Question

Magnetic moment of a revolving electron of charge $e$ and mass $m$ in terms of angular momentum $L$ of the electron is:

• A. $\frac{eL}{8m}$
• B. $\frac{eL}{4m}$
• C. $\frac{eL}{2m}$
• D. $\frac{eL}{m}$

Hint:

The ratio of magnetic moment to angular momentum ($\frac{\mu}{L} = \frac{e}{2m}$) is a universal value known as the gyromagnetic ratio for an electron. Memorizing this constant ratio allows you to instantly solve for either variable on entrance exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to derive or identify the standard formula for the orbital magnetic dipole moment ($\mu$) of an electron moving in a circular orbit, expressed in terms of its orbital angular momentum ($L$).

Step 2: Key Formula or Approach:
1.

Magnetic Moment ($\mu$): A circulating charge loop generates a magnetic moment defined by the current and area: $$\mu = I \cdot A$$ 2.

Angular Momentum ($L$): The mechanical orbital angular momentum of a mass moving in a circle of radius $r$ at speed $v$ is: $$L = mvr$$ We link these two expressions by eliminating the speed parameter $v$.

Step 3: Detailed Explanation:
Consider an electron of charge $e$ revolving in a circular path of radius $r$ with a constant speed $v$.
The time period $T$ for one complete revolution is $T = \frac{2\pi r}{v}$.
The equivalent electric current $I$ created by this orbital path is: $$I = \frac{e}{T} = \frac{ev}{2\pi r}$$ The area enclosed by the circular orbit loop is $A = \pi r^2$.
Substitute these current and area expressions into the magnetic moment definition: $$\mu = I \cdot A = \left(\frac{ev}{2\pi r}\right) \cdot (\pi r^2) = \frac{evr}{2}$$ Now, notice that the product $vr$ can be substituted from the angular momentum definition ($L = mvr \implies vr = \frac{L}{m}$): $$\mu = \frac{e}{2} \left(\frac{L}{m}\right) = \frac{eL}{2m}$$ This matches the standard expression in option (C).

Step 4: Final Answer:
The magnetic moment is $\frac{eL}{2m}$, which corresponds precisely to option (C).