An alpha particle travels at a velocity of magnitude 630 m/s through a uniform m

Question

An alpha particle travels at a velocity of magnitude 630 m/s through a uniform magnetic field of magnitude 0.035 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 48°. 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?

Chapter 28, Problem 004 An alpha particle travels at a velocity of magnitude 630 m/s through a uniform magnetic field of magnitude 0.035 T. (An alpha particle has a charge of charge of +3.2 x 1019 C and a mass 6.6 x 10 27 kg) The angle between the particle's direction of motion and the magnetic field is 48°. 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? (a) Number Units SE (b) Number Units

Explanation / Answer

(a) Magnetic force, Fb = q (v x B) = q v B sin(theta)

Fb = (3.2 x 10^-19) (630) (0.035) sin48

Fb = 5.244 x 10^-16 N .......Ans

(b) a = Fb / m = (5.244 x 10^-16 ) / (6.6 x 10^-27)

a = 7.94 x 10^8 m/s^2 ........Ans

(c) This acceleration will always be perpendicular to the velocity vector.

hence it will change direction only not magntude.

so speed of particle will remain the same.

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