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

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 (A)  (qvR/2)  (B)  qvR2 (C)  (qvR2/2) (D)  qvR

Tardigrade Admin · Accepted Answer

As revolving charge is equivalent to a current, so

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{q v}{2 π R}

Now, magnetic moment associated with charged particle is given by

μ = I A = I \times π R^{2}

or

μ = \frac{q v}{2 π R} \times π R^{2}

= \frac{1}{2} q v R

Q. 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

$\frac{q v R}{2}$

$q v R^{2}$

$\frac{q v R ^{2}}{2}$

$q v R$

Q. 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

2qvR​

qvR2

2qvR2​

qvR

As revolving charge is equivalent to a current, so I=qf=q×2πω​ But ω=Rv​ where R is radius of circle and v is uniform speed of charged particle. Therefore, I=2πRqv​ Now, magnetic moment associated with charged particle is given by μ=IA=I×πR2 or μ=2πRqv​×πR2 =21​qvR

Q. 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

$\frac{q v R}{2}$

$q v R^{2}$

$\frac{q v R ^{2}}{2}$

$q v R$