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)

$a^{2}eB$

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
$J=r\times p$
$=a\times mv$
$J=a\times m\left(\frac{aeB}{m}\right)$
$J=a^{2}eB$