A proton is moving in 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 charge on proton is `e`

Options

A. `(Be)/(a^(2))`
B. `eB^(2)a`
C. `a^(2)eB`
D. `aeB`

Answer

C

Explanation

Under uniform magnetic field, force `evB` acts on proton and provides the necessary centripetal force `mv^(2)//a`.
`:.(mv^(2))/a=evB` or `v=(aeB)/m`…..i
Now, angular momentum
`L=rxxp`
Here `L=axxmv`
Putting value of `v` from Eq (i) we get
`L=axxm((aeB)/m)`
or `L=a^(2)eB`