Question

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity \[\omega \]. This can be considered as equivalent to a loop carrying a steady current \[\dfrac{Q\omega }{2\pi }\]. 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 proportionally constant \[\gamma \]. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is
\[A)-\gamma BQ{{R}^{2}}\]
\[B)\dfrac{-\gamma BQ{{R}^{2}}}{2}\]
\[C)\gamma \dfrac{BQ{{R}^{2}}}{2}\]
\[D)\gamma BQ{{R}^{2}}\]

Answer 1

Hint: It is given that change in angular momentum is proportional to the change in magnetic moment. Also change in angular momentum is related to the torque. First, find the induced electric field and then the force acting on the particle. This force is proportional to the torque. Upon substituting the torque in angular momentum equation and integrating it, we can determine the change in angular momentum and magnetic moment.

Formula used:
\[\Delta L=\int{\tau dt}\]
\[\int{E.dr}=A\dfrac{dB}{dt}\]
\[F=qE\]

Complete answer:
Let, \[M\] be the magnetic moment and \[L\] be the angular momentum.
It is given that,
\[M\propto L\], \[M=\gamma L\]
Then,
\[\Delta M=\gamma \Delta L\] ------------A
We have, change in angular momentum,\[\Delta L=\int{\tau dt}\] -------- 1
Where,
\[\tau \]is the torque acting on the particle
For an induced electric field, \[\int{E.dr}=A\dfrac{dB}{dt}\]
Where,
\[\dfrac{dB}{dt}\]is the rate of change of magnetic flux
\[A\]is the area
\[E\int{dr}=A.B\]
Here, the point charge Q is moving in a circular orbit of radius \[R\],
\[E\times 2\pi R=\pi {{R}^{2}}B\]
Therefore,
\[E=\dfrac{RB}{2}\] ----------- 2
We have,
\[F=qE\]
Where,
\[F\]is the force on electric charge \[q\] due to induced electric field \[E\]
Now, due to the induced electric field, force on charge Q is given by,
\[F=QE\] ------------- 3
Substitute 3 in equation 2. We get,
\[F=\dfrac{QRB}{2}\] ----------- 4
We have,
Torque, \[\tau =FR\] ---------- 5
Substitute 4 in equation 5, then,
\[\tau =\dfrac{Q{{R}^{2}}B}{2}\] -------- 6
Substitute 6 in equation 1. Then,
\[\Delta L=\int\limits_{0}^{1}{\dfrac{Q{{R}^{2}}B}{2}}dt\]
\[\Delta L=\dfrac{Q{{R}^{2}}B}{2}\times 1\]
\[\Delta L=\dfrac{Q{{R}^{2}}B}{2}\] ------------7
Substitute the above equation in A. Then,
\[\Delta M=\gamma \dfrac{Q{{R}^{2}}B}{2}\]

So, the correct answer is “Option C”.

Note:
When a charge moves in a magnetic field, it experiences a force. The magnetic force is perpendicular to the plane formed by the velocity vector and magnetic field. The directions are given by the right hand rule. The direction of force on a negative charge is opposite to that of a positive charge. If velocity of the charge or particle happens to be aligned parallel to the magnetic field, or is zero, the magnetic force will be zero.