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

Hint: 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