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(D) The magnetic moment $M$ of a current loop is given by $M = iA$,where $i$ is the current and $A$ is the area of the loop.For a particle of charge $

Vedclass · Accepted Answer

$-\frac{m v^{2} \vec{B}}{2 B^{2}}$

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(D) The magnetic moment $M$ of a current loop is given by $M = iA$,where $i$ is the current and $A$ is the area of the loop.For a particle of charge $q$ moving in a circle of radius $r$ with speed $v$,the time period $T$ is $T = \frac{2 \pi r}{v}$.The equivalent current is $i = \frac{q}{T} = \frac{qv}{2 \pi r}$.The area of the loop is $A = \pi r^2$.Thus,the magnitude of the magnetic moment is $M = iA = \left( \frac{qv}{2 \pi r} ight) 	imes (\pi r^2) = \frac{qvr}{2}$.For a charged particle moving in a magnetic field,the radius of the circular path is $r = \frac{mv}{qB}$.Substituting $r$ into the expression for $M$,we get $M = \frac{qv}{2} 	imes \left( \frac{mv}{qB} ight) = \frac{mv^2}{2B}$.From the right-hand rule,the direction of the magnetic moment is opposite to the direction of the magnetic field for a positively charged particle moving in a circle. Therefore,in vector form,$\overrightarrow{M} = -\frac{mv^2}{2B} \hat{B}$.Since $\hat{B} = \frac{\vec{B}}{B}$,we have $\overrightarrow{M} = -\frac{mv^2}{2B} \left( \frac{\vec{B}}{B} ight) = -\frac{mv^2 \vec{B}}{2B^2}$.

$A$ charged particle going around in a circle can be considered to be a current loop. $A$ particle of mass $m$ carrying charge $q$ is moving in a plane with speed $v$ under the influence of magnetic field $\overrightarrow{ B }$. The magnetic moment of this moving particle is:

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