An electron with energy 0.1,ke,V moves at right angle to earth's magnetic field of 1 × 10^{-4},

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4.Moving Charges and Magnetism

easy

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

A

$1.6 \times 10^5\,Hz$

B

$5.6 \times 10^5\,Hz$

C

$2.8 \times 10^6\,Hz$

D

$1.8 \times 10^6\,Hz$

(JEE MAIN-2022)

Solution

$f =\frac{1}{ T }=\frac{ eB }{2 \pi m }$

$=\frac{1.6 \times 10^{-19} \times 10^{-4}}{2 \pi \times 9 \times 10^{-31}}=2.8 \times 10^{6}\,Hz$

Standard 12

Physics

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(AIPMT-2009)