Question

A proton travelling at 1.6 x 10^7 m/s entres a region with a uniform magnetic field of strength 4 T. If the proton's initial velocity vector makes an angel of 45 degree with the magnetic field, what is the speed of the proton 1s after entering the magnetic field?

Answer 1

Magnetic Force, Fm = qvBsinx ; x is the angle between B and v

This force act perpendicular to the plane containing both B and v. ( v X B )

So the force will be perpendicular to the velocity.

There force the acceleration in this direction will add some velocity to this direction as well.

Acceleration, a= Fm/ mass of proton = 1.602*10^-19 * 1.6*10⁷ * 4* sin(45) / 1.6*10^-27

a = 4.53*10¹⁵ m/s²

Now, Velocity in new direction (Vz) due to magnetic force will be a*time = 4.53*10¹⁵ m/s

Now resultant velocity will be equal to ((Vz)² + (v)² )^0.5 which is approximately equal to 4.53*10¹⁵ m/s