Question

Chapter 28, Problem 004

An alpha particle travels at a velocity of magnitude 760 m/s through a uniform magnetic field of magnitude 0.042 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 62°. 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?

Answer 1

velocity v =760m/s
uniform magnetic field B = 0.042T

charge q = 3.2 10^-19 C

mass m = 6.6 10^-27 kg

The angle between and is? = 62°.

(a) the magnitude of the force _B acting onthe particle due to the field F = Bvq sin ?
                                                                       F = 9.01* 10 ^-18 N

(b) the magnitude of the acceleration of the particle due to is a = F / m

                            a = 13.66*10⁸ m / s ²

c) remain the same

Lorenz force has null work. According to kinetic energy theorem the kinetic energy variation (thus velocity change) is equal to the work done by the force. If this work is null then speed variation is null and so it's constant. Work is null since force and path are perpendicular (due to cross product) and thus scalar product is null.