JEE Physics Moving Charges and Magnetism MCQs Set 14

Answer: (d) \( \frac{\pi^2}{8\sqrt2} \)

Explanation:
1. Circle: \( R=\frac{l}{2\pi},\ B_A=\frac{\mu_0I}{2R}=\frac{\pi\mu_0I}{l} \).
2. Square: \( a=\frac l4,\ B_B=4\cdot\frac{\mu_0I}{4\pi(a/2)}\cdot2\sin45^\circ=\frac{2\sqrt2\mu_0I}{\pi a}=\frac{8\sqrt2\mu_0I}{\pi l} \).
3. \( \frac{B_A}{B_B}=\frac{\pi^2}{8\sqrt2} \)

Teacher's Note:
1. Use the same wire length to express both R and a in terms of l.

Answer: (d) \( \frac{2\pi m}{qB} \)

Explanation:
1. \( T=\frac{2\pi r}{v}=\frac{2\pi r}{qBr/m}=\frac{2\pi m}{qB} \)

Teacher's Note:
1. The period is independent of speed and radius.

Answer: (a) \( S(R+G)=RG \)

Explanation:
1. Initially \( I=\frac{V_E}{R+G} \). With the shunt, the current through G is halved: \( \frac{V_E}{R+\frac{GS}{G+S}}\cdot\frac{S}{G+S}=\frac{V_E}{2(R+G)} \).
2. Simplifying: \( 2S(R+G)=R(G+S)+GS\Rightarrow S(R+G)=RG \)

Teacher's Note:
1. The galvanometer current falls to half, so the shunt branch must carry the other half in the new circuit.

Answer: (c) 400 A

Explanation:
1. \( B=\frac{\mu_0I}{2\pi a}\Rightarrow I=\frac{2\pi aB}{\mu_0}=\frac{2\pi(10^{-3})(0.08)}{4\pi\times10^{-7}}=400\ \text{A} \)

Teacher's Note:
1. The field at the surface of a long wire is μ0I/(2πa).

Answer: (c) 24

Explanation:
1. \( I=ef=1.6\times10^{-19}\times10^{16}=1.6\times10^{-3}\ \text{A} \)
2. \( m=I\pi r^2=1.6\times10^{-3}\times\pi\times(0.5\times10^{-10})^2=12.56\times10^{-24}\ \text{A m}^2 \), so n = 24.

Teacher's Note:
1. An orbiting electron is a current loop of current ef and area πr².

Answer: (c) 400 N m

Explanation:
1. Area \( A=\pi r^2=4\ \text{m}^2 \), so magnetic moment \( m=IA=4\ \text{A m}^2 \) along the z-axis.
2. The field is perpendicular to m, so \( \tau=mB\sin90^\circ=4\times100=400\ \text{N m} \)

Teacher's Note:
1. Torque is m B sin of the angle between the moment and the field; here the field lies in the plane of the loop.

Answer: (c) 48

Explanation:
1. The right angle is at P; sides PQ and PR pass through P and give no field. Distance of P from QR: \( h=\frac{3x\cdot4x}{5x}=2.4x \).
2. The foot of the perpendicular divides QR into 1.8x and 3.2x, so \( \sin\alpha_1=0.6,\ \sin\alpha_2=0.8 \).
3. \( B=\frac{\mu_0I}{4\pi h}(0.6+0.8)=\frac{7\mu_0I}{48\pi x} \), so k = 48.

Teacher's Note:
1. Only the side QR (not through P) contributes; use the finite straight wire formula.

Answer: (b) 1

Explanation:
1. \( E=\frac{600}{3\times10^{-3}}=2\times10^5\ \text{V/m} \)
2. \( B=\frac Ev=\frac{2\times10^5}{2\times10^4}=10\ \text{T} \), so x = 1.

Teacher's Note:
1. Undeviated motion means the electric and magnetic forces balance: eE = evB.

Answer: (d) 6

Explanation:
1. Element \( dq=\frac QL dx \) at distance x is a ring of current \( dI=\frac{dq\,\omega}{2\pi} \) and moment \( dm=dI\,\pi x^2=\frac{\omega}{2}\frac QLx^2dx \).
2. \( m=\frac{\omega Q}{2L}\cdot\frac{L^3}{3}=\frac{Q\omega L^2}{6} \), so λ = 6.

Teacher's Note:
1. Integrate the magnetic moments of thin rotating rings along the rod.

Answer: (b) 100 A

Explanation:
1. Equilibrium per unit length: \( \frac{\mu_0I_1I_2}{2\pi d}=\lambda g \).
2. \( I_2=\frac{0.2\times0.02}{2\times10^{-7}\times200}=100\ \text{A} \)

Teacher's Note:
1. The magnetic repulsion per unit length balances the weight per unit length.

Answer: (c) 19

Explanation:
1. Current ratio: galvanometer 0.05, shunt 0.95, and the two are in parallel: \( 0.05\,G=0.95\,S \).
2. \( S=\frac{G}{19}\Rightarrow k=19 \)

Teacher's Note:
1. In a parallel combination currents divide inversely as resistances.

Answer: (a) 1

Explanation:
1. At a/2 (inside): \( B=\frac{\mu_0I}{4\pi a} \); at 2a (outside): \( B=\frac{\mu_0I}{4\pi a} \).
2. The fields are equal, so the ratio is 1 : 1 and λ = 1.

Teacher's Note:
1. Inside the field grows with r; outside it decreases as 1/r.

Answer: (b) 30°

Explanation:
1. Charge 2e = 3.2 × 10⁻¹⁹ C: \( v=\sqrt{\frac{2qV}{m}}=10^6\ \text{m/s} \).
2. \( r=\frac{mv}{qB}=0.2\ \text{m} \).
3. \( \sin\theta=\frac dr=\frac{0.1}{0.2}\Rightarrow\theta=30^\circ \)

Teacher's Note:
1. Geometry of the circular arc: the thickness of the field region is r sin θ.

Answer: (c) 420

Explanation:
1. \( B_A=\frac{\mu_0NI}{2R}=\frac{\mu_0(24)(10)}{2(0.25)}=480\mu_0 \)
2. \( B_B=\frac{\mu_0(18)(15)}{2(0.15)}=900\mu_0 \)
3. Opposite directions: \( 900\mu_0-480\mu_0=420\mu_0 \)

Teacher's Note:
1. Subtract the fields because the currents circulate in opposite senses.