An electron with energy 0.1 keV 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$-$31 kg)
Solution
<p>$T = {{2\pi m} \over {Bq}}$</p>
<p>$\Rightarrow$ Frequency $f = {{Bq} \over {2\pi m}}$</p>
<p>$$ = {{{{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}}} \over {2\pi \times 9 \times {{10}^{ - 31}}}}$$</p>
<p>$\simeq 2.8 \times {10^6}$ Hz</p>
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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