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ExamsJEE MainPhysics

Paragraph: A point charge Q revolves in a circular path of radius R in the x-y plane with angular speed omega. This orbiting charge is equivalent to a current loop carrying steady current Qomega/(2pi). A uniform magnetic field directed along +z is then turned on, growing steadily from 0 to B over a time of one second, while the orbit radius is assumed unchanged. The growing field induces an emf around the orbit (the emf equals the work done by the induced electric field per unit positive charge taken once around the loop). For the orbiting charge, the magnetic dipole moment is proportional to the angular momentum, the constant of proportionality being gamma. The change in the magnetic dipole moment of the orbit at the end of the one-second interval is

  1. -gamma*B*Q*R²
  2. -gamma*B*Q*R²/2
  3. gamma*B*Q*R²/2
  4. gamma*B*Q*R²

Correct answer: -gamma*B*Q*R²/2

Solution

The increasing B creates an induced electric field along the orbit. This field tangentially accelerates the charge, changing its angular momentum and therefore (via M = gamma*L) its magnetic moment. Computing the impulse of the induced tangential force gives the change in L, and the sign is negative because the induced effect opposes the change (Lenz's law) for a positive charge in this geometry.

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