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An electron having charge $1.6 \times {10^{ - 19}}\,C$ and mass $9 \times {10^{ - 31}}\,kg$ is moving with $4 \times {10^6}\,m{s^{ - 1}}$ speed in a magnetic field $2 \times {10^{ - 1}}\,tesla$ in a circular orbit. The force acting on electron and the radius of the circular orbit will be
$12.8 \times {10^{ - 13}}\,N,\,1.1 \times {10^{ - 4}}\,m$
$1.28 \times {10^{ - 14}}\,N,\,1.1 \times {10^{ - 3}}\,m$
$1.28 \times {10^{ - 13}}\,N,\,1.1 \times {10^{ - 3}}\,m$
$1.28 \times {10^{ - 13}}\,N,\,1.1 \times {10^{ - 4}}\,m$
Solution
(d) $F = evB$
$ = 1.6 \times {10^{ – 19}} \times 4 \times {10^6} \times 2 \times {10^{ – 1}}$$ = 1.28 \times {10^{ – 13}}\,N$
Also $\frac{{m{v^2}}}{r} = evB \Rightarrow r = \frac{{mv}}{{eB}}$
$ \Rightarrow \;r = \frac{{9 \times {{10}^{ – 31}} \times 4 \times {{10}^6}}}{{1.6 \times {{10}^{ – 19}} \times 2 \times {{10}^{ – 1}}}} = 1.1 \times {10^{ – 4}}\,m$
