MCQ
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 $\overrightarrow{ B }$. The magnetic moment of this moving particle
  • A
    $-\frac{ mv ^{2} \overrightarrow{ B }}{ B ^{2}}$
  • B
    $-\frac{m v^{2} \vec{B}}{2 \pi B^{2}}$
  • C
    $\frac{m v^{2} \vec{B}}{2 B^{2}}$
  • $-\frac{m v^{2} \vec{B}}{2 B^{2}}$

Answer

Correct option: D.
$-\frac{m v^{2} \vec{B}}{2 B^{2}}$
d
Magnetic moment

$M=i A$

$M=\left(\frac{q}{T}\right) \times \pi r^{2}=\frac{q \pi r^{2}}{\left(\frac{2 \pi r}{v}\right)}=\frac{q v r}{2}$

$M=\frac{q v}{2} \times \frac{v m}{q B}$

$M =\frac{ mv ^{2}}{2 B }$

As we can see from the figure, direction of magnetic moment $(M)$ is opposite to magnetic field.

$\overrightarrow{ M }=-\frac{ mv ^{2}}{2 B } \hat{ B }$

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

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