An electron enters a magnetic field of 0.47 t with a velocity perpendicular to the direction of the field. at what frequency does the electron traverse a circular path? (the mass of an electron is 9.1 × 10-31 kg, and the charge of an electron is 1.6 × 10-19 c.)

frequency = 1.3158  × Hz

Explanation:

given data

magnetic field B = 0.47 T

mass of an electron m = 9.1 × kg

charge of an electron e = 1.6 ×  C

to find out

frequency does the electron traverse a circular path

solution

we know that Magnetic force is equal to centripetal force

= qvB   ................1

qB =

qB =

put here value we get time period

1.6 ×  × 0.47 =

solve it we get here

time period = 7.5994  ×

so frequency will be

frequency =   .............2

frequency =

frequency = 1.3158  × Hz

f = 1.31 x 10^10 Hz

Explanation:

magnetic field, B = 0.47 T

mass, m = 9.1 x 10^-31 kg

q = 1.6 x 10^-19 C

the time period is given by

T = 7.599 x 10^-11 second

frequency is defined as the reciprocal of time period.

f = 1.31 x 10^10 Hz

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