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(B) The magnetic dipole moment $M$ is given by the product of current $i$ and the area $A$ of the loop.$M = i A$For a particle with charge $q$ moving

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$evr$

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(B) The magnetic dipole moment $M$ is given by the product of current $i$ and the area $A$ of the loop.$M = i A$For a particle with charge $q$ moving with frequency $f$,the equivalent current is $i = qf$.For an $\alpha$ particle,the charge is $q = 2e$.The frequency $f$ is related to the velocity $v$ and radius $r$ by $f = \frac{v}{2 \pi r}$.Thus,the current is $i = (2e) 	imes \left( \frac{v}{2 \pi r} ight) = \frac{ev}{\pi r}$.The area of the circular path is $A = \pi r^2$.Substituting these into the formula for $M$:$M = \left( \frac{ev}{\pi r} ight) 	imes (\pi r^2) = evr$.

An $\alpha$ particle is revolving in a circle of radius $r$ with frequency $f$. Find the value of its magnetic dipole moment.

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