Question

An electron is moving in a circular path under the influence of a transverse magnetic field of $3\cdot 57\times {10}^{-2}T$. If the value of e/m is $1\cdot 76\times {10}^{11}$ C/kgg. the frequency of revolution of the electron is.

• A. $6\cdot 28$ MHz
• B. 1 GHZ
• C. 100 MHz
• D. $62\cdot 8$ MHz

Answer 1

The correct option is B 1 GHZ

Given : $\frac{q}{m}=1.76\times {10}^{11}$ $C/kg$ $B=3.57\times {10}^{-2}$ $T$

Frequency of revolution $\nu =\frac{Bq}{2\pi m}$

$\therefore$ $\nu =\frac{3.57\times {10}^{-2}\times 1.76\times {10}^{11}}{2\pi }=1\times {10}^9$ Hz $=1$ GHz