# NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 4: Moving Charges and Magnetism

NCERT Solutions for CBSE Class 12 Physics are available here. You can also download these NCERT solutions in PDF format with the help of download link available in this article. Some important topics of this chapter are Magnetic Force, Motion in a Magnetic Field, Motion in Combined Electric and Magnetic Fields, Magnetic Field due to a Current Element, Biot-Savart Law, Magnetic Field on the Axis of a Circular Current Loop, Ampere’s Circuital Law, Solenoid and Toroid, Force between two Parallel Currents, The Ampere, Force between two Parallel Currents, Torque on Current Loop, Magnetic Dipole, Moving Coil Galvanometer, Cyclotron. Most of the questions given in this chapter are based on these topics. These NCERT Solutions are important for CBSE 12th Physics board exam 2018.

*NCERT Solutions for Class 12 Physics ‒ Chapter 4: Moving Charges and Magnetism are given below*

**Question4.1:** A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

**Solution4.1:**

Given:

N = 100

R = 8cm

I = 0.4 A

By the formula 𝐵 = (μ_{o}*IN*)/(2*R*)

Substitution values, we have,

B = 3.1x10^{‒4} T.

**NCERT Exemplar: CBSE Class 12 Physics – Chapter 4 Moving Charges and Magnetism**

**Question4.2: ** A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

**Solution4.2:**

Given:

*I *= 35 A

*x *= 20 cm

Let us consider an Amperian loop of radius equal to *x*, i.e. 20 cm

Therefore by formula

𝐵 =(μ_{o}*I*)/(2πR)

Solving for B

B = 3.5 × 10^{ ‒5} T.

**Question4.3:** A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

**Solution4.3:**

Given:

*I* = 50 A

*R* = 2.5 m

Using the above formula

B = 4 × 10^{‒6} T

Direction is vertical up.

**Question4.4:** A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

**Solution4.4:**

Given:

I = 90 A

R = 1.5 m

Considering Amperian loop and using the above formula

B = [(2 × 10^{‒7}) × (90)] / 1.5

Or B = 1.2 × 10^{‒5} T

Direction is towards south.

**Question4.5:** What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

**Solution4.5:**

Given:

I = 8 A

Angle = θ = 30^{o}

B = 0.15 T

*l* = 1m

The force is given by

F = I . *l* .B. sin θ

Or F = 8 x 1 x 0.15 x 0.5

Or F = 0.6 N / m.

**CBSE Class 12th Physics Notes: All Chapters**

**Question4.6:** A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

**Solution4.6:**

Given:

*l *= 3 cm

I = 10 A

B = 0.27 T

θ = 90^{o}

Then the force

F = I. *l* B. sin θ

Substituting the values, we have

F = 0.081 N.

**Question4.7:** Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

**Solution4.7:**

Given:

*I _{a} *= 8 A

*I _{b} *= 5 A

*d* = 4 cm

*L* = 10 cm

Force of wire A

F = (μ_{o}) × [(*I _{a} I_{b}* L)/(2π

*d*)]

Or

F = 2 × 10^{‒5} N.

**Question4.8:** A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre. 4.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

**Solution4.8:**

Given:

*L* = 80 cm

*N* = 2000

r = 0.9

I = 8 A

Enclosed current I_{e} = 8 x 2000 = 16000 A

Since, BL =μ_{o} I_{e}

Putting the values

B = *μ*_{o} (16000/0.8) = 0.025 T

**Question4.9:** A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

**Solution4.9:**

L = 10 cm

N = 20

I = 12 A

θ = 30^{o}

B = 0.8 T

We know

Torque = *m B *sin θ

Or T = (*NIA*) *B* sin θ

Or T = 20 x 12 x 0.1 x 0.1 x 0.8 x 0.5

Or T = 0.96 Nm.

**Question4.10:** Two moving coil meters, M1 and M2 have the following particulars:

R_{1 }= 10 Ω, N_{1} = 30, A_{1} = 3.6 × 10^{–3} m^{2}, B_{1} = 0.25 T, R_{2} = 14 Ω, N_{2} = 42, A_{2} = 1.8 × 10^{–3} m^{2}, B_{2} = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_{2} and M_{1}.

**Solution4.10:**

Details of M_{1}

R_{1} = 10 ohms

N_{1} = 30

A_{1 }= 3.6x10^{-3} m^{2}

B_{1} = 0.25 T

Details of M_{2}

R_{2} = 14 ohms

N_{2} = 42

A_{2} = 1.8x10^{-3} m^{2}

B_{2} = 0.5 T

(a) Current sensitivity of M_{1} = (N_{1}.A_{1}.B_{1})_{ }/ k

Current sensitivity of M_{2} = (N_{2}.A_{2}.B_{2}) / k

Ratio of current sensitivity = N_{2}A_{2}B_{2} / N_{1}A_{1}B_{1} (since k is identical)

= 0.0378 / 0.027

= 1.4

Thus ratio of current sensitivity of M_{2} and M_{1} is 1.4

(b) Voltage sensitivity of M_{1} = (N_{1}.A_{1}.B_{1})/ (k.R_{1})

Voltage sensitivity of M_{2} = (N_{2}.A_{2}.B_{2})/ (k.R_{2})

Ratio of voltage sensitivity = (N_{2}A_{2}B_{2}R_{1})/(N_{1}A_{1}B_{1}R_{2})= 1.

Thus ratio of voltage sensitivity of M_{2 }and M_{1} is 1.

**Question4.11:** In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10–19 C, me = 9.1×10–31 kg).

**Solution4.11:**

Since the electron has velocity perpendicular to magnetic field B, the centripetal force (*q*.*V* × *B*) comes in action and a circular motion is described by electron perpendicular to the magnetic field.

Its radius is given by

r = (*mv* /(*qB*)

Here

m = 9.1 × 10^{‒31} kg

v = 4.8 × 10^{6} m / s

q = 1.6 × 10^{-19} C

B = 6.5 × 10^{-4} T

Substituting these values we get

*r* = 0.042 m.

**Question4.12:** In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

**Solution4.12:**

Frequency of revolution, ω = *v */ *r*

Or 2𝜋*f* = 4.8x106 / 0.042

Or* f* = 18 MHz

Frequency is independent of velocity.

4.13 (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.)

**Solution4.13:**

N = 30

r = 8 cm

I = 6 A

B = 1T

θ = 60^{o}

(*a*) We know

Torque = N I A B sinθ

Substituting the values

Torque = 3.13 Nm

(*b*) No, the torque does not depend on configuration of loop.

**Moving Charges and Magnetism - CBSE Class 12th NCERT Solution**