NCERT Questions for Class 12 Physics Chapter 4 – Moving Charges and Magnetism

Important Questions with Solutions of Class 12 Physics Chapter 4 – Moving Charges and Magnetism

1) An alpha element & proton travel in the sheet plane where uniform magnetic field B is directed parallel to the sheet plane. When the 2 elements have the same linear motion, calculate the radii ratio in which they travel in the field.

Ans – As the radius of the path will be R= mv/Bq

R ∝ 1/q

Here, R_α & R_p are the radii of the α-particle & the proton. Also, q_α & q_p are their charges. R_α:R_p = 1:2

Thus, the desired ratio is 1:2.

2) Create an equation for the force acting on a particle of charge q moving with speed in a magnetic field B. Express in the face of this force.

(a) K.E. of the elements won’t change.

Ans – As you know, the magnetic force can be denoted as,

As the direction of the force is perpendicular to the plane containing

F = qvBsin90° = qvB

Here we witness force & displacement perpendicular to each other

Thus, W=FScosθ  

⇒ W = FScos90° = 0

⇒ KE = 0

Hence, we know the kinetic energy constant at the given state.

b) When the instantaneous energy is zero.

Ans: We have an immediate expression of strength that p = Fvcosθ

When power and velocity are relative to each other p = Fvcos90° = 0

Hence, we see that the instantaneous potential is zero.

3) A horizontal overhead power cable contains a current of 90 A from east to west. What will be the magnitude & direction of the magnetic field when water flows 1.5 m down the path?

Ans – As you know,

Current in the power line, I = 90A

There is a point downstream of the remote power line r = 1.5m Thus, the magnetic field at that point is denoted as,

where µ₀ the permeability of the free space is 4π x 10^-7 T m A^-1

⇒ B = 1.2 x 10^-5 T

Since, the current flows from east to west & the mentioned point is down the power line, incorporating Maxwell’s right-hand rule we obtain the magnetic field direction in the south.

4) A 3.0 cm wire containing a current of 10A is injected into a solenoid perpendicular to its axis. Given that the magnetic field in the solenoid is 0.27 T, what is the magnetic force on the wire?

Ans – As you know,

The length of wire, I = 3cm = 0.03m

Current flowing in the wire I = 10 A

Magnetic field, B = 0.27T 

The angle between current & magnetic field 0 = 90°

The magnetic force applied to the wire can be given as F = BIlsinθ

Substituting the values given, we obtain

⇒ F = 0.27 x 10 x 0.03sin90°

⇒ F = 8.1 x 10^-2 N

Thus, the magnetic potential on the wire is 8.1 x 10^-2 N & the force is measured using Fleming’s left-hand rule.

5) Two long & parallel straight wires A & B have currents of 8.0 A & 5.0 A in the same direction and are distinguished by a distance of 4.0 cm. Calculate the force in a 10 cm section of wire A.

Ans – As you know,

Current flowing in wire A, I_A = 8.0A

Current flowing in wire B, I_B = 5.0A

Spacing between two wires r = 4.0 cm = 0.04m

The A wire one-half length I = 10cm = 0.1m

The force exerted on a length I due to a magnetic field is

Where µ₀ = permeability of free space is 4π × 10^-7 TmA^-1

⇒ B = 2 x 10^-5 N

The magnitude of the force is 2 x 10^-3 N. This is the normal gravitational force for A to B since the direction of the current in the wires is the equivalent.

6) A square coil of 10cm diameter makes 20 turns & has a current of 12A. The coil is suspended vertically & is normal to the coil plane. It creates an angle of 30° in the direction of a uniform horizontal magnetic field of magnitude 0.80T. Calculate the magnitude of the torque developed by the coil.

Ans – As you know,

The length of one side of a rectangular coil, I = 10cm = 0.1m

Area of square, A = I² = (0.1)² = 0.01m²

Current flowing in the coil, I = 12A

Number of turns on the coil, n = 20

The angle of the coil plane with magnetic field θ = 30°

Magnetic field strength B = 0.80T

The highest magnetic torque received by the coil in the magnetic field is denoted as,

τ = nIABsinθ

If we use the values given replacement we get,

τ = 20 x 0.8 x 12 x 0.01 x sin30°

τ = 0.96 Nm

Hence the maximum torque developed by the coil is 0.96 Nm.

7) a) A circular coil of 30 turns & a radius of 8.0 cm contains a current of 6.0A and is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0T. The field lines create an angle with the coil. What will be the applied magnitude of the counter-torque to prevent the coil from spinning?

Ans – As you know,

The number of turns on a circular coil, n = 30

Radius of coil, r = 8.0cm = 0.08m

Area of coil = πr² = π(0.08)² = 0.0201 m²

Current flowing in a coil is given by I = 6.0A

Magnetic field strength, B = 1T

The angle between field lines & normal to coil surface, θ = 60°

The coil is subjected to torque in the magnetic field & it rotates. The counter-torque is used to prevent the coil from rotating and is denoted as,

T = nIBAsinθ …(1) 

⇒ T = 30 x 6 x 1 x 0.0201 x sin60°

⇒ T = 3.133 Nm

Hence, the applied counter torque against coil change is 3.133 Nm.

b) Would you modify your answer when the sphere in (a) was replaced by a thinner coil with some irregular shape connecting the same area? (Given that, the rest of the values remain unchanged).

Ans – From relation (1) we can infer that the magnitude of the applied torque does not depend on the size of the coil. Depending on the location of the coil. Thus, if the circular coil in the above case is replaced by a flat coil of some irregular shape encompassing the same area, the latter remains unchanged.

8) Answer the following questions:

a) Magnetic field varying in magnitude from point to point but maintaining a constant direction (east to west) during the orbit. A charged element enters a chamber & moves unchanged along a straight line at constant velocity. How will you define the initial velocity of that element?

Ans – The initial velocity of the element can be parallel/antiparallel to the magnetic field. Since it travels in a straight line without any local distractions.

b) The charged particles enter a stable and inhomogeneous state magnetic field of varying magnitude & direction. It follows complicated paths and results. Do you believe that the final velocity will equal the initial velocity if it does not collide with the environment?

Ans – Yes, the final acceleration of a charged particle will be equal to the initial acceleration because the magnetic field can change the direction of the acceleration, but not its magnitude.

c) An electron flowing from west to east passes through a chamber with a uniform electrical charge from north to south. Describe the direction in which a uniform magnetic field must be maintained to restrict the electron from changing its orthogonal axis.

Ans – An electron moving from west to east passes through a chamber with equal electrical charge in a north-south direction. When the electric field acting on it is equal & opposite to the magnetic field, this flow of electrons won’t be disturbed. The magnetic field would move to the right. Also, by Fleming’s left-hand law, the magnetic field must be applied downwards.

9) A solenoid is 60cm long & 4.0cm in radius having 3 levels of 300-turn windings. A 2.0cm long wire of mass 2.5g is normal to the axis of the solenoid (near its centre). The wire & the axis of the solenoid are in the horizontal plane. The wire is connected to the external battery by 2 wires parallel to the solenoid axis that contains a 6.0A current to the wire. Calculate the value of current present in the windings of the solenoid (imagine it conducts well) in the solenoid that can support the weight of the wire. g = 9.8ms^-2.

Ans – Since solenoid length is denoted as, L = 60cm = 0.6m

Radius of solenoid, r = 4cm = 0.04m

Given that there are 3 layers of winding making 300 turns each Total power, n = 3 x 300 = 900

Wire length I = 2cm = 0.02m

Thickness of the wire, m = 2.5g = 2.5 x 10^-3 kg

Current flowing through the wire, i = 6A

Acceleration due to gravity, g = 9.8ms^-2

The magnetic field generated inside the solenoid, B = µ₀nI/L 

Where µ₀ the permeability of free space µ₀ = 4π x 10^-7 TmA^-1

The current flowing through the windings of the solenoid, I

The magnetic force is given by the relation,

Now, we have the force on the string with the string equal in weight

⇒ I = 108A

The current flowing through the solenoid is 108A.