Question

A proton is moving with velocity 10/4 m/s parallel to the magnetic field of intensity S Tesla the force on the proton is

Answer 1

Hello dear ☺️

Here's the solution ⤵️

The answer will be: 2.4 * 10 raise to power -12 N


The magnitude of the magnetic force on a charg particle is F =  |q| v B sin θ

Here q = charge on a proton = 1.602 x 10−19 C

v = 2 x 107 m/s

θ = 30°

B = 1.5 T

Substituting all these values in the expression for F, we get 

F = 1.602 x 10−19 C x 2 x 107 m/s x 1.5 T sin 30°
   = 2.403 x 10−12 N

Hope it is helpful to you ✌️