CBSE · NCERT · Class 12 Physics · Chapter 4

NCERT Solutions: Class 12 Physics Chapter 4 - Moving Charges and Magnetism

13 textbook Q&A13 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Moving Charges and Magnetism, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 13
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1Exercises13 questions
Q.4.1A 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?v
Solution

For an $N$-turn circular coil, $B=N\mu_0I/(2R)$. With $N=100$, $I=0.40\,\text{A}$ and $R=0.080\,\text{m}$, $B=100(4\pi\times10^{-7})(0.40)/(2\times0.080)=3.1\times10^{-4}\,\text{T}$.

Answer:

$3.1\times10^{-4}\,\text{T}$.

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Q.4.2A 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?v
Solution

For a long straight wire, $B=\mu_0I/(2\pi r)$. Thus $B=(4\pi\times10^{-7})(35)/(2\pi\times0.20)=3.5\times10^{-5}\,\text{T}$.

Answer:

$3.5\times10^{-5}\,\text{T}$.

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Q.4.3A 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.v
Solution

Magnitude is $B=\mu_0I/(2\pi r)=2\times10^{-7}(50)/2.5=4.0\times10^{-6}\,\text{T}$. Using the right-hand rule with current from north to south, the magnetic field at a point east of the wire is vertically upward.

Answer:

$4.0\times10^{-6}\,\text{T}$, vertically upward.

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Q.4.4A 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?v
Solution

The magnitude is $B=\mu_0I/(2\pi r)=2\times10^{-7}(90)/1.5=1.2\times10^{-5}\,\text{T}$. With current from east to west, the right-hand rule gives the field below the wire towards south.

Answer:

$1.2\times10^{-5}\,\text{T}$, towards south.

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Q.4.5What 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?v
Solution

Force per unit length is $F/l=IB\sin\theta$. Thus $F/l=8(0.15)\sin30^\circ=0.60\,\text{N m}^{-1}$.

Answer:

$0.60\,\text{N m}^{-1}$.

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Q.4.6A 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?v
Solution

The wire is perpendicular to the solenoid axis and hence perpendicular to the magnetic field, so $F=IlB$. With $l=0.030\,\text{m}$, $I=10\,\text{A}$ and $B=0.27\,\text{T}$, $F=10(0.030)(0.27)=8.1\times10^{-2}\,\text{N}$.

Answer:

$8.1\times10^{-2}\,\text{N}$.

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Q.4.7Two 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.v
Solution

Force on length $l$ of one parallel wire is $F=\mu_0I_1I_2l/(2\pi d)$. Substituting $I_1=8.0\,\text{A}$, $I_2=5.0\,\text{A}$, $l=0.10\,\text{m}$ and $d=0.040\,\text{m}$ gives $F=2\times10^{-7}(8)(5)(0.10)/0.040=2.0\times10^{-5}\,\text{N}$. Same-direction currents attract, so the force on A is towards B.

Answer:

$2.0\times10^{-5}\,\text{N}$, towards wire B.

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Q.4.8A 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.v
Solution

Total turns are $5\times400=2000$, and length is $0.80\,\text{m}$, so $n=2000/0.80=2500\,\text{m}^{-1}$. For a long solenoid, $B=\mu_0nI=(4\pi\times10^{-7})(2500)(8.0)=2.5\times10^{-2}\,\text{T}$.

Answer:

$2.5\times10^{-2}\,\text{T}$.

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Q.4.9A 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?v
Solution

Torque on a current loop is $\tau=NIAB\sin\theta$. Here $A=(0.10)^2=0.010\,\text{m}^2$, $N=20$, $I=12\,\text{A}$, $B=0.80\,\text{T}$ and $\theta=30^\circ$. Thus $\tau=20(12)(0.010)(0.80)(0.5)=0.96\,\text{N m}$.

Answer:

$0.96\,\text{N m}$.

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Q.4.10Two moving coil meters, M1 and M2 have the following particulars: R1= 10 Ω, N1= 30, A1 = 3.6 × 10–3 m2, B1 = 0.25 T R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10–3 m2, B2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.v
Solution

Current sensitivity is proportional to $NAB/k$. Since $k$ is the same, $S_{I2}/S_{I1}=(N_2A_2B_2)/(N_1A_1B_1)=42(1.8\times10^{-3})(0.50)/[30(3.6\times10^{-3})(0.25)]=1.4$. Voltage sensitivity is current sensitivity divided by resistance, so $S_{V2}/S_{V1}=1.4(R_1/R_2)=1.4(10/14)=1.0$.

Answer:

(a) $1.4$; (b) $1.0$.

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Q.4.11In 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 = 1.5 × 10–19 C, me = 9.1×10–31 kg)v
Solution

Because $\mathbf{v}\perp\mathbf{B}$, the magnetic force $e vB$ is perpendicular to the velocity at every instant and does no work; it only changes the direction of motion, producing circular motion. Equating $evB=mv^2/r$ gives $r=mv/(eB)$. With $B=6.5\times10^{-4}\,\text{T}$, $r=(9.1\times10^{-31})(4.8\times10^6)/[(1.5\times10^{-19})(6.5\times10^{-4})]=4.5\times10^{-2}\,\text{m}$.

Answer:

The magnetic force is always perpendicular to the velocity and acts as centripetal force; the radius is $4.5\times10^{-2}\,\text{m}$.

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Q.4.12In 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.v
Solution

The cyclotron frequency is $f=eB/(2\pi m)$. Using the printed constants, $f=(1.5\times10^{-19})(6.5\times10^{-4})/[2\pi(9.1\times10^{-31})]=1.7\times10^7\,\text{Hz}$. The speed cancels in the derivation because $r=mv/(eB)$ and $f=v/(2\pi r)$.

Answer:

$1.7\times10^7\,\text{Hz}$; it does not depend on the electron's speed.

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Q.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 60° 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.)v
Solution

The counter torque must equal the magnetic torque $\tau=NIAB\sin\theta$. The area is $A=\pi R^2=\pi(0.080)^2=2.01\times10^{-2}\,\text{m}^2$. Hence $\tau=30(6.0)(2.01\times10^{-2})(1.0)\sin60^\circ=3.1\,\text{N m}$. The torque on a planar current loop in a uniform field depends on $NIA$ and the angle between area vector and field, not on the detailed shape, so the answer would not change.

Answer:

(a) $3.1\,\text{N m}$. (b) No, if the area and all other particulars remain the same.

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