An alpha particle travels at a velocity of magnitude 790 m/s through a uniform m

Question

An alpha particle travels at a velocity of magnitude 790 m/s through a uniform magnetic field of magnitude 0.046 T. (An alpha particle has a charge of charge of +3.2 × 10-19 C and a mass 6.6 × 10-27 kg) The angle between the particle's direction of motion and the magnetic field is 72°. What is the magnitude of (a) the force acting on the particle due to the field, and (b) the acceleration of the particle due to this force? (c) Does the speed of the particle increase, decrease, or remain the same?

Explanation / Answer

Given,

Velocity, v = 790 m/s

Charge, q = 3.2 x 10^-19 C

a) Force, F = q (V X B)

= qvb sin?

= 3.2 x 10-19 x 790 x 0.046 x sin(72)

= 1.106 x 10^-17 N

b) Acceleration = Force/mass

= 1.106 x 10^-17 / 6.6 x 10^-27

= 1.68 x 10^9 m/s^2

c) Speed of Particle remains same as magnetic force acts perpendicular.

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