A beam of Proton moves in a uniform magnetic field, the force acting on proton is______

A beam of proton ( q = 1.6 X 10 ^-19 C ) moves at 3 X 10 ⁵ m/s through a uniform magnetic field with magnitude 4 T, that is directed along Z axis. The velocity of each proton lies in X-Y plane at an angle 30^o to the Z axis. find the force on proton.

Solution :

Given:

Charge on proton                              :         q        =       1.6 X 10 ^-19 C

Velocity of proton                             :         v        =       3 X 10 ⁵ m/s

Magnitude of magnetic induction     :         B       =       4 T

Angle between v and B                     :         θ        =       30^o

The force acting on proton due to external magnetic field can be given as

F = q ( v X B )

F = q v B sin θ

Putting values we get,

F = 1.6 X 10 ^-19 ( 3 X 10 ⁵     X  4 ) sin 30

F = 9.6 N

The force acting on proton ( q = 1.6 X 10 ^-19 C ) moving with velocity  3 X 10 ⁵ m/s through a uniform magnetic field of magnitude 4 T making an angle of 30^o with the direction of magnetic induction is  

                    F = 9.6 N