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

Answer:

frequency = 1.3158  × $10^{10}$ Hz

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

$\frac{mv^2}{r}$ = qvB   ................1

qB = $\frac{mv}{r}$

qB = $\frac{m*2* pi\ }{T}$

put here value we get time period

1.6 × $10^{-19}$  × 0.47 = $\frac{9.1*10^{-31}*2* pi\ }{T}$

solve it we get here

time period = 7.5994  × $10^{-11}$

so frequency will be

frequency = $\frac{1}{T}$   .............2

frequency = $\frac{1}{7.5994*10^{-11}}$

frequency = 1.3158  × $10^{10}$ Hz

Answer 2

Answer:

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

$T=\frac{2\pi m}{Bq}$

$T=\frac{2\times 3.14\times 9.1\times 10^{-31}}{0.47\times 1.6\times 10^{-19}}$

T = 7.599 x 10^-11 second

frequency is defined as the reciprocal of time period.

f = 1.31 x 10^10 Hz