When a charged particle in motion is exposed to a magnetic field that is perpendicular to its direction of travel, we know that the particle will move in a circular motion. When a particle with charge q and mass m is exposed to a perpendicular magnetic field B and moves with velocity v, the force experienced by the particle is:

= q(v×B)As the cross product of two vectors is always perpendicular to both the vectors, so the resulting force will always be perpendicular to the velocity of the particle.

Thus, it will lead the particle into a circular motion.

= q(v×B)F = qvBsinθWhere θ is the angle between v and B

Now, for the velocity of the particle and magnetic field to be perpendicular, sinθ = 1 Or, F = qvB

Now, the magnetic force will provide the necessary centripetal force for the circular motion.

The centripetal force acting on a particle of mass m moving with velocity v in a circular path of radius

r is: Fc = mv2/r

Equating the two forces, qvB = mv2/r = mv/qB

Now, for the given particle, its velocity vector is always perpendicular to the magnetic field, so it will undergo circular motion.

Also, it enters from P and comes out at Q with velocity normal to the initial, making its path a quarter circle.

The radius of this quarter circle will be equal to R So, using the above result, r = mv/qB for this particle, we have:

R = mv/qB

B = mv/qR

Now, the distance travelled by the particle is one-fourth of the circumference of a complete circle of radius R

So, distance travelled: 1/4(2πR) =πR/2

Speed of particle remains the same as v

So time taken is distance divided by speed. T=D/s

T = πR/2/v

T = πR/2v

We have: B = mv/qR

So, v = BqR/m

Substituting v = BqR/m in T = πR/2v

We get, T = πR/2×BqR/m

**T = πm/2Bq**