JEE Physics Moving Charges and Magnetism MCQs Set 22

Answer: (c) zero at every point

Explanation:
1. Take a circular Amperian loop inside the pipe: it encloses no current.
2. So \( B\cdot2\pi r=0 \) and the field is zero everywhere inside.

Teacher's Note:
1. Ampere's law with a loop inside the shell encloses no current.
2. The same holds for a hollow cylinder.

Answer: (b) outside the cable

Explanation:
1. Outside the cable the Amperian loop encloses net current \( I-I=0 \), so the field is zero.
2. Inside the inner conductor, in the gap and in the outer conductor the loop encloses a nonzero current.

Teacher's Note:
1. Zero net enclosed current gives zero field.
2. This is why coaxial cables do not radiate magnetic fields.

Answer: (a) \( \sqrt3\,\dfrac{\mu_0 I}{2R} \)

Explanation:
1. Each ring gives \( \dfrac{\mu_0I}{2R} \) along its own axis.
2. The three axes are mutually perpendicular, so \( B=\sqrt{3}\,\dfrac{\mu_0I}{2R} \).

Teacher's Note:
1. Add perpendicular fields by Pythagoras in three dimensions.
2. Equal components give a factor \( \sqrt3 \).

Answer: (d) \( \dfrac{4}{\sqrt7} \)

Explanation:
1. A small coil acts as a dipole: \( B=\dfrac{\mu_0M}{4\pi r^3}\sqrt{1+3\cos^2\theta} \).
2. On the axis (\( \theta=0 \)): factor 2. At 60°: \( \sqrt{1+3/4}=\sqrt7/2 \).
3. Ratio \( =2\Big/\dfrac{\sqrt7}{2}=\dfrac{4}{\sqrt7} \).

Teacher's Note:
1. Use the general dipole field formula at angle θ from the axis.
2. Axial value is twice the equatorial value.

Answer: (b) \( \mu_0\sigma v_0 \)

Explanation:
1. The rotating charged surface is a surface current of \( K=\sigma v_0 \) per unit length.
2. This is a solenoid: \( B=\mu_0K=\mu_0\sigma v_0 \) inside.

Teacher's Note:
1. Moving surface charge is a surface current.
2. The field inside is uniform, outside it is zero.

Answer: (d) \( \dfrac{\mu_0 i c}{2\pi(b^2-a^2)} \)

Explanation:
1. Current density \( J=\dfrac{i}{\pi(b^2-a^2)} \). Superpose a full rod of density \( J \) and a filled hole of density \( -J \).
2. At the hole axis, the full rod gives \( B_1=\dfrac{\mu_0Jc}{2} \); the filled hole gives zero at its own centre.
3. So \( B=\dfrac{\mu_0 i c}{2\pi(b^2-a^2)} \).

Teacher's Note:
1. Use the principle of superposition with a negative current density in the hole.
2. Field inside a solid wire is \( \mu_0Jr/2 \).

Answer: (a) \( \dfrac{m\cot\theta}{qB} \)

Explanation:
1. The block slides with \( v=g\sin\theta\,t \).
2. The magnetic force \( qvB \) is perpendicular to the plane and pointing away from it. Contact is lost when \( qvB=mg\cos\theta \).
3. So \( t=\dfrac{m\cos\theta}{qB\sin\theta}=\dfrac{m\cot\theta}{qB} \).

Teacher's Note:
1. Contact is lost when the normal reaction becomes zero.
2. The magnetic force grows with speed.

Answer: (a) \( \dfrac{25}{2\alpha E_0} \)

Explanation:
1. The magnetic force does no work. Only the electric force changes the kinetic energy: \( qE_0x_0=\tfrac12mv^2 \).
2. Speed squared is \( 16+9=25 \), so \( x_0=\dfrac{25}{2\alpha E_0} \).

Teacher's Note:
1. Work-energy theorem: magnetic force does zero work.
2. Use the speed from the velocity components.

Answer: (b) \( t=\dfrac{\sqrt3\,mv_0}{qE_0} \)

Explanation:
1. The magnetic field only rotates the velocity component perpendicular to it, keeping that component's magnitude \( v_0 \).
2. Along \( x \) the electric field accelerates the particle: \( v_x=\dfrac{qE_0}{m}t \).
3. Speed: \( v_0^2+v_x^2=4v_0^2\Rightarrow v_x=\sqrt3v_0 \), so \( t=\dfrac{\sqrt3mv_0}{qE_0} \).

Teacher's Note:
1. With E parallel to B the two motions separate.
2. The perpendicular speed stays constant.

Answer: (b) \( 1.96\times10^{-5} \) T

Explanation:
1. \( B=B_H\tan30^\circ=0.34\times10^{-4}\times0.577 \).
2. So \( B\approx1.96\times10^{-5} \) T.

Teacher's Note:
1. The tangent law relates the two perpendicular fields.
2. tan 30° = 0.577.

Answer: (d) \( 2\sqrt{\dfrac{\tau}{\sqrt3\,BI}} \)

Explanation:
1. Field in the plane of the coil, so \( \tau=IAB \) (maximum).
2. Area of the triangle \( A=\dfrac{\sqrt3}{4}\ell^2 \).
3. So \( \ell^2=\dfrac{4\tau}{\sqrt3BI} \) and \( \ell=2\sqrt{\dfrac{\tau}{\sqrt3BI}} \).

Teacher's Note:
1. The coil's area decides the torque, not its shape.
2. Take the square root at the end.

Answer: (a) \( 3.18\times10^{-4} \) J

Explanation:
1. \( W=MB(\cos30^\circ-\cos45^\circ) \).
2. Numerically \( 4\times5\times10^{-4}\,(0.866-0.707)\approx3.18\times10^{-4} \) J.

Teacher's Note:
1. Work done by the external agent equals the increase in potential energy.
2. Potential energy is \( -MB\cos\theta \).

Answer: (a) the magnetic equator

Explanation:
1. \( B_V=B\sin\delta \), where \( \delta \) is the angle of dip.
2. The dip is zero at the magnetic equator, so \( B_V=0 \) there.

Teacher's Note:
1. Dip is zero at the equator and 90° at the poles.
2. The horizontal component is zero at the magnetic poles.

Answer: (d) 4 times the original

Explanation:
1. Length \( 2\pi R=N\cdot2\pi R' \) with \( N=2 \) gives \( R'=R/2 \).
2. Field \( B'=\dfrac{\mu_0NI}{2R'}=\dfrac{\mu_0\cdot2I}{2(R/2)}=4\cdot\dfrac{\mu_0I}{2R} \).

Teacher's Note:
1. Number of turns doubles and radius halves.
2. Both effects increase the field.

Answer: (b) \( 4\times10^{-5} \) T

Explanation:
1. Distance of the centroid from each side: \( d=\dfrac{a}{2\sqrt3}=1.3\times10^{-2} \) m.
2. One side gives \( \dfrac{\mu_0I}{4\pi d}(\sin60^\circ+\sin60^\circ) \) and the three add (same sense).
3. Total \( =3\times\dfrac{10^{-7}\times1\times1.732}{1.3\times10^{-2}}=4\times10^{-5} \) T.

Teacher's Note:
1. The perpendicular distance to each side is the inradius.
2. All three sides give fields in the same direction.

Answer: (c) \( 1.0\times10^{-4} \) T

Explanation:
1. Sides: \( a=d\sin15^\circ=4.14 \) cm and \( b=d\cos15^\circ=15.45 \) cm.
2. Each side contributes \( \dfrac{\mu_0I}{4\pi h}\cdot2\sin\alpha \) with \( h \) the half of the other side and \( \sin\alpha=\dfrac{L/2}{\sqrt{L^2/4+h^2}} \).
3. Adding the four sides (all in the same sense) gives \( B\approx1.0\times10^{-4} \) T.

Teacher's Note:
1. All four sides contribute in the same direction at the centre.
2. Use the finite-wire formula for each side.

Answer: (d) \( \sqrt2\,B \)

Explanation:
1. The two fields are perpendicular to each other (along the two axes).
2. \( B_{net}=\sqrt{B^2+B^2}=\sqrt2B \).

Teacher's Note:
1. Perpendicular vectors add by Pythagoras.
2. Equal magnitudes give a factor \( \sqrt2 \).

Answer: (d) (1), (2) and (3) only

Explanation:
1. Inside: \( B=\dfrac{\mu_0Ir}{2\pi R^2}\propto r \), so (1) is true and the field at the centre is zero, so (3) is true and (4) false.
2. Outside: \( B=\dfrac{\mu_0I}{2\pi r}\propto1/r \), so (2) is true.

Teacher's Note:
1. Linear rise inside, inverse fall outside, maximum at the surface.
2. Zero at the axis by symmetry.

Answer: (a) either the y axis or the z axis

Explanation:
1. The force must be perpendicular to the velocity, so the field must be perpendicular to the x axis.
2. A field along \( y \) or along \( z \) turns the electron through a semicircle in the xz or xy plane, respectively, reversing its velocity.
3. A field along \( x \) gives no force.

Teacher's Note:
1. Any field perpendicular to the velocity makes the particle turn in a circle.
2. Applying it for half a period reverses the direction.