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A particle having a charge of $10.0\,\mu C$ and mass $1\,\mu g$ moves in a circle of radius $10\,cm$ under the influence of a magnetic field of induction $0.1\,T$. When the particle is at a point $P$, a uniform electric field is switched on so that the particle starts moving along the tangent with a uniform velocity. The electric field is......$V/m$
$0.1$
$1$
$10$
$100$
Solution
(c) When the particle moves along a circle in the magnetic field $B$, the magnetic force is radially inward. If an electric field of proper magnitude is switched on which is directed radially outwards, the particle may experience no force. It will then move along a straight line with uniform velocity. This will be the case when $qE = \varepsilon vB \Rightarrow E = vB$
also $r = \frac{{mv}}{{qB}} \Rightarrow v = \frac{{qBr}}{m}$
So $E = \frac{{q{B^2}r}}{m}$
$ = \frac{{(10 \times {{10}^{ – 6}}) \times {{(0.1)}^2} \times 10 \times {{10}^{ – 2}}}}{{1 \times {{10}^{ – 3}} \times {{10}^{ – 6}}}} = 10\;V/m$
