Question

The ratio between the magnetic dipole moment of a revolving electron in a circular orbit to its angular momentum is (where \( e \) is the charge and \( m \) is the mass of the electron):

• A. \( \frac{e^2}{2m} \)
• B. \( \frac{e^2}{m} \)
• C. \( \frac{e}{2m} \)
• D. \( \frac{e}{m^2} \)
• E. \( \frac{e}{2m^2} \)

Hint:

In problems involving revolving electrons and magnetic dipole moment, remember that the ratio of the magnetic dipole moment to the angular momentum is \( \frac{e}{2m} \), derived from the relationship between the current, area, and angular momentum.

Answer 1

The Correct Option is C

The magnetic dipole moment \( \mu \) of an electron moving in a circular orbit is given by: \[ \mu = I \cdot A \] where \( I \) is the current due to the motion of the electron, and \( A \) is the area of the orbit. For a revolving electron, the current \( I \) is given by the charge per unit time, which is: \[ I = \frac{e}{T} \] where \( T \) is the time period of the electron’s revolution. The area \( A \) of the orbit is given by: \[ A = \pi r^2 \] where \( r \) is the radius of the orbit. Now, the angular momentum \( L \) of the electron is given by: \[ L = mvr \] where \( v \) is the velocity of the electron and \( r \) is the radius of the orbit. Thus, the ratio between the magnetic dipole moment and angular momentum is: \[ \frac{\mu}{L} = \frac{\frac{e}{T} \cdot \pi r^2}{mvr} \] We know that \( v = \frac{2\pi r}{T} \), so: \[ \frac{\mu}{L} = \frac{\frac{e}{T} \cdot \pi r^2}{m \cdot \frac{2\pi r}{T} \cdot r} = \frac{e}{2m} \] Thus, the correct answer is option (C), \( \frac{e}{2m} \).