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← JEE Advanced 2013 · Paper 2
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JEE Advanced 2013 · Paper 2, Physics question 10

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Paragraph for Questions 9 and 10

A point charge QQ is moving in a circular orbit of radius RR in the xx-yy plane with an angular velocity ω\omega. This can be considered as equivalent to a loop carrying a steady current Qω2π\dfrac{Q\omega}{2\pi}. A uniform magnetic field along the positive zz-axis is now switched on, which increases at a constant rate from 0 to BB 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\gamma.

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

Choose one option
Answer and worked solution

Official answer: B

Torque from the induced field. The tangential force QE=QBR2QE=\frac{QBR}2 acts for 1 s with lever arm RR, changing the angular momentum by QBR22\frac{QBR^2}2.

Direction. By Lenz's law the induced field opposes the increase of flux, so the change in magnetic moment points opposite to B⃗\vec B:
Δμ=−γBQR22.\Delta\mu=-\gamma\frac{BQR^2}2.

(B)