Question

A proton is moving in a uniform magnetic field $B$ in a circular path of radius $a$ in a direetion perpendicular to $z$ -axis along which field $B$ exists. Calculate the angular momentum, if the radius is a charge on proton is $e$.

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

Answer 1

Answer: (C)

Magnetic force provides: the necessary centripetal force to the proton to move on circular path.
Under uniform magnetic field, force $e v B$ acts on proton and provides the necessary centripetal force $m v^{2} / a$
$\therefore \frac{m v^{2}}{a}=e v B$
or$v=\frac{a e B}{m}\,\,\,...(i)$
Now, angular momentum
$J=r \times p $
Here, $J=a \times m v$
Putting value of $v$ from Eq. (i), we get
$J =a \times m\left(\frac{a e B}{m}\right) $
$=a^{2} e B$