An alpha particle travels at a velocity of magnitude 510 m/s through a uniform m

Question

An alpha particle travels at a velocity of magnitude 510 m/s through a uniform magnetic field of magnitude 0.036 T. (An alpha partidle has a charge of charge of +3.2 x 1019 C and a mass 6.6 x 10-27 kg) The angle between the partidle's direction of motion and the magnetic field is 76°. 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

V = 510 m/s

B = 0.036T

q = 3.2 x 10-19 C

m = 6.6 x 10-27 kg

= 76º

Part A:

F = q (V x B)

where V x B is the cross product

F = q*V*B*sin

F = 3.2 x 10-19*510*0.036*sin76o

F = 5.7 x 10-18

Part B:   

F = ma

a = (5.7 x 10-18)/( 6.6 x 10-27 kg)

a = 8.63 x 108 m/s2

Part C:

Speed will remain the same because Force cannot change the speed & kinetic energy, just direction. Because the force is always perpendicular to the direction of motion

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