An alpha particle travels at a velocity vector v of magnitude 660 m/s through a uniform magnetic field vector B of magnitude 0.060 T. (An alpha particle has a charge of + 3.2 10-19 C and a mass of 6.6 10-27 kg.) The angle between vector v and vector B is 57°. What happens to the speed?

Given Information:

Velocity of alpha particle = v = 660 m/s

Magnetic field = B = 0.060 T

Angle = θ = 57°

Charge on alpha particle = 3.2x10⁻¹⁹ C

Mass of alpha particle = 6.6x10⁻²⁷ kg

Required Information:

What happens to speed = ?

Answer:

Magnitude of speed remains same

Explanation:

The force acting on the alpha particle that is traveling at a velocity v through a magnetic field B is given by

F = qvBsin (θ)

Where q is the charge on alpha particle and θ is the angel between velocity v and magnetic field B.

The force F will change the direction of the velocity of alpha particle but the magnitude of the velocity remains same since the force is perpendicular to the velocity, which means that the acceleration direction is perpendicular to the velocity, so velocity direction changes but magnitude doesn't.

The force will be

F = 3.2x10⁻¹⁹*660*0.060*sin (57)

F = 1.063x10 ⁻¹⁷ N