2007 AIPMT-PRE Moving Charges and MagnetismMagnetic Dipole Moment of Current Loop Easy Question 4 of 5 in Magnetic Dipole Moment of Current Loop

A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given by

A $qvR^2$
B $\dfrac{qvR^2}{2}$
C $qvR$
D $\dfrac{qvR}{2}$

The answer is D.

Solution

A charge moving in a circle is equivalent to a current loop.
Time period of revolution: $T = \dfrac{2\pi R}{v}$
Equivalent current: $I = \dfrac{q}{T} = \dfrac{qv}{2\pi R}$
Magnetic moment: $\mu = IA$, with area $A = \pi R^2$
$\mu = \dfrac{qv}{2\pi R}(\pi R^2)$
$\mu = \dfrac{qvR}{2}$

Concept Magnetic moment of a circulating charge

Magnetic Dipole Moment of Current Loop in past papers

5 questions from this chapter have appeared across 5 exam years.