Question

At one instant, an electron (charge = –1.6 x 10–19 C) is moving in the xy plane, the components of its velocity being vx = 5.0 x 105 m/s and vy = 3.0 x 105 m/s. A magnetic field of 0.80 T is in the positive y direction. At that instant, what is the magnitude of the magnetic force on the electron?

Answer 1

Answer:

The direction of force is along negative Z axis and the magnitude of force is

6.4 x 10^-14 N.

Explanation:

q = - 1.6 x 10^-19 c

vx = 5 x 10^5 m/s, vy = 3 x 10^5 m/s, B = 0.8 T along Y axis

The velocity vector is given by

v = 5 x 10^5 i + 3 x 10^5 j

B = 0.8 j

Force on a charged particle place in a magnetic field is given by

F = q (v x B)

F = -1.6 x 10^-19 {(5 x 10^5 i + 3 x 10^5 j) x (0.8 j)}

F = - 1.6 x 10^-19 (5 x 0.8 x 10^5 k)

F = - 6.4 x 10^-14 k

The direction of force is along negative Z axis and the magnitude of force is

6.4 x 10^-14 N.