An insulating rod of length L carries charge q distributed uniformly on it. The rod is pivoted at one end and is rotated at a frequency f about a fixed perpendicular axis. Then the magnetic dipole moment of the system is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{3}\pi qfL^{2} \)
A rotating charge is a current. A charge dq going round a circle of radius x at frequency f crosses any point of its path f times a second, so it is equivalent to a current loop of current \( dI = f\,dq \) and area \( \pi x^{2} \), with magnetic moment \( dM = dI\cdot\pi x^{2} \). Since the charge is spread along the rod, add the contributions of all the small elements.
Take an element of length dx at distance x from the pivot. The charge per unit length is \( q/L \), so
\( \displaystyle dq = \frac{q}{L}\,dx \)
\( \displaystyle dM = (f\,dq)\,\pi x^{2} = \frac{\pi qf}{L}\,x^{2}\,dx \)
Integrating from the pivot (\( x = 0 \)) to the free end (\( x = L \)):
\( \displaystyle M = \frac{\pi qf}{L}\int_{0}^{L}x^{2}\,dx = \frac{\pi qf}{L}\cdot\frac{L^{3}}{3} \)
\( \displaystyle M = \frac{1}{3}\pi qfL^{2} \)