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 circular path with a speed v under the influence of magnetic field $\vec{B}$ . The magnetic moment of this moving particle is

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

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

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

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

answer is D.

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Detailed Solution

$I = \frac{a}{t} = \frac{qv}{2 πr}$
Question Image
$M = IA = \frac{qv}{2 πr} \times {πr}^{2} = \frac{1}{2} qvr = \frac{1}{2} \times \frac{mv}{qB} = \frac{{mv}^{2}}{2 B}$
$Direction of \vec{M} is opposite to that of \vec{B} \vec{M} = \frac{{mv}^{2}}{2 B} (\frac{\vec{B}}{B}) = - \frac{{mv}^{2} \vec{B}}{2 B^{2}}$