## In This post you will get answer of Q.

A proton beam enters a magnetic field of $10^{-4},Wb,m^{-2}$ normally. If the specific charge of the proton is $10^{11}, C ,kg^{-1}$ and its velocity is $10^9 m,s^{-1}$ then the radius of the ?circle decribed will be

Solution:

## Given, $beta=10^{-4} ,Wb ,m ^{-2}$

$frac{q}{m} =10^{11} C ,kg ^{-1} $

$v =10^{9} ,ms ^{-1}$

We know that, radius of the circle,

$I =frac{m v}{q B} $

$I =frac{v}{left(frac{q}{m}right) B} $

$r =frac{10^{9}}{left(10^{11}right)left(10^{-4}right)} $

$I =100, m$