Bar Magnet & Magnetic Moment question from NEET 2007 (AIPMT)

Answer: (d)

1. Equivalent current $I = \dfrac{q}{T} = \dfrac{qv}{2\pi R}$.
2. Area of the orbit $A = \pi R^2$.
3. $\mu = IA = \dfrac{qv}{2\pi R}\cdot \pi R^2 = \dfrac{qvR}{2}$
4. So the answer is (d) $qvR/2$.