Welcome to the Chapter 4 - Moving Charges and Magnetism, Class 12 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 4 - Moving Charges and Magnetism. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Moving Charges and Magnetism and excel in their exams. By going through these Moving Charges and Magnetism question answers, you can strengthen your foundation and improve your performance in Class 12 Physics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.
Number of turns on the circular coil, n = 100
Radius of each turn, r = 8.0 cm = 0.08 m
Current flowing in the coil, I = 0.4 A
Magnitude of the magnetic field at the centre of the coil is given by the relation,
| B | = μ0 2πnI / 4π r
Where,
μ0 = Permeability of free space
= 4π × 10–7 T m A–1
| B | = 4π x 10-7 x 2π x 100 x 0.4 / 4π x 0.08
= 3.14 x 10-4 T
Hence, the magnitude of the magnetic field is 3.14 × 10–4 T.
For moving coil meter M1:
Resistance, R1 = 10 Ω
Number of turns, N1 = 30
Area of cross-section, A1 = 3.6 × 10-3 m2
Magnetic field strength, B1 = 0.25 T
Spring constant K1 = K
For moving coil meter M2:
Resistance, R2 = 14 Ω
Number of turns, N2 = 42
Area of cross-section, A2 = 1.8 × 10-3 m2
Magnetic field strength, B2 = 0.50 T
Spring constant, K2 = K
(a) Current sensitivity of M1 is given as:
And, current sensitivity of M2 is given as:
so, Ratio 
Hence, the ratio of current sensitivity of M2 to M1 is 1.4.
(b) Voltage sensitivity for M2 is given as:
And, voltage sensitivity for M1 is given as:
Magnetic field strength, B = 6.5 G = 6.5 × 10-4 T
Speed of the electron, v = 4.8 × 106 m/s
Charge on the electron, e = 1.6 × 10-19 C
Mass of the electron, me = 9.1 × 10-31 kg
Angle between the shot electron and magnetic field, θ = 90°
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB sinθ
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r. Hence, centripetal force exerted on the electron,
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
Hence, the radius of the circular orbit of the electron is 4.2 cm.
Magnetic field strength, B = 6.5 × 10 - 4 T
Charge of the electron, e = 1.6 × 10 - 19 C
Mass of the electron, me = 9.1 × 10 - 31 kg
Velocity of the electron, v = 4.8 × 106 m/s
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = v
Angular frequency of the electron = w = 2πv
Velocity of the electron is related to the angular frequency as:
v = rw
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as:
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.
(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
(a) Number of turns on the circular coil, n = 30
Radius of the coil, r = 8.0 cm = 0.08 m
Area of the coil
Current flowing in the coil, I = 6.0 A
Magnetic field strength, B = 1 T
Angle between the field lines and normal with the coil surface,
θ = 60°
The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,
T = n IBA sinθ … (i)
= 30 × 6 × 1 × 0.0201 × sin60°
= 3.133 N m
(b) It can be inferred from relation (i) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.
Radius of coil X, r1 = 16 cm = 0.16 m
Radius of coil Y, r2 = 10 cm = 0.1 m
Number of turns of on coil X, n1 = 20
Number of turns of on coil Y, n2 = 25
Current in coil X, I1 = 16 A
Current in coil Y, I2 = 18 A
Magnetic field due to coil X at their centre is given by the relation,
Where,
µ0 = Permeability of free space = 4π x 10-7TmA-1
Magnetic field due to coil Y at their centre is given by the relation,
Hence, net magnetic field can be obtained as:
Magnetic field strength, B = 100 G = 100 × 10- 4 T
Number of turns per unit length, n = 1000 turns m- 1
Current flowing in the coil, I = 15 A
Permeability of free space, µ0 = 4π x 10-7TmA-1
Magnetic field is given by the relation,
If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.
Radius of circular coil = R
Number of turns on the coil = N
Current in the coil = I
Magnetic field at a point on its axis at distance x is given by the relation,
Where,
µ0 = Permeability of free space
(a) If the magnetic field at the centre of the coil is considered, then x = 0.
This is the familiar result for magnetic field at the centre of the coil.
(b) Radii of two parallel co-axial circular coils = R
Number of turns on each coil = N
Current in both coils = I
Distance between both the coils = R
Let us consider point Q at distance d from the centre.
Then, one coil is at a distance of 
from point Q.
therefore, Magnetic field at point Q is given as:
Also, the other coil is at a distance of 
from point Q.
Magnetic field due to this coil is given as:
Total magnetic field,
Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.
Inner radius of the toroid, r1 = 25 cm = 0.25 m
Outer radius of the toroid, r2 = 26 cm = 0.26 m
Number of turns on the coil, N = 3500
Current in the coil, I = 11 A
(a) Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.
(b) Magnetic field inside the core of a toroid is given by the relation,
B = 
Where,
µ0 = Permeability of free space = 4π × 10-7 T m A-1
l = length of toroid
(c) Magnetic field in the empty space surrounded by the toroid is zero.
(a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.
(b) Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude.
(c) An electron travelling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain undeflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming's left hand rule, magnetic field should be applied in a vertically downward direction.
Magnetic field strength, B = 0.15 T
Charge on the electron, e = 1.6 × 10-19 C
Mass of the electron, m = 9.1 × 10-31 kg
Potential difference, V = 2.0 kV = 2 × 103 V
Thus, kinetic energy of the electron = eV
Where,
v = velocity of the electron
(a) Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.
Magnetic force on the electron is given by the relation,
B ev
Centripetal force 
From equations (1) and (2), we get
Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
(b) When the field makes an angle θ of 30° with initial velocity, the initial velocity will be,
From equation (2), we can write the expression for new radius as:
Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.
Current in the wire, I = 35 A
Distance of a point from the wire, r = 20 cm = 0.2 m
Magnitude of the magnetic field at this point is given as:
B 
Where,
μ0= Permeability of free space = 4π × 10–7 T m A–1
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10–5 T.
Magnetic field, B = 0.75 T
Accelerating voltage, V = 15 kV = 15 × 103 V
Electrostatic field, E = 9 × 105 V m - 1
Mass of the electron = m
Charge of the electron = e
Velocity of the electron = v
Kinetic energy of the electron = eV
Since the particle remains undeflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.
Putting equation (2) in equation (1), we get
This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He++, Li++, etc.
Length of the rod, l = 0.45 m
Mass suspended by the wires, m = 60 g = 60 × 10-3 kg
Acceleration due to gravity, g = 9.8 m/s2
Current in the rod flowing through the wire, I = 5 A
(a) Magnetic field (B) is equal and opposite to the weight of the wire i.e.,
A horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up in order to get zero tension in the wire. The magnetic field should be such that Fleming's left hand rule gives an upward magnetic force.
(b) If the direction of the current is revered, then the force due to magnetic field and the weight of the wire acts in a vertically downward direction.
∴Total tension in the wire = BIl + mg
= 0.26 x 5 x 0.45 + (60 x 10-3) x 9.8
= 1.176 N
Current in both wires, I = 300 A
Distance between the wires, r = 1.5 cm = 0.015 m
Length of the two wires, l = 70 cm = 0.7 m
Force between the two wires is given by the relation,
Where,
µ0 = Permeability of free space = 4π x 10-7TmA-1
Since the direction of the current in the wires is opposite, a repulsive force exists between them.
Magnetic field strength, B = 1.5 T
Radius of the cylindrical region, r = 10 cm = 0.1 m
Current in the wire passing through the cylindrical region, I = 7 A
(a) If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region.
Thus, l = 2r = 0.2 m
Angle between magnetic field and current, θ = 90°
Magnetic force acting on the wire is given by the relation,
F = BIl sin θ
= 1.5 × 7 × 0.2 × sin 90°
= 2.1 N
Hence, a force of 2.1 N acts on the wire in a vertically downward direction.
(b) New length of the wire after turning it to the Northeast-Northwest direction can be given as: :
Angle between magnetic field and current, θ = 45°
Force on the wire,
F = BIl1 sin θ
= BIl
= 1.5 x 7 x 0.2
= 2.1 N
Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angleθbecause l sinθ is fixed.
(c) The wire is lowered from the axis by distance, d = 6.0 cm
Suppose wire is passing perpendicularly to the axis of cylindrical magnetic field then lowering 6 cm means displacing the wire 6 cm from its initial position towards to end of cross sectional area.
Thus the length of wire in magnetic field will be 16 cm as AB= L =2x =16 cm
Now the force,
F = iLB sin90° as the wire will be perpendicular to the magnetic field.
F= 7 × 0.16 × 1.5 =1.68 N
The direction will be given by right hand curl rule or screw rule i.e. vertically downwards.
Magnetic field strength, B = 3000 G = 3000 × 10-4 T = 0.3 T
Length of the rectangular loop, l = 10 cm
Width of the rectangular loop, b = 5 cm
Area of the loop,
A = l × b = 10 × 5 = 50 cm2 = 50 × 10-4 m2
Current in the loop, I = 12 A
Now, taking the anti-clockwise direction of the current as positive and vise-versa:
(a) Torque, 
From the given figure, it can be observed that A is normal to the y-z plane and B is directed along the z-axis.
The torque 1.8 x 10-2 is N m along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.
(b) This case is similar to case (a). Hence, the answer is the same as (a).
(c) Torque 
From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis.
The torque 1.8 x 10-2 is N m along the negative x direction and the force is zero.
(d) Magnitude of torque is given as:
Torque is 1.8 x 10-2 N m at an angle of 240° with positive x direction. The force is zero.
(e) Torque
Hence, the torque is zero. The force is also zero.
(f) Torque
Hence, the torque is zero. The force is also zero.
In case (e), the direction of 
and 
is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.
Whereas, in case (f), the direction of and is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.
Number of turns on the circular coil, n = 20
Radius of the coil,r = 10 cm = 0.1 m
Magnetic field strength, B = 0.10 T
Current in the coil, I = 5.0 A
(a) The total torque on the coil is zero because the field is uniform.
(b) The total force on the coil is zero because the field is uniform.
(c) Cross-sectional area of copper coil, A = 10-5 m2
Number of free electrons per cubic meter in copper, N = 1029 /m3
Charge on the electron, e = 1.6 × 10-19 C
Magnetic force, F = Bevd
Where,
vd = Drift velocity of electrons
Hence, the average force on each electron is 5 x 10-25 N.
Length of the solenoid, L = 60 cm = 0.6 m
Radius of the solenoid,r = 4.0 cm = 0.04 m
It is given that there are 3 layers of windings of 300 turns each.
Total number of turns, n = 3 × 300 = 900
Length of the wire, l = 2 cm = 0.02 m
Mass of the wire, m = 2.5 g = 2.5 × 10-3 kg
Current flowing through the wire, i = 6 A
Acceleration due to gravity, g = 9.8 m/s2
Magnetic field produced inside the solenoid, 
Where,
µ0 = Permeability of free space = 4π x 10-7TmA-1
I = Current flowing through the windings of the solenoid
Magnetic force is given by the relation,
Also, the force on the wire is equal to the weight of the wire.
Hence, the current flowing through the solenoid is 108 A.
Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection, Ig = 3 mA = 3 × 10-3 A
Range of the voltmeter is 0, which needs to be converted to 18 V.
therefore, V = 18 V
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
Hence, a resistor of resistance 5998 Ω is to be connected in series with the galvanometer.
Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection,
= 4 mA = 4 × 10-3 A
Range of the ammeter is 0, which needs to be converted to 6 A.
Current, I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
Hence, a 10mΩ shunt resistor is to be connected in parallel with the galvanometer.
Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
Magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation, B
Where,
= Permeability of free space = 4π × 10–7 T m A–1
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell's right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.
Current in the power line, I = 90 A
Point is located below the power line at distance, r = 1.5 m
Hence, magnetic field at that point is given by the relation,
Where,
µ0 = Permeability of free space = 4π × 10–7 T m A–1
The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell's right hand thumb rule, the direction of the magnetic field is towards the South.
Current in the wire, I = 8 A
Magnitude of the uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, Ω¸ = 30°.
Magnetic force per unit length on the wire is given as:
f= BI sinΩ¸
= 0.15 x 8 x 1 x sin30°
= 0.6 N m-1
Hence, the magnetic force per unit length on the wire is 0.6 N m-1.
Length of the wire, l = 3 cm = 0.03 m
Current flowing in the wire, I = 10 A
Magnetic field, B = 0.27 T
Angle between the current and magnetic field, Ω¸ = 90°
Magnetic force exerted on the wire is given as:
F= BIlsinΩ¸
= 0.27 x 10 x 0.03 sin90°
= 8.1 x 10-2N
Hence, the magnetic force on the wire is 8.1 x 10-2 N. The direction of the force can be obtained from Fleming's left hand rule.
Current flowing in wire A, IA = 8.0 A
Current flowing in wire B, IB = 5.0 A
Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, l = 10 cm = 0.1 m
Force exerted on length l due to the magnetic field is given as:
Where,
µ0 = Permeability of free space = 4π × 10-7 T m A-1
The magnitude of force is 2 × 10-5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.
Length of the solenoid, l = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
Total number of turns on the solenoid, N = 5 × 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, I = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
Where,
µ0 = Permeability of free space = 4π × 10-7 T m A-1
Hence, the magnitude of the magnetic field inside the solenoid near its centre is 2.512 × 10-2 T.
Length of a side of the square coil, l = 10 cm = 0.1 m
Current flowing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with magnetic field, θ = 30°
Strength of magnetic field, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
T = n BIA sinθ
Where,
A = Area of the square coil
l × l = 0.1 × 0.1 = 0.01 m2
∴ T = 20 × 0.8 × 12 × 0.01 × sin30°
= 0.96 N m
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.
, approximately.