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)

Question

Maritza Fernandez

Physics

11 April, 16:35

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)

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Answers (1)

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Aldo Andersen · Answer 1

1.85 x 10^10 cycles per second

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