First slide

Magnetic Dipole moment

Question

A charged particle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic moment $μ$ is given by

Moderate

A

$\frac{qvR}{2}$
B

$q v R^{2}$
C

$\frac{{qvR}^{2}}{2}$
D

$q v R$

Solution

As revolving charge is equivalent to a current loop, so electric current corresponds to the revolution of electron is $I = q f = q \times \frac{ω}{2 π}$
But $ω = \frac{v}{R},$ where R is radius of circle and v is uniform speed of charged particle.
Therefore, $I = \frac{qv}{2 πR}$
Now, magnetic moment associated with charged particle is given by
$or μ = \frac{qv}{2 πR} \times {πR}^{2} = \frac{1}{2} qvR$

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