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An electron enters a magnetic field of 0.47 t with a velocity perpendicular to the

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.)
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shradhwaip2426
shradhwaip2426
4,4(45 marks)

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

georgesarkes12
georgesarkes12
4,8(15 marks)

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

neekobecky599
neekobecky599
4,7(19 marks)

Thats not a nice question to ask on a learning app pls take down this question

Explanation:

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