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  • An electron with energy moves at right angle to the earth's magnetic field of <img alt="1 \times 10^{-4} \mathrm{Wbm}

An electron with energy 0.1 \mathrm{keV} moves at right angle to the earth's magnetic field of 1 \times 10^{-4} \mathrm{Wbm}^{-2}. The frequency of revolution of the electron will be
(Take mass of electron =9.0 \times 10^{-31} \mathrm{~kg} )
 

Option: 1

1.6 \times 10^{5} \mathrm{~Hz}
 


Option: 2

5.6 \times 10^{5} \mathrm{~Hz}
 


Option: 3

2.8 \times 10^{6} \mathrm{~Hz}


Option: 4

1.8 \times 10^{6} \mathrm{~Hz}


Answers (1)

best_answer

\mathrm{K E \text { of etelectron } =0.1 \mathrm{KeV} }

                            \mathrm{B=10^{-4} \frac{\mathrm{wb}}{\mathrm{m}^2} }

\mathrm{f =\frac{q B}{2 \pi \mathrm{m}}=\frac{1.6 \times 10^{-19} \times 10^{-4}}{6.28 \times 9.1 \times 10^{-31}} }

   \mathrm{=\frac{1.6}{6.28 \times 9.1} \times 10^{+8} }

\mathrm{f =2.8 \times 10^6 \mathrm{~Hz}}

Hence (3) is correct option

Posted by

Anam Khan

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