JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2022
- Easy
- MCQ
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 \(=9.0 \times 10^{-31}\,kg\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(2.8 \times 10^6\,Hz\)
Step-by-step Solution
Detailed explanation
\(f =\frac{1}{ T }=\frac{ eB }{2 \pi m }\)
\(=\frac{1.6 \times 10^{-19} \times 10^{-4}}{2 \pi \times 9 \times 10^{-31}}=2.8 \times 10^{6}\,Hz\)
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