Question

The magnetic moment of an electron revolving in an orbit of 0.5 m radius with a velocity of \( 8 \times 10^7 \, \text{m/s} \) is (in \( \text{Am}^2 \))

• A. \( 3.2 \times 10^{-12} \, \text{Am}^2 \)
• B. \( 0.4 \times 10^{-12} \, \text{Am}^2 \)
• C. \( 6.4 \times 10^{-12} \, \text{Am}^2 \)
• D. \( 1.6 \times 10^{-1} \, \text{Am}^2 \)
• E. \( 0.8 \times 10^{-12} \, \text{Am}^2 \)

Hint:

For calculating the magnetic moment of an electron, remember to use \( \mu = I \times A \), and for current \( I \), use \( I = \frac{e v}{2 \pi r} \).

Answer 1

The Correct Option is A

The magnetic moment \( \mu \) of an electron moving in a circular orbit is given by the formula: \[ \mu = I \times A \] Where: - \( I = \frac{e v}{2 \pi r} \) is the current due to the electron's motion, with \( e \) being the charge of the electron, \( v \) the velocity of the electron, and \( r \) the radius of the orbit. - \( A = \pi r^2 \) is the area of the orbit. Substitute the known values: - \( e = 1.6 \times 10^{-19} \, \text{C} \) - \( v = 8 \times 10^7 \, \text{m/s} \) - \( r = 0.5 \, \text{m} \) The current \( I \) is: \[ I = \frac{1.6 \times 10^{-19} \times 8 \times 10^7}{2 \pi \times 0.5} \] \[ I = \frac{1.28 \times 10^{-11}}{3.1416 \times 0.5} \] \[ I = 8.16 \times 10^{-12} \, \text{A} \] Now, the magnetic moment is: \[ \mu = I \times A = 8.16 \times 10^{-12} \times \pi \times (0.5)^2 \] \[ \mu = 8.16 \times 10^{-12} \times 0.7854 \] \[ \mu = 3.2 \times 10^{-12} \, \text{Am}^2 \] Thus, the magnetic moment is \( 3.2 \times 10^{-12} \, \text{Am}^2 \). Hence, the correct answer is (A) \( 3.2 \times 10^{-12} \, \text{Am}^2 \).