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 $\frac{Q\omega }{2\pi }$. 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 applications 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 $\gamma$.

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

• A. $2\gamma BQ{R}^2$
• B. $-\gamma \frac{BQ{R}^2}{2}$
• C. $\frac{\gamma BQ{R}^2}{2}$
• D. $\gamma BQ{R}^2$

Answer 1

The correct option is B

$-\gamma \frac{BQ{R}^2}{2}$

$\frac{M}{L}=\frac{Q}{2m}\therefore M=\left(\frac{Q}{2m}\right)L\Rightarrow M\alpha L$
where $\gamma =\frac{Q}{2m}$
$\left(\frac{Q}{2m}\right)\left(I\omega \right)=\left(\frac{Q}{2\omega }\right)\left(m{R}^2\omega \right)=\frac{Q\omega R^2}{2}$
Induced electric field is opposite so,
${\omega }^{\prime }=\omega -\alpha t\alpha =\frac{\tau }{I}=\frac{\left(QE\right)R}{m{R}^2}=\frac{\left(Q\right)\left(\frac{BR}{2}\right)}{m{R}^2}=\frac{QB}{2m}\therefore {\omega }^{\prime }\omega -\frac{QB}{2m}=\omega -\frac{QB}{2m}{M}_f=\frac{Q{\omega }^{\prime }{R}^2}{2}=Q(\omega -\frac{QB}{2m})\frac{R^2}{2}\therefore \Delta M=M_f-M_i=-\frac{QB{R}^2}{4m}m=-\gamma \frac{QB{R}^2}{2}$