First slide

Force on a charged particle moving in a magnetic field

Question

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{- 2}$ T. If the value of $e / m$ is $1.76 \times 10^{11} C k g^{- 1}$ , the frequency of revolution of the electron is

Easy

A

1 GHz
B

100 MHz
C

62.8 MHz
D

6.28 MHz

Solution

Here, $B = 3.57 \times 10^{- 2} T, \frac{e}{m} = 1.76 \times 10^{11} {Ckg}^{- 1}$
Frequency of revolution of the electron,
$v = \frac{1}{T} = \frac{v}{2 π r}$ ……..(1)
$Also, \frac{m v^{2}}{r} = e v B \Rightarrow \frac{v}{r} = \frac{e B}{m}$ ……….(2)
From eqns. (i) and (ii)
$v = \frac{1}{2 π} \times \frac{e B}{m} = \frac{1}{2 \times 3.14} \times 1.76 \times 10^{11} \times 3.57 \times 10^{- 2}$
$= 10^{9} Hz = 1 GHz$

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