Question

A particle of charge '\(q\)' and mass '\(m\)' moves in a circular orbit of radius '\(r\)' with angular speed '\(ω\)'. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends upon

• A. q and m
• B. \(ω\) and m
• C. \(ω\) and q
• D. \(ω\) , q and m

Hint:

For a charge in a circular orbit, M/L = q/2m, which depends on q and m only.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A charge \(q\) moving in a circle is a current loop. Its magnetic moment is the product of current and area, and its angular momentum is \(m v r\).

Step 2: Key Formula or Approach:
Current: \(I = \dfrac{q}{T} = \dfrac{q\omega}{2\pi}\). Area: \(A = \pi r^2\).

Step 3: Detailed Explanation:
Magnetic moment:
\[ M = IA = \frac{q\omega}{2\pi}\cdot\pi r^2 = \frac{q\omega r^2}{2} \]
Angular momentum:
\[ L = mvr = m\omega r^2 \]
Ratio:
\[ \frac ML = \frac{q\omega r^2/2}{m\omega r^2} = \frac{q}{2m} \]
The factors \(\omega\) and \(r\) cancel. So the ratio depends only on \(q\) and \(m\). Options (B), (C) and (D) include \(\omega\), which does not appear in the final ratio.

Final Answer:
The ratio depends only on \(q\) and \(m\), option (A). \[ \boxed{\frac{M}{L}=\frac{q}{2m} \text{ (A)}} \]