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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
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2957 Upvotes
Verified Answer
The correct answer is:
$\frac{q v R}{2}$
As revolving charge is equivalent to a current, so
$\begin{aligned}
& I=q f=q \times \frac{\omega}{2 \pi} \\
But \quad & \omega=\frac{v}{R}
\end{aligned}$
where $R$ is radius of circle and $v$ is uniform speed of charged particle.
Therefore,
$I=\frac{q v}{2 \pi R}$
Now, magnetic moment associated with charged particle is given by
or
$\begin{aligned}
\mu & =I A=I \times \pi R^2 \\
or \quad \mu & =\frac{q v}{2 \pi R} \times \pi R^2 \\
& =\frac{1}{2} q v R
\end{aligned}$
$\begin{aligned}
& I=q f=q \times \frac{\omega}{2 \pi} \\
But \quad & \omega=\frac{v}{R}
\end{aligned}$
where $R$ is radius of circle and $v$ is uniform speed of charged particle.
Therefore,
$I=\frac{q v}{2 \pi R}$
Now, magnetic moment associated with charged particle is given by
or
$\begin{aligned}
\mu & =I A=I \times \pi R^2 \\
or \quad \mu & =\frac{q v}{2 \pi R} \times \pi R^2 \\
& =\frac{1}{2} q v R
\end{aligned}$
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