An alpha particle travels at a velocity v of magnitude 620 m/s through a uniform

Question

An alpha particle travels at a velocity v of magnitude 620 m/s through a uniform magnetic field B of magnitude 0.095 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 v and B is 46°.

(a) What is the magnitude of the force FB acting on the particle due to the field?
N
(b) What is the magnitude of the acceleration of the particle due to FB?
m/s2 (c) Does the speed of the particle increase, decrease, or remain equal to 620 m/s?

The speed of the particle decreases.The speed of the particle increases.    The speed of the particle remains the same.

Explanation / Answer

a) FB =q ( v x B)

|FB| = |q||v| |B| sin@

|FB| = 3.2 x 10^-16 x 620 x 0.095 x sin46 = 1.36 x 10^-14 N

b) a = FB/m = (1.36 x 10^-14 ) / (6.6 x 10^-27) = 2.05 x 10^12 m/s^2

c)force will be always perpendicular to the velocity,.

(perpendicular force only can change direction not magnitude)

hence particle will perform cicrcular motion with constant speed.

so speed remains same.

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