Question

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.)

Answer 1

frequency = 1.3158 ×
`10^(10)` Hz

Step-by-step explanation:

given data

magnetic field B = 0.47 T

mass of an electron m = 9.1 ×
`10^(-31)` kg

charge of an electron e = 1.6 ×
`10^(-19)` C

to find out

frequency does the electron traverse a circular path

solution

we know that Magnetic force is equal to centripetal force

`(mv^2)/(r)` = qvB ................1

qB =
`(mv)/(r)`

qB =
`(m*2* pi\ )/(T)`

put here value we get time period

1.6 ×
`10^(-19)` × 0.47 =
`(9.1*10^(-31)*2* pi\ )/(T)`

solve it we get here

time period = 7.5994 ×
`10^(-11)`

so frequency will be

frequency =
`(1)/(T)` .............2

frequency =
`(1)/(7.5994*10^(-11))`

frequency = 1.3158 ×
`10^(10)` Hz

Answer 2

f = 1.31 x 10^10 Hz

Step-by-step 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=(2\pi m)/(Bq)`

`T=(2* 3.14* 9.1* 10^(-31))/(0.47* 1.6* 10^(-19))`

T = 7.599 x 10^-11 second

frequency is defined as the reciprocal of time period.

f = 1.31 x 10^10 Hz