Alpha Particle An alpha particle travels at a velocity v of magnitude 420 m/s th

Question

Alpha Particle An alpha particle travels at a velocity v of magnitude 420 m/s through a uniform magnetic field B of magnitude 0.040 T. (An alpha particle has a charge of + 3.2 times 10^-19 C and a mass of 6.6 times 10^-27 kg.) The angle between v and B is 56 degree. (a) What is the magnitude of the force F_g acting on the particle due to the field? (b) What is the magnitude of the acceleration of the particle due to F_g? (c) Does the speed of the particle increase, decrease, or remain equal to 420 m/s? The speed of the particle remains the same. The speed of the particle decreases. The speed of the particle increases.

Explanation / Answer

a)The magnitude of the force will be:

F = q v B sin(theta)

F = 3.2 x 10^-19 x 420 x 0.04 x sin56 = 4.46 x 10^-18 N

Hence, F = 4.46 x 10^-18 N

b)We kow that,

F = ma => a = F/m

a = 4.46 x 10^-18/6.6 x 10^-27 = 6.76 x 10^8 m/s^2

Hence, a = 6.76 x 10^8 m/s^2

c)Increases.

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