Question:

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

Updated On: Jul 5, 2022
  • $\frac{q vR }{2}$
  • $q v R ^{2}$
  • $\frac{q vR ^{2}}{2}$
  • $qvR$
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The Correct Option is A

Solution and Explanation

As revolving charge is equivalent to a current, so $I=q f=q \times \frac{\omega}{2 \pi}$ But $\omega=\frac{v}{R}$ where $R$ is radius of circle and $v$ is uniform speed of charged particle. Therefore, $I=\frac{q v}{2 \pi R}$ Now, magnctic moment associated with charged particle is given by $\mu=I A=I \times \pi R^{2}$ $\mu =\frac{q v}{2 \pi R} \times \pi R^{2}$ or $=\frac{1}{2} q v R$
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Concepts Used:

Electric Dipole

An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance of 2a. The direction from q to -q is said to be the direction in space.

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