JEE PYQ: Magnetic Effects Of Current Question 22
Question 22 - 2020 (06 Sep 2020 Shift 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 $\vec{B}$. The magnetic moment of this moving particle is:
(1) $\frac{mv^2 \vec{B}}{2B^2}$
(2) $-\frac{mv^2 \vec{B}}{2\pi B^2}$
(3) $-\frac{mv^2 \vec{B}}{B^2}$
(4) $-\frac{mv^2 \vec{B}}{2B^2}$
Show Answer
Answer: (4) $-\frac{mv^2 \vec{B}}{2B^2}$
Solution
$M = \frac{1}{2}qvr = \frac{mv^2}{2B}$. Direction opposite to $\vec{B}$: $\vec{M} = -\frac{mv^2}{2B^2}\vec{B}$.
BETA
