Question

A proton experiences a force of 4 Newton when it enters perpendicular to the direction of the magnetic field with a speed 100 m/s. What is the force experienced by the proton if it enters parallel to the direction of the magnetic field with a speed of 200 m/s?

Answer 1

Answer:

F = 0 N

Explanation:

Force on a moving charge in constant magnetic field is given by the formula

$F = q(\vec v\times \vec B)$

so here it depends on the speed of charge, magnetic field and the angle between velocity of charge and the magnetic field

here when charge is moving with speed 100 m/s in a given magnetic field then the force on the charge is given as

$F = 4 N$

now when charge is moving parallel to the magnetic field with different speed then in that case

$\vec v \time \vec B = 0$

so here we have

F = 0