
ε?
R = 10.0 Ω
l1 = 58.3 cm
i
R, E1 = iR
X
. you
Balance point for this resistor,
Hence, potential drop across
The relation connecting emf and balance point is,
Therefore, the value of the unknown resistance,
If we fail to find a balance point with the given cell of emf,
across R and X must be reduced by putting a resistance in series with it. Only if the
potential drop across R or X is smaller than the potential drop across the potentiometer
wire AB, a balance point is obtained.
Question 3.24:
Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance
of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of
9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length
of the potentiometer wire. Determine the internal resistance of the cell.
Answer
l2 = 68.5 cm
X, E2 = iX
X, is 11.75 Ω.
ind ε, then the potential drop
B, ,
Internal resistance of the cell =
Balance point of the cell in open circuit,
An external resistance (R) is connected to the circuit with
New balance point of the circuit,
Current flowing through the circuit =
The relation connecting resistance and emf is,
Therefore, the internal resistance of the cell
r
l1 = 76.3 cm
) R = 9.5 Ω
l2 = 64.8 cm
I
is 1.68Ω.

Chapter - 5 Magnetism And Matter Class 12

Question 5.1:
Answer the following questions regarding earth’s magnetism:
A vector needs three quantities for its specification. Name the three independent
quantities conventionally used to specify the earth’s magnetic field.
The angle of dip at a location in southern India is about 18º.
Would you expect a greater or smaller dip angle in Britain?
If you made a map of magnetic field lines at Melbourne in Australia, would the lines
seem to go into the ground or come out of the ground?
In which direction would a compass free to move in the vertical plane point to, if located
right on the geomagnetic north or south pole?
The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic
moment 8 × 1022 J T−1 located at its centre. Check the order of magnitude of this number
in some way.
(f ) Geologists claim that besides the main magnetic N-S poles, there are several local
poles on the earth’s surface oriented in different directions. How is such a thing possible
at all?
Answer
The three independent quantities conventionally used for specifying earth’s magnetic
field are:
Magnetic declination,
Angle of dip, and
Horizontal component of earth’s magnetic field
(b)The angle of dip at a point depends on how far the point is located with respect to the
North Pole or the South Pole. The angle of dip would be greater in Britain (it is about
70°) than in southern India because the location of Britain on the globe is closer to the
magnetic North Pole.
(c)It is hypothetically considered that a huge bar magnet is dipped inside earth with its
north pole near the geographic South Pole and its south pole near the geographic North
Pole.
Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic
south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at
Melbourne, Australia would seem to come out of the ground.
(d)If a compass is located on the geomagnetic North Pole or South Pole, then the compass
will be free to move in the horizontal p
magnetic poles. In such a case, the compass can point in any direction.
(e)Magnetic moment, M = 8 × 10
Radius of earth, r = 6.4 × 106
Magnetic field strength,
Where,
= Permeability of free space =
This quantity is of the order of magnitude of the observed field on earth.
(f)Yes, there are several local poles on earth’s surface oriented in different directions. A
magnetised mineral deposit is an example of a local N
Question 5.2:
Answer the following questions:
The earth’s magnetic field varies from point to point in space.
Does it also change with time? If so, on what time scale does it change appreciably?
The earth’s core is known to contain iron. Yet geologists do not regard this as a
source of the earth’s magnetism. Why?
The charged currents in the outer conducting regions of the earth’s core are
thought to be responsible for earth’s magnetism. What migh
the source of energy) to sustain these currents?
The earth may have even reversed the direction of its field several times during its
history of 4 to 5 billion years. How can geologists know about the earth’s field in
such distant past?
plane while earth’s field is exactly vertical to the
1022 J T−1
6 m
N-S pole.
might be the ‘battery’ (i.e.,
lane t
The earth’s field departs from its dipole shape substantially at large distances
(greater than about 30,000 km). What agencies may be responsible for this
distortion?
(f ) Interstellar space has an extremely weak magnetic field of the order of 10
such a weak field be of any significant consequence? Explain.
[Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions
above are tentative or unknown. Brief answers wherever possible are given at the end. For
details, you should consult a good text on geomagnetism.]
Answer
Earth’s magnetic field changes with time. It takes a few hundred years to change by an
appreciable amount. The variation in earth’s magnetic field with the time cannot be
neglected.
(b)Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence, this
is not considered as a source of earth’s magnetism.
(c)Theradioactivity in earth’s interior is the source of energy that sustains the currents in
the outer conducting regions of
responsible for earth’s magnetism.
(d)Earth reversed the direction of its field several times during its history of 4 to 5 billion
years. These magnetic fields got weakly recorded in rocks during t
can get clues about the geomagnetic history from the analysis of this rock magnetism.
(e)Earth’s field departs from its dipole shape substantially at large distances (greater than
about 30,000 km) because of the presence of the ion
gets modified because of the field of single ions. While in motion, these ions produce the
magnetic field associated with them.
(f)An extremely weak magnetic field can bend charged particles moving in a circle. This
may not be noticeable for a large radius path. With reference to the gigantic interstellar
space, the deflection can affect the passage of charged particles.
Question 5.3:
A short bar magnet placed with its axis at 30º with a uniform external
0.25 T experiences a torque of magnitude equal to 4.5 × 10
magnetic moment of the magnet?
earth’s core. These charged currents are considered to be
their solidification. One
ionosphere. In this region, earth’s field
ay magnetic field of
10−2 J. What is the magnitude of
10−12 T. Can
heir osphere.
Answer
Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T
Angle between the bar magnet and the external magnetic field,
Torque is related to magnetic moment (
T = MB sin θ
Hence, the magnetic moment of the magnet is 0.36 J T
Question 5.4:
A short bar magnet of magnetic moment m = 0.32 J T
field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would
correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of
the magnet in each case?
Answer
Moment of the bar magnet, M
External magnetic field, B = 0.15 T
= 4.5 × 10−2 J
et θ = 30°
M) as:
T−1
.
T−1 is placed in a uniform magnetic
= 0.32 J T−1
(a)The bar magnet is aligned along the magnetic field. This system is considered as being
in stable equilibrium. Hence, the angle
is 0°.
Potential energy of the system
(b)The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable
equilibrium.
θ = 180°
Potential energy = − MB cos
Question 5.5:
A closely wound solenoid of 800 turns and area of
current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is
its associated magnetic moment?
Answer
Number of turns in the solenoid,
Area of cross-section, A = 2.5 × 10
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops
along its axis, i.e., along its length.
The magnetic moment associated with the given current
as:
M = n I A
θ, between the bar magnet and the magnetic
θ
cross section 2.5 × 10−4 m
n = 800
10−4 m2
current-carrying solenoid is calculated
, field
m2 carries a
= 800 × 3 × 2.5 × 10−4
= 0.6 J T−1
Question 5.6:
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform
horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the
solenoid when its axis makes an angle of 30° with the direction of applied field?
Answer
Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 T
The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
Question 5.7:
A bar magnet of magnetic moment 1.5 J T
magnetic field of 0.22 T.
What is the amount of work required by an external torque to turn the magnet
align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field
direction?
What is the torque on the magnet in cases (i) and (ii)?
Answer
T−1
, rque T−1 lies aligned with the direction of a uniform
so as to
(a)Magnetic moment, M = 1.5 J T−1
Magnetic field strength, B = 0.22 T
(i)Initial angle between the axis and the magnetic field, θ1 = 0°
Final angle between the axis and the magnetic field, θ2 = 90°
The work required to make the magnetic moment normal to the direction of magnetic
field is given as:
(ii) Initial angle between the axis and the magnetic field, θ1 = 0°
Final angle between the axis and the magnetic field, θ2 = 180°
The work required to make the magnetic moment opposite to the direction of magnetic
field is given as:
(b)For case (i):
∴Torque,
For case (ii):
∴Torque,
Question 5.8:
A closely wound solenoid of 2000 turns and area of cross
a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
What is the magnetic moment associated with the solenoid?
What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5
× 10−2 T is set up at an angle of 30º with the axis of the solenoid?
Answer
Number of turns on the solenoid,
Area of cross-section of the solenoid,
Current in the solenoid, I = 4 A
(a)The magnetic moment along the axis of the solenoid is calculated as:
M = nAI
= 2000 × 1.6 × 10−4 × 4
= 1.28 Am2
(b)Magnetic field, B = 7.5 × 10
Angle between the magnetic field and the axis of the solenoid,
Torque,
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the
solenoid is
cross-section 1.6 × 10−4
n = 2000
A = 1.6 × 10−4 m2
10−2 T
θ = 30°
m2, carrying
Question 5.9:
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its
plane normal to an external field of magnitude 5.0 × 10−2 T. The coil is free to turn about
an axis in its plane perpendicular to the field direction. When the coil is turned slightly
and released, it oscillates about its stable equilibrium with a frequency of 2.0 s−1. What is
the moment of inertia of the coil about its axis of rotation?
Answer
Number of turns in the circular coil, N = 16
Radius of the coil, r = 10 cm = 0.1 m
Cross-section of the coil, A = πr2 = π × (0.1)2 m2
Current in the coil, I = 0.75 A
Magnetic field strength, B = 5.0 × 10−2 T
Frequency of oscillations of the coil, v = 2.0 s−1
∴Magnetic moment, M = NIA
= 16 × 0.75 × π × (0.1)2
= 0.377 J T−1
Where,
I = Moment of inertia of the coil
Hence, the moment of inertia of the coil about its axis of rotation is
Question 5.10:
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has
its north tip pointing down at 22º with the horizontal. The
earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the
earth’s magnetic field at the place.
Answer
Horizontal component of earth’s magnetic field,
Angle made by the needle wit
Earth’s magnetic field strength =
We can relate B and BHas:
Hence, the strength of earth’s magnetic field at the given location is 0.377 G.
horizontal component of the
BH = 0.35 G
with the horizontal plane = Angle of dip =
B
Question 5.11:
At a certain location in Africa, a compass
north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian
points 60º above the horizontal. The horizontal component of the earth’s field is measured
to be 0.16 G. Specify the dire
Answer
Angle of declination,θ = 12°
Angle of dip,
Horizontal component of earth’s magnetic field,
Earth’s magnetic field at the given location =
We can relate B and BHas:
Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian,
making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.
Question 5.12:
A short bar magnet has a magnetic moment of 0.48 J T
magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the
centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the
magnet.
Answer
points 12º west of the geographic north. The
direction and magnitude of the earth’s field at the location.
BH = 0.16 G
B
T−1. Give the direction and
ction ive
Magnetic moment of the bar magnet,
Distance, d = 10 cm = 0.1 m
The magnetic field at distance
relation:
Where,
= Permeability of free space =
The magnetic field is along the S
The magnetic field at a distance of 10 cm (i.e.,
magnet is given as:
The magnetic field is along the N
Question 5.13:
M = 0.48 J T−1
d, from the centre of the magnet on the axis is given by the
− N direction.
d = 0.1 m) on the equatorial line of the
− S direction.
,
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic
north-south direction. Null points are found on the axis of the magnet at 14 cm from the
centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the
is zero. What is the total magnetic field on the normal bisector of the magnet at the same
distance as the null−point (i.e., 14 cm) from the centre of the magnet? (At
field due to a magnet is equal and opposite to the horizontal
magnetic field.)
Answer
Earth’s magnetic field at the given place,
The magnetic field at a distance
Where,
= Permeability of free space
M = Magnetic moment
The magnetic field at the same distance
Total magnetic field,
Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.
Question 5.14:
−null points
component of earth’s
H = 0.36 G
d, on the axis of the magnet is given as:
eld d, on the equatorial line of the magnet is given as:
angle of dip
points,
,
If the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points
be located?
Answer
The magnetic field on the axis of the magnet at a distance d1 = 14 cm, can be written as:
Where,
M = Magnetic moment
= Permeability of free space
H = Horizontal component of the magnetic field at d1
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial
line.
Hence, the magnetic field at a distance d2, on the equatorial line of the magnet can be
written as:
Equating equations (1) and (2), we get:
The new null points will be located 11.1 cm on the normal bisector.
Question 5.15:
A short bar magnet of magnetic moment 5.25 × 10
perpendicular to the earth’s field direction. At what distance from the centre of the
magnet, the resultant field is inclined at 45º with earth’s field on
its normal bisector and (b) its axis. Magnitude of the earth’s field at the
be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Answer
Magnetic moment of the bar magnet,
Magnitude of earth’s magnetic field at a place,
The magnetic field at a distance
given by the relation:
Where,
= Permeability of free space = 4π
10−2 J T−1 is placed with its axis
place is given to
M = 5.25 × 10−2 J T−1
H = 0.42 G = 0.42 × 10−4 T
R from the centre of the magnet on the normal bisector is
× 10−7 Tm A−1
When the resultant field is inclined at 45° with earth’s field,
The magnetic field at a distanced
The resultant field is inclined at 45° with earth’s field.
Question 5.16:
Answer the following questions:
Why does a paramagnetic sample display greater
magnetising field) when cooled?
Why is diamagnetism, in contrast, almost independent of temperature?
B = H
ic from the centre of the magnet on its axis is given as:
magnetisation (for the same
If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or
(slightly) less than when the core i
Is the permeability of a ferromagnetic material independent of the magnetic field? If not,
is it more for lower or higher fields?
Magnetic field lines are always nearly normal to the surface of a ferromagnet at every
point. (This fact is analogous
of a conductor at every point.) Why?
(f ) Would the maximum possible magnetisation of a paramagnetic sample be of the same
order of magnitude as the magnetization of a ferromagnet?
Answer
(a)Owing to therandom thermal motion of molecules, the alignments of dipoles get
disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a
paramagnetic sample displays greater magnetisation when cooled.
(b)The induced dipole moment
magnetising field. Hence, the internal motion of the atoms (which is related to the
temperature) does not affect the diamagnetism of a material.
(c)Bismuth is a diamagnetic substance. Hence, a toroid wi
magnetic field slightly greater than a toroid whose core is empty.
(d)The permeability of ferromagnetic materials is not independent of the applied
magnetic field. It is greater for a lower field and vice versa.
(e)The permeability of a ferromagnetic material is not less than one. It is always greater
than one. Hence, magnetic field lines are always nearly normal to the surface of such
materials at every point.
(f)The maximum possible magnetisation of a paramagnetic
order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising
fields for saturation.
Question 5.17:
Answer the following questions:
is empty?
to the static electric field lines being normal to the surface
in a diamagnetic substance is always opposite to the
with a bismuth core has a
sample can be of the same
th
Explain qualitatively on the basis of domain picture the irreversibility in the
magnetisation curve of a ferromagnet.
The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel
piece. If the material is to go through repeated cycles of magnetisation, which piece will
dissipate greater heat energy?
‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing
memory?’ Explain the meaning of this statement.
What kind of ferromagnetic material is used for coating magnetic tapes in a cassette
player, or for building ‘memory stores’ in a modern computer?
A certain region of space is to be shielded from magnetic fields.
Suggest a method.
Answer
The hysteresis curve (B-H curve) of a ferromagnetic material is shown in the following
figure.
It can be observed from the given curve that magnetisation persists even when the
external field is removed. This reflects the irreversibility of a ferromagnet.
(b)The dissipated heat energy is directly proportional to the area of a hysteresis loop. A
carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat
energy.
(c)The value of magnetisation is memory or record of hysteresis loop cycles of
magnetisation. These bits of information correspond to the cycle of magnetisation.
Hysteresis loops can be used for storing information.
(d)Ceramic is used for coating magnetic tapes in cassette players and for building
memory stores in modern computers.
(e)A certain region of space can be shielded from magnetic fields if it is surrounde
soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.
Question 5.18:
A long straight horizontal cable carries a current of 2.5 A in the direction 10º south of
west to 10° north of east. The magnetic meridian of
the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the
angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable).
(At neutral points, magnetic field due
the horizontal component of earth’s magnetic field.)
Answer
Current in the wire, I = 2.5 A
Angle of dip at the given location on earth,
Earth’s magnetic field, H = 0.33 G = 0.33 × 10
The horizontal component of earth’s magnetic field is given as:
HH = H cos
The magnetic field at the neutral point at a distance
relation:
Where,
the place happens to be 10º west of
, to a current-carrying cable is equal and opposite to
= 0°
10−4 T
R from the cable is given by the
surrounded by
= Permeability of free space =
Hence, a set of neutral points parall
distance of 1.51 cm.
Question 5.19:
A telephone cable at a place has four long straight horizontal wires carrying a current of
1.0 A in the same direction east to west. The earth’s magnetic field
and the angle of dip is 35º. The magnetic declination is nearly zero. What are the resultant
magnetic fields at points 4.0 cm below the cable?
Answer
Number of horizontal wires in the telephone cable,
Current in each wire, I = 1.0 A
Earth’s magnetic field at a location,
Angle of dip at the location, δ = 35°
Angle of declination, θ ∼ 0°
For a point 4 cm below the cable:
Distance, r = 4 cm = 0.04 m
The horizontal component of earth’s magnetic
Hh = Hcosδ − B
Where,
parallel to and above the cable are located at a normal
at the place is 0.39 G,
n = 4
H = 0.39 G = 0.39 × 10−4 T
field can be written as:
el
B = Magnetic field at 4 cm due to current I in the four wires
= Permeability of free space = 4π × 10−7 Tm A−1
= 0.2 × 10−4 T = 0.2 G
∴ Hh = 0.39 cos 35° − 0.2
= 0.39 × 0.819 − 0.2 ≈ 0.12 G
The vertical component of earth’s magnetic field is given as:
Hv = Hsinδ
= 0.39 sin 35° = 0.22 G
The angle made by the field with its horizontal component is given as:
The resultant field at the point is given as:
For a point 4 cm above the cable:
Horizontal component of earth’s magnetic field:
Hh = Hcosδ + B
= 0.39 cos 35° + 0.2 = 0.52 G
Vertical component of earth’s magnetic field:
Hv = Hsinδ
= 0.39 sin 35° = 0.22 G
Angle, θ
And resultant field:
Question 5.20:
A compass needle free to turn
of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45º with
the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to
east.
Determine the horizontal component of the earth’s magnetic field at the location.
The current in the coil is reversed, and the coil is rotated about its vertical axis by an
angle of 90º in the anticlockwise sense looking from above. Predict the direction of the
needle. Take the magnetic declination at the places to be zero.
Answer
Number of turns in the circular coil,
Radius of the circular coil, r = 12 cm = 0.12 m
Current in the coil, I = 0.35 A
Angle of dip, δ = 45°
The magnetic field due to current
Where,
= 22.9°
in a horizontal plane is placed at the centre of circular coil
ne eedle. N = 30
I, at a distance r, is given as:
= Permeability of free space = 4π
= 5.49 × 10−5 T
The compass needle points from West to East. Hence, the horizontal component of
earth’s magnetic field is given as:
BH = Bsin δ
= 5.49 × 10−5 sin 45° = 3.88 × 10
When the current in the coil is reversed and the coil is rotated about its vertical axis by an
angle of 90 º, the needle will reverse its original direction. In this case, the needle will
point from East to West.
Question 5.21:
A magnetic dipole is under the influence of two magnetic fields. The angle between the
field directions is 60º, and one of the fields has a magnitude of 1.2 × 10
comes to stable equilibrium at an angle of 15º with this fie
other field?
Answer
Magnitude of one of the magnetic fields,
Magnitude of the other magnetic field =
Angle between the two fields,
At stable equilibrium, the angle between the dipole and f
Angle between the dipole and field
At rotational equilibrium, the torques between both the fields must balance each other.
∴Torque due to field B1 = Torque due to field
× 10−7 Tm A−1
10−5 T = 0.388 G
10−2 T. If the dipole
field, what is the magnitude of the
B1 = 1.2 × 10−2 T
B2
θ = 60°
field B1, θ1 = 15°
B2, θ2 = θ − θ1 = 60° − 15° = 45°
B2
ld,
MB1 sinθ1 = MB2 sinθ2
Where,
M = Magnetic moment of the dipole
Hence, the magnitude of the other magnetic field is 4.39 × 10
Question 5.22:
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected
to a horizontal magnetic field of 0.04 G
down deflection of the beam over a distance of 30 cm (
in this exercise are so chosen that the answer will give you an idea of the effect of earth’s
magnetic field on the motion of the electron beam from the electron gun to the screen in a
TV set.]
Answer
Energy of an electron beam, E
Charge on an electron, e = 1.6 × 10
E = 18 × 103 × 1.6 × 10−19 J
Magnetic field, B = 0.04 G
Mass of an electron, me = 9.11 × 10
Distance up to which the electron beam travels,
We can write the kinetic energy of the electron beam as:
10−3 T.
normal to the initial direction. Estimate the up or
me= 9.11 × 10−19 C). [
otion = 18 keV = 18 × 103 eV
10−19 C
10−19 kg
d = 30 cm = 0.3 m
Note: Data
The electron beam deflects along a circular path of radius,
The force due to the magnetic field balances the centripetal force of the path.
Let the up and down deflection of the electron beam be
Where,
θ = Angle of declination
Therefore, the up and down deflection of the beam is 3.9 mm.
Question 5.23:
A sample of paramagnetic salt contains 2.0 × 10
1.5 × 10−23 J T−1. The sample is placed under a homogeneous magnetic field of 0.64 T,
and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal
r.
1024 atomic dipoles each of dipole moment
.
to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and
a temperature of 2.8 K? (Assume Curie’s law)
Answer
Number of atomic dipoles, n = 2.0 × 1024
Dipole moment of each atomic dipole, M = 1.5 × 10−23 J T−1
When the magnetic field, B1 = 0.64 T
The sample is cooled to a temperature, T1 = 4.2°K
Total dipole moment of the atomic dipole, Mtot = n × M
= 2 × 1024 × 1.5 × 10−23
= 30 J T−1
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment,
When the magnetic field, B2 = 0.98 T
Temperature, T2 = 2.8°K
Its total dipole moment = M2
According to Curie’s law, we have the ratio of two magnetic dipoles as:
Therefore, is the total dipole moment of the sample for a magnetic field of
0.98 T and a temperature of 2.8 K.
Question 5.24:
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic
core of relative permeability 800. What is the m
magnetising current of 1.2 A?
Answer
Mean radius of a Rowland ring,
Number of turns on a ferromagnetic core,
Relative permeability of the core material,
Magnetising current, I = 1.2 A
The magnetic field is given by the relation:
B
Where,
μ0 = Permeability of free space = 4π
Therefore, the magnetic field in the core is 4.48 T.
Question 5.25:
The magnetic moment vectors μ
momentum S and orbital angular momentum
by quantum theory (and verified experimentally to a high accuracy) to be given by:
μs= –(e/m) S,
μl = –(e/2m)l
magnetic field B in the core for a
r = 15 cm = 0.15 m
N = 3500
× 10−7 Tm A−1
μsand μlassociated with the intrinsic spin angular
l, respectively, of an electron are predicted
,
Which of these relations is in accordance with the result expected classically? Outline the
derivation of the classical result.
Answer
The magnetic moment associated with the intrinsic spin angular momentum ( ) is given
as
The magnetic moment associated with the orbital angular momentum ( ) is given as
For current i and area of cross-section A, we have the relation:
Where,
e = Charge of the electron
r = Radius of the circular orbit
T = Time taken to complete one rotation around the circular orbit of radius r
Angular momentum, l = mvr
Where,
m = Mass of the electron
v = Velocity of the electron
Dividing equation (1) by equation (2), we get:
Therefore of the two relations,is in accordance with classical physics
 

French Revolution Class 9 NCERT Solutions

Question:- 1. Describe the circumstances leading to the outbreak of revolutionary protest in France.

Answer:- Following are some of the causes which had a cumulative effect to result in revolution in France:

(a) The war with Britain for an independent America: This war led to mounting debt on the French monarchy. This necessitated imposition of new taxes on the public.

(b) Privilege based on birth: People got privileges and position based on their lineage and not on their merit. This led to resentment among common people.

(c) Concentration of power among the privileged: People belonging to the first and second estate had all the power and money. Masses were at the mercy of this privileged class.

(d) Subsistence Crisis:- Rising population and less grain production resulted in demand supply gap of bread, which was the staple diet. Wages did not keep pace with rising prices. It was becoming difficult for people.

(e) Growing Middle Class: Because of increased overseas trade a new class emerged. This class was wealthy not because of birth but because of its ability to utilize opportunities. People of the middle class started raising their voice for an end to privileges based on lineage.

All of this led to a general sense of resentment among people. Certain thinkers of the period spread awareness through various media. Some from the privileged classes also advocated a switch to democracy. So, finally there was revolution in France.

Question:- 2. Which groups of French society benefited from the revolution? Which groups were forced to relinquish power? Which sections of society would have been disappointed with the outcome of the revolution?

Answer: Peasants and artisans of French society benefited from the revolution. Clergy, nobles and church had to relinquish power. It is obvious that those who had to forego power and privileges would have been disappointed. People from the first and the second estate must have been a disappointed lot.

Question: 3. Describe the legacy of the French Revolution for the peoples of the world during the nineteenth and twentieth centuries.

Answer: The ideas of liberty and democratic rights were the most important legacy of the French Revolution. These spread from France to the rest of Europe during the nineteenth century, where feudal systems were abolished. Further these ideas spread to different colonies of the European nations. Colonised people interpreted and moulded these ideas according to respective needs. This was probably like seed for an end of colonization in many countries. By the mid of 20th century major part of the world adopted democracy as the preferred mode of rule and the French Revolution can be termed as the initiation point for this development.

Question: 4. Draw up a list of democratic rights we enjoy today whose origins could be traced to the French Revolution.

Answer: The following fundamental rights, given in the Indian constitution can be traced to the French Revolution:

• The right to equality
• The right to freedom of speech and expression
• The right to freedom from exploitation
• The right to constitutional remedies

Question:- 5. Would you agree with the view that the message of universal rights was beset with contradictions? Explain.

Answer: The major contradiction in the message of universal rights as per the French Constitution of 1791 was the total ignorance of women. All rights were given to men. Apart from that the presence of huge number of people as passive citizens, without voting rights, was like not putting into practice what you preach. In other words it can be said that although the declaration of universal rights was a good starting point but it left much to be desired.

Question: 6. How would you explain the rise of Napoleon?

Answer: After France became a republic in 1792, the then ruler, Robespeirre, gave more privileges to the wealthier section of society. Further, he was a sort of autocrat himself. This led to reign of terror for the following many years. After Robespeirre’s rule came to an end a directory was formed to avoid concentration of power in one individual. Members of the directory often fought among themselves leading to total chaos and political instability. This created a political vaccum in France. This was a conducive situation and Napoleon Bonaparte took the reign of power as a military dictator.

EXTRA QUESTIONS For EXTRA MARKS

Q1: Important dates related to French Revolution
Answer:

• 1774 : Louis XVI becomes King of France.
• 1789 : Third Estate forms National Assembly. The Bastille is stormed., Session of Estates General
• 1791 : A constitution is framed to limit the powers of the monarch.
• 1792-93 : France becomes a republic. The king is executed. Directory rules France.
• 1804 : Napoleon becomes emperor of France.
• 1815 : Napoleon was defeated at Waterloo.
• 1848 : Slavery was abolished in all French colonies.
• 1946 : French women got the right to vote.

Q2: What principles were supported by the revolution?
Answer: Liberty, Equality and Fraternity

Run Quiz

Q3: Important Terms to remember.

Answer:

1. First Estate: French society was divided into classes called Estates, The First Estate consisted of the Clergy which held vast land, wealth and were exempted from taxes.
2. Second Estate: It consisted of the aristocracy and controlled all the top positions in the government. parliament and in the army and navy. They were also exempted from taxation and led an extravagant life.
3. Third Estate: This comprised everyone who was neither nobility nor clergy and constituted almost 97% of the population. The wealthy upper middle class (merchants, bankers, doctors. lawyers), lower middle class, shopkeepers, craftsmen. and peasants comprised the Third Estate. This class lacked political power, social status and was heavily taxed.
4. The Declaration of Rights of Man and Citizen: In 1789. the French National Assembly adopted a set of basic principles called the Declaration of the Rights of Man and Citizen. Proposed by Lafayette and based on the ideas of Locke, Montesquieu and Jefferson, this document stated that “men are born and remain free and equal in rights” and that the “source of power resides in the people”. It guaranteed all Frenchmen the basic rights of liberty, security, equal justice, fair taxes, religion, fair speech, and thought.
5. Livre: Unit of currency used in France till 1794.
6. Clergy: Group of persons invested with special functions in the Church.
7. Tithe: A tax levied by the Church equal to one-tenth of the agricultural produce.

Q4: Write a short note on 'Reign of Terror'.
Answer:  The term 'Reign of terror' referred to policy of severe control and punishment. This policy was adopted by Robespierre during his reign from 1793 to 1794. During this period, all of those whom he considered being enemies of the republic (e.g. nobles, clergy, members of other political parties) were arrested. They were tried by a revolutionary tribunal and eventually
more then 15,000 person including Queen of France were guillotined or executed. Meat and bread
were rationed. Peasants were forced to sell their grains at the prices fixed by the government. All citizens were forced to eat the equality bread i.e., a loaf made of whole wheat. Churches
were shutdown and their buildings were converted into barracks or offices.

Q5: What role did women of France of third Estate play in French revolution?
Answer: Women of third estate of France played an crucial role in the revolution:

• During Revolutionary years, women started their own political clubs and newspapers. They set up about sixty women clubs in different cities of France.
• The most famous among them was the Society of Revolutionary and Republican Women.
• They also demanded to enjoy the same political rights as men, i.e., right to vote, to be elected to the Assembly and to hold political office.

However, the women were disappointed because they were not given same political rights as enjoyed by the men of France.

Q6: In context of France the volunteers from Marseilles sang the Marseillaise a patriotic song when they marched into Paris. Who composed this song ?

(a) Maximilian Robespierre

(b) Marie Antoinette

(c) Roget de L'lsle

(d) Mirabeau

Answer: (c) Roget de L'lsle

Q7: What did the Red Cap worn by Sans Culottes in France symbolize? 
(a) Liberty 

(b) Brotherhood
(c) Love 

(d) Equality

Answer: (a) Liberty

Q8: What was the legacy of the French Revolution?
Answer:

1. It led to the decade of political changes in Europe.
2. Three founding words of the French revolution i.e. 'liberty, equality and fraternity' reflected the coming of new democratic and social order in Europe and rest of the world.
3. It inspired the Germans, Italians and Austrians to overthrow their oppressive regimes.
4. It inspired struggling nations of Asia and Africa. E.g. India's struggle for Independence was inspired by the thoughts of french philosophers like Voltaire and Rousseau.
5. French revolution put into the practice the idea that sovereignty comes from the people from below not from the above.
6. It marked the beginning for the first time in history about active and institutionalized mass participation in the government. It inculcated the spirit of nationalism among the people.

Q9: Which of the following refers to the political body representing the three estates of pre-revoluitonary France?

(a) Parliament of France

(b) National Assembly

(c) Estates General

(d) Estates Committee

Answer: (c) Estates General

Q10: The word 'Guillotine' during French revolution era refers to

(a) Beheading a person 

(b) Awarding a person

(c) Taxing a person 

(d) Threatening a person

Answer: (a) Beheading a person 

Q11: What is the importance of the document 'Declaration of Rights of Man and Citizen'?

OR

Why 'Declaration of Rights of Man and Citizen' is considered as a revolutionary document in french revolution?

Answer: Declaration of the Rights of Man and Citizen is regarded as  a revolutionary document because:

1. It abolished the privileges and power given to the French feudal classes i.e. First Estate and Second Estate.
2. It provided equal distribution of the burdens of  taxation and rights to public property among all citizens.
3. The Declaration emphasized equality before law and freedom of speech and press.Every citizen has right to speak, print and express.

Q12 (CBSE 2010): Explain how did the freedom of speech and expression under the revolutionary government in France promote the ideals of Liberty and Equality into everyday practice.

Answer:

1. The Declaration of Rights of Man and Citizen emphasized that freedom of speech and expression be established as established as ‘natural and inalienable’ right.
2. The ideas of these philosophers were discussed intensively in salons and coffee-houses and spread among people through books and newspapers. 
3. Newspapers, pamphlets, books and printed pictures flooded the towns of France from where they traveled rapidly into the countryside.
4. Freedom of the press also meant that opposing views of events could be expressed.
5. Plays, songs and festive processions attracted large numbers of illiterate people to help them grasp and identify with ideas such as liberty or justice that political philosophers wrote about.

Q13: What was the important aim of setting up Directory? What was the final outcome after appointing the Directory?

Answer:  After the fall of Jacobin government, the two elected legislative Council appointed an executive of five members called Directory. The main objective of the directory was to work as a safeguard against the concentration of power in a one-man executive as under the Jacobins.

However, the Directors often clashed with the legislative councils, who then sought to dismiss them. The political instability of the Directory paved the way for the rise of a military dictator, Napoleon Bonaparte.

Q14: Who authored the book 'The Spirit of the Laws'? What was proposed in this book?
Answer: Montesquieu was the author of The Spirit of the Laws. In the book he proposed a division of power within the government between the legislative, the executive and the judiciary.

Q15: Which cities became prosperous because of slave trade?
Answer: Port cities like Bordeaux and Nantes prospered due to the flourishing slave trade.

Q16: What impact did French Revolution have on slave trade?
Answer:

1. During and after revolution, there was little criticism of slavery in France. 
2. It was totally ironical that National assembly held long debates about the rights of men but did not pass any laws to abolish slavery. 
3. In 1794, a convention was passed to free all slaves in French overseas possessions. However it turned out to be a short term measure. Napoleon reintroduced slavery after 10 years of ban.
4. Slavery was finally abolished in French colonies in 1848.

Q17: What was the role of French philosophers and revolutionary thinkers in the French Revolution?
Answer: The 18th century witnessed the emergence of revolutionary thinkers such as John Locke, Jean Jacques Rousseau and Montesquieu. They didn't play any active role in the events of the revolution but their ideas inspired the revolutionary movement. Their revolutionary ideas encouraged people to fight for their rights.

1. The idea of a society based on freedom, equal laws and opportunities for all were put forward by philosophers such as John Locke and Jean Jacques Rousseau. In his book Two Treatises of Government, Locke sought to refute the doctrine of the divine and absolute right of the monarch.
2. Rousseau carried his radical ideas forward by proposing a form of government based on social contract between people and their representatives.
3. Montesquieu, in his book, The Spirit of the Laws proposed a division of power within the government between the legislative, the executive and the judiciary.
4. Voltaire's ideas revolved around individual liberties. He believed that people should have freedom of expression.

The ideas of these philosophers were discussed and debated in salons and coffee houses and spread among people through books and newspapers. This founded the way to revolution of 1789.

Q18: The French revolution took place on _________

(a) July 14, 1789

(b) July 14, 1788

(c) July 14, 1786

(d) July 14, 1785

Answer: (a) July 14, 1789 (Storming of the Bastille)

Q19: On what charges was Louis XVI guillotined?

(a) Cruelty

(b) Treason

(c) Absolute Role

(d) misgovernance

Answer: (b) Treason

Q20: The tax called tithe was collected from French Peasants by

(a) The Church

(b) The emperor

(c) The Nobles

(d) Chief of the Army

Answer: (a) The Church

Q21: Why was the subsistence crisis caused in France?

(a) The wages of the people were low

(b) There was widespread unemployment

(c) Increase in population led to rapid increase in the demand of food grains.

(d) The government imposed various taxes.

Answer: (c) Increase in population led to rapid increase in the demand of food grains.

Note: A ‘subsistence crisis’ is defined as an economic crisis which threatens the food supplies or, more precisely, the survival prospects of larger population.

Q22: The Bastille was hated by all in France because:

(a) it stood for the despotic power of the king.

(b) it was a fortress prison

(c) Prison In charge tortured the inmates.

(d) it housed dreaded criminals.

Answer: (a) it stood for the despotic power of the king.

Note: A despot is a ruler or other person who holds absolute power, typically exercising it cruelly.

Q23: What were the different political groups operating in the Legislative Assembly during French revolution?
Answer: Following political groups were active within the Legislative Assembly during French Revolution:

1. The Girondins:  It was the largest group who wanted war against all despots.
2. The Jacobins: The most influential and ruthless group who were responsible for the reign of terror. They did not favour war.
3. The Feuillants: They proposed constitutional monarchy.

Q24: What was the impact of the French Revolution in France?
Answer: The impact of the French Revolution in France can be summarized as:

1. End of Monarchy: It marked the end of absolute monarchy and paved way for a republic government.
2. Laid foundation of Democratic Principles: It upheld the theory of sovereignty and laid foundation of the democratic principles i.e. government decisions should be based on the consent of her citizens.
3. Ideas of Liberty, Equality and Fraternity: The slogans of Liberty, Equality and Fraternity became the founding milestones for a democratic nations.
4. Declaration of Rights and Equality: The declaration provided equality among the citizens for all public offices, freedom of speech and expression was granted and freedom from arrest without  a proven cause was implemented.
5. Abolition of Censorship: Mass media was allowed Freedom of press was granted.
6. Taxes to be paid by all: Taxes would be borne by all the people irrespective of their status. 
7. Religious Freedom: It guaranteed free exercise of religious worship and abolished the taxes collected by the churches.
8. Reforms and New Initiatives: New initiatives and reforms were carried out in education, administration and judiciary domain.

Q25: Who were allowed to vote for the formation of the National Assembly?

Answer:

• Only men above 25 years of age and those who paid taxes equal to at least 3 days of a labourers wage were given the status of active citizens. And they were entitled to vote.
• The remaining men and women were classified as passive citizens and were not allowed to vote.
• To qualify as an elector and then as a member of the assembly, a man had to belong to the highest bracket of tax payers.

Q26: What was Rousseau's concept about the state?
Answer: Rousseau, a thinkers, is regarded as an inspiration behind the French Revolution.

1. According to him, the society creates a Political state for a stable life.
2. The state is not a divine creation but is the outcome of a sort of unwritten social contract.
3. If the state failed to abide by the terms of the contract, people have natural and moral rights to overthrow the state authority.

Q27: Who was the president of USA during French revolution?

Answer: George Washington became the first president of USA at that time.

Q28: Arrange the following events of French Revolution in chronological order (what happened next).
(i) Louis XVI is guillotined.
(ii) the Oath of the Tennis Court
(iii) The Bastille falls
(iv) French Republic is declared.
(v) The Great Fear

Answer: The correct sequence is (ii) - (iii) - (v) - (iv) - (i)

Q29: What was The Great Fear in French revolution?

Answer: After the storming of the Bastille, the National Assembly was busy at Versailles drafting a
constitution, the rest of France seethed with turmoil.

1. In the countryside rumours spread from village to village that the royalists of the manor had hired bands of brigands who were on their way to destroy the ripe crops. 
2. Rumours also spread that Kings men were planning to kill National assembly representatives. 
3. Caught in a frenzy of fear, peasants in several districts seized hoes and pitchforks and attacked chateaux. They looted hoarded grain and burnt down documents containing records of manorial dues. 
4. A large number of nobles fled from their homes, many of them migrating to neighbouring countries. 

This turmoil at that time is termed as The Great Fear.

Q30: When did the Assembly pass a decree abolishing the feudal system of obligations and taxes?

Answer: 4 August 1789

Q31: Who among the following proposed a division of power within the government?
(a) John Locke
(b) Jean Jacque Rousseau
(c) Voltaire
(d) Montesquieu

Answer: (d) Montesquieu


Q32: When did Louis XVI called for The Estates General? When was it convened last time?

Answer:  Louis XVI called for The Estates General on May 5, 1789. It was called after a gap of 175 years since 1614

Q33: From where Jacobin's club got its name? Who was the leader of Jacobin's club?

Answer: It got its name from the former convent of St Jacob in Paris.Maximillian Robespierre was the leader of Jacobins.

Q34: What was the outcome of Battle of Waterloo? When did this battle occur?

Answer: The Battle of Waterloo was fought on 1815 between Napolean Forces and allied army of Prussia, UK and Netherlands. In this battle Napolean was defeated.