Question

The magnetic moment of an electron moving in a circular orbit of radius R with a time period T is

• A. \(\frac{2\pi Re}{T} \)
• B. \(\frac{\pi eR}{T} \)
• C. \(\frac{\pi eR^2}{T} \)
• D. \(\pi R^2 eT \)

Hint:

To find the magnetic moment of a charged particle in a circular orbit: 1. Equivalent Current: The orbiting charge \(q\) with time period \(T\) creates an equivalent current \(I = q/T\). 2. Area of Loop: For a circular orbit of radius \(R\), the area is \(A = \pi R^2\). 3. Magnetic Moment: The magnetic moment is \(\mu = IA\). For an electron (charge \(e\)), this simplifies to \(\mu = \frac{\pi eR^2}{T}\).

Answer 1

The Correct Option is C

Step 1: Define the magnetic moment of a current loop.
The magnetic moment (\(\mu\)) of a current loop is defined as the product of the current (\(I\)) flowing through the loop and the area (\(A\)) enclosed by the loop: \[ \mu = IA \] Step 2: Determine the current due to the orbiting electron.
An electron orbiting in a circular path constitutes a current. Current is defined as the amount of charge passing a point per unit time.
In this case, the charge is that of a single electron, \(e\).
The time taken for one complete revolution (one full cycle) is the time period \(T\).
Therefore, the equivalent current \(I\) due to the electron's motion is: \[ I = \frac{\text{charge}}{\text{time period}} = \frac{e}{T} \] Step 3: Determine the area of the circular orbit.
The electron moves in a circular orbit of radius \(R\). The area of a circle is given by: \[ A = \pi R^2 \] Step 4: Substitute the expressions for current and area into the magnetic moment formula.
Now, substitute the expressions for \(I\) and \(A\) into the magnetic moment formula \(\mu = IA\): \[ \mu = \left(\frac{e}{T}\right) (\pi R^2) \] Rearranging the terms, we get: \[ \mu = \frac{\pi eR^2}{T} \] The final answer is \( \boxed{\frac{\pi eR^2}{T}} \).