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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

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

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

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

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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