JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Electromagnetic Induction

Paragraph for Questions 9 and 10

A point charge

Q

is moving in a circular orbit of radius

R

in the

x - y

plane with an angular velocity

ω

. This can be considered as equivalent to a loop carrying a steady current

\frac{Q ω}{2 π}

. A uniform magnetic field along the positive

z

-axis is now switched on, which increases at a constant rate from 0 to

B

in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant

γ

.

The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is

(A) $- γ B Q R^{2}$
(B) $- γ \frac{B Q R ^{2}}{2}$ ✓ Correct
(C) $γ \frac{B Q R ^{2}}{2}$
(D) $γ B Q R^{2}$

Answer: (B)

Solution

The induced electric field

E = \frac{B R}{2}

exerts a force

F = QE = \frac{QB R}{2}

on the charge.

Torque on the charge:

τ = F R = \frac{QB R ^{2}}{2}

Change in angular momentum:

Δ L = \int_{0}^{1} τ d t = \frac{QB R ^{2}}{2} \times 1 = \frac{QB R ^{2}}{2}

Since

M = γ L

, the change in magnetic moment:

Δ M = γ Δ L = γ \frac{QB R ^{2}}{2}

The sign depends on the direction of motion. Since the direction is not specified, the magnitude is

γ \frac{B Q R ^{2}}{2}

, and the official answer is (B) with negative sign, indicating opposition to the applied field (Lenz's law).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Single correct option

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Electromagnetic Induction

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