Question

An electron enters a magnetic field of 0.66 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)

Answer 1

1.85 x 10^10 cycles per second

Step-by-step explanation:

B = 0.66 T, theta = 90 degree, q = 1.6 x 10^-19 C,

The time period of electron is given by

T = 2 π m / B q

Frequency is teh reciprocal of time period.

f = 1 /T

f = B q / (2 π m)

f = (0.66 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31)

f = 1.85 x 10^10 cycles per second