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An electron (mass = $9.1 \times {10^{ - 31}}$ $kg$; charge = $1.6 \times {10^{ - 19}}$ $C$) experiences no deflection if subjected to an electric field of $3.2 \times {10^5}$ $V/m$, and a magnetic fields of $2.0 \times {10^{ - 3}} \,Wb/m^2$. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius.......$m$
$45$
$4.5$
$0.45$
$0.045$
Solution
(c) For no deflection in mutually perpendicular electric and magnetic field $v = \frac{E}{B} = \frac{{3.2 \times {{10}^5}}}{{2 \times {{10}^{ – 3}}}} = 1.6 \times {10^8}\,m/s$.
If electric field is removed then due to only magnetic field radius of the path described by electron $r = \frac{{mv}}{{qB}} = \frac{{9.1 \times {{10}^{ – 31}} \times 1.6 \times {{10}^8}}}{{1.6 \times {{10}^{ – 19}} \times 2 \times {{10}^{ – 3}}}} = 0.45\,m$
