Question

A charge $q$ is spread uniformly over an insulated loop of radius$r$. If it is rotated with an angular velocity $\omega$ with respect to normal axis then the magnetic moment of the loop is-
A. $\dfrac{1}{2}q\omega {r^2}$
B. $\dfrac{4}{3}q\omega {r^2}$
C. $\dfrac{3}{2}q\omega {r^2}$
D. $q\omega {r^2}$

Answer 1

When a charge rotates in the loop with certain angular velocity, about an axis, a charge rotating about an axis gives a magnetic moment. A moving charge gives result to electric current.
The velocity with which particles moves in circular path give the expression for the angular momentum is written as,
$L = m{r^2}\omega$
Here, $m$ is the mass of a particle, $r$ is the radius of the circular loop and $\omega$ is the angular velocity.

Complete step by step answer:
Understand that, calculate the magnetic moment of the loop using the expression for the magnetic moment.
Write the expression for the magnetic moment$\mu$,
$\mu = \dfrac{q}{{2m}}L$
Here $q$ is the charge in the loop, $m$ is the mass and $L$ is the angular moment of the charge in the loop.
Substitute $m{r^2}\omega$ for $L$

\mu = \dfrac{q}{{2m}}\left( {m{r^2}\omega } \right) \\\ = \dfrac{1}{2}q{r^2}\omega \\\ $$. **So, the correct answer is “Option A”.** **Note:** The magnetic moment has been calculated using the expression for the ration magnetic moment and angular momentum. Then the expression is rearranged for a magnetic moment. And substitute the value of the angular momentum in the obtained expression to obtain the final expression for the magnetic moment. A circulating current in a loop with a certain enclosed area gives a magnetic moment. The expression for the magnetic moment$$\mu $$, is written as $$\mu = \dfrac{q}{{2m}}L$$ Here $$q$$ is the charge in the loop, $$m$$ is the mass and $$L$$ is the angular moment of the charge in the loop