The angular momentum of an electron moving in the circular path of diameter \( D \) in uniform magnetic field \( B_0 \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( (e\,B_0\,D^2)/4 \)
The electron moves in a circle because the magnetic force supplies the centripetal force. Let the radius be \( r = D/2 \).
\( \displaystyle e v B_0 = \frac{m v^{2}}{r} \;\Rightarrow\; m v = e B_0 r \)
The angular momentum about the centre of the circle is \( L = m v r \).
\( \displaystyle L = (m v)\,r = e B_0 r^{2} = e B_0\left(\frac{D}{2}\right)^{2} = \frac{e B_0 D^{2}}{4} \)