An alpha particle travels at a velocity of magnitude 760 m/s through a uniform magnetic field of magnitude 0.034 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 51°. 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

Question

Ronin Choi

Physics

15 June, 13:53

An alpha particle travels at a velocity of magnitude 760 m/s through a uniform magnetic field of magnitude 0.034 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 51°. 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

+4

Answers (1)

Know the Answer?

Shawn Chang · Answer 1

(a) 6.42 x 10^-18 N

(b) 9.73 x 10^8 m/s^2

Explanation:

v = 760 m/s, B = 0.034 T, m = 6.6 x 10^-27 kg, q = 3.2 x 10^-19 C, theta = 51 degree

(a) F = q v B Sin theta

F = 3.2 x 10^-19 x 760 x 0.034 x Sin 51

F = 6.42 x 10^-18 N

(b) Acceleration, a = Force / mass

a = (6.42 x 10^-18) / (6.6 x 10^-27)

a = 0.973 x 10^9

a = 9.73 x 10^8 m/s^2