Question

A proton is moving in a uniform magnetic field $B$ in a circular path of radius a in a direction perpendicular to $Z$ axis along which field $B$ exists. Calculate the angular momentum. If the radius is a and charge on proton is $e$.

• A. $ \frac{Be}{{{a}^{2}}} $
• B. $ e{{B}^{2}}a $
• C. $ {{a}^{2}}eB $
• D. $ aeB $

Answer 1

Answer: (C)

Under uniform magnetic field, force $ evB $
acts on proton and provides the necessary centripetal force
$ \frac{m{{v}^{2}}}{a} $
$ \frac{m{{v}^{2}}}{a}=evB $
$ v=\frac{aeB}{m} $
Now, angular momentum
$ J=r\times p $
$ =a\times mv $
$ J=a\times m\left( \frac{aeB}{m} \right) $
$ J={{a}^{2}}eB $