JEE Physics Moving Charges And Magnetism MCQs Set 11

Answer: (a) 47 A

Explanation:
1. Weight of the upper wire \( mg=1.2\times10^{-3}\times10=0.012\ \text{N} \).
2. Magnetic repulsion between the parallel wires (opposite currents): \( F=\frac{\mu_0I^2\ell}{2\pi d}=2\times10^{-7}\frac{I^2(0.2)}{0.0075} \).
3. Equate: \( 5.33\times10^{-6}I^2=0.012\Rightarrow I\approx47\ \text{A} \).

Teacher's Note:
1. For equilibrium the repulsive force balances the weight.
2. Use the force between two parallel currents.

Answer: (b) zero

Explanation:
1. Each element of the coil feels a force \( I\,d\vec\ell\times\vec B_0 \), all in the plane of the coil and directed radially outward (or inward).
2. The elements on opposite sides cancel, so the net force on a closed loop in a uniform field is zero.

Teacher's Note:
1. The net force on a closed current loop in a uniform field is always zero.
2. Only the torque may be non-zero.

Answer: (a) tends to get shorter

Explanation:
1. Neighbouring turns carry current in the same direction, and parallel currents attract.
2. So the turns attract each other and the spring tends to contract.

Teacher's Note:
1. Parallel currents attract; antiparallel currents repel.
2. Adjacent turns of a coil are parallel current elements.

Answer: (d) F/18

Explanation:
1. \( F\propto\frac{i_1i_2}{d} \).
2. New force \( =F\times\frac13\times\frac13\times\frac12=\frac F{18} \).

Teacher's Note:
1. The force is proportional to the product of currents and inversely to the distance.
2. Include both current factors.

Answer: (c) \( 0.628\ \text{A m}^2 \)

Explanation:
1. \( M=NIA=10\times2\times\pi(0.1)^2=0.628\ \text{A m}^2 \).

Teacher's Note:
1. Magnetic moment = number of turns × current × area.
2. Area of the coil is πr².

Answer: (b) \( Ni\pi r^2\times\frac{\mu_0I}{2R} \)

Explanation:
1. Field at the centre of the big ring \( B=\frac{\mu_0I}{2R} \) along its axis.
2. The moment of the small ring \( M=Ni\pi r^2 \) lies in the plane of the big ring, perpendicular to B, so \( \tau=MB\sin90^\circ=Ni\pi r^2\frac{\mu_0I}{2R} \).

Teacher's Note:
1. Perpendicular planes mean the angle between M and B is 90°.
2. Torque = M B sin θ.

Answer: (a) \( 2\pi NiR^2B \)

Explanation:
1. Potential energy \( U=-MB\cos\theta \) with \( M=Ni\pi R^2 \).
2. \( W=U_{180^\circ}-U_{0^\circ}=MB+MB=2MB=2\pi NiR^2B \).

Teacher's Note:
1. Work done equals the change in potential energy.
2. From 0° to 180° the change is 2MB.

Answer: (d) 1

Explanation:
1. For N turns the radius is \( r=\frac{\ell}{2\pi N} \).
2. Torque \( \tau_{max}=NiAB=Ni\pi\left(\frac{\ell}{2\pi N}\right)^2B=\frac{i\ell^2B}{4\pi N} \).
3. This is maximum for the smallest N, that is N = 1.

Teacher's Note:
1. The area falls as 1/N², faster than N rises.
2. A single large loop gives the greatest torque for a given length of wire.

Answer: (c) \( \frac{\mu_0i}{16r} \)

Explanation:
1. Field at the centre of an arc: \( B=\frac{\mu_0i}{4\pi r}\theta \).
2. \( B=\frac{\mu_0i}{4\pi r}\cdot\frac\pi4=\frac{\mu_0i}{16r} \).

Teacher's Note:
1. A full circle (θ = 2π) gives μ0i/2r.
2. The field is proportional to the angle subtended.

Answer: (b) \( 6.28\times10^{-5}\ \text{T} \)

Explanation:
1. Circumference of one turn \( \pi\times0.2=0.628\ \text{m} \), so the number of turns \( N=\frac{6.28}{0.628}=10 \).
2. \( B=\frac{\mu_0NI}{2r}=\frac{4\pi\times10^{-7}\times10\times1}{2\times0.1}=6.28\times10^{-5}\ \text{T} \).

Teacher's Note:
1. Find the number of turns from the wire length.
2. The coil radius is 0.1 m.

Answer: (a) 1 : 1 : 2

Explanation:
1. Kinetic energy gained \( K=qV \).
2. Charges: proton \( e \), deuteron \( e \), α-particle \( 2e \), so \( K:K:K=1:1:2 \).

Teacher's Note:
1. Kinetic energy depends on the charge, not the mass, for a given voltage.
2. An α-particle has twice the charge of a proton.

Answer: (a) 0.1

Explanation:
1. Electric field between the plates \( E=\frac Vd=\frac{600}{3\times10^{-3}}=2\times10^5\ \text{V/m} \).
2. For no deflection the magnetic force balances the electric force: \( evB=eE \).
3. \( B=\frac Ev=\frac{2\times10^5}{2\times10^6}=0.1\ \text{Wb/m}^2 \).

Teacher's Note:
1. Undeflected motion means a velocity selector: E = vB.
2. E = V/d for the parallel plates.

Answer: (d) \( 7.44\times10^{-12}\ \text{N} \)

Explanation:
1. Kinetic energy \( =5\times10^6\times1.6\times10^{-19}=8\times10^{-13}\ \text{J} \).
2. Speed \( v=\sqrt{\frac{2K}m}=\sqrt{\frac{1.6\times10^{-12}}{1.67\times10^{-27}}}\approx3.1\times10^7\ \text{m/s} \).
3. The velocity is perpendicular to B: \( F=qvB=1.6\times10^{-19}\times3.1\times10^7\times1.5\approx7.4\times10^{-12}\ \text{N} \).

Teacher's Note:
1. Convert MeV to joules first.
2. The proton moves perpendicular to the field, so sin θ = 1.

Answer: (c) \( 7\times10^6\ \text{V} \)

Explanation:
1. Radius \( r=\frac{mv}{qB}\Rightarrow v=\frac{qBr}{m}=\frac{3.2\times10^{-19}\times1.2\times0.45}{6.8\times10^{-27}}\approx2.5\times10^7\ \text{m/s} \).
2. Kinetic energy \( K=\tfrac12mv^2=\frac{q^2B^2r^2}{2m}\approx2.2\times10^{-12}\ \text{J} \).
3. \( V=\frac Kq=\frac{2.2\times10^{-12}}{3.2\times10^{-19}}\approx7\times10^6\ \text{V} \).

Teacher's Note:
1. The kinetic energy gained equals qV.
2. Find the speed from the radius of the circular path.

Answer: (b) 0.125 N

Explanation:
1. The flexible loop becomes a circle: \( 2\pi r=0.5\Rightarrow r=0.0796\ \text{m} \).
2. Tension \( T=BIr=1.0\times1.57\times0.0796=0.125\ \text{N} \).

Teacher's Note:
1. The outward magnetic force on each element is balanced by the tension.
2. For a circular element, T = BIr.

Answer: (a) \( \frac{Bid}{m}t \)

Explanation:
1. The force on the wire \( F=Bid \) is constant and horizontal, along the rails.
2. Acceleration \( a=\frac{Bid}{m} \), so \( v=at=\frac{Bid}{m}t \) (directed away from the battery).

Teacher's Note:
1. A constant current in a uniform field gives a constant force.
2. Starting from rest, v = at.