A particle having a charge $20\mathrm{\mu C}$ and mass $20\mathrm{\mu g}$ moves along a circle of radius 5.0 cm under the action of a magnetic field $\mathrm{B}=1.0\mathrm{T}$ . When the particle is at a point P, a uniform electric field is switched on and it is found that the particle continues on the tangent through P with a uniform velocity. Find the electric field

 
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
$\mathrm{qE}=\mathrm{qvB}\text{ or }\mathrm{E}=\mathrm{vB}$
The radius of the circle in a magnetic field is given by $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$
Or, $\mathrm{v}=\frac{\mathrm{rqB}}{\mathrm{m}}$
$=\frac{\left(5.0\times {10}^{-2}\mathrm{m}\right)\left(20\times {10}^{-6}\mathrm{C}\right)(1.0\mathrm{T})}{20\times {10}^{-9}\mathrm{kg}}=50{\mathrm{ms}}^{-1}$

The required electric field is
=  $\mathrm{E}=\mathrm{vB}=\left(50{\mathrm{ms}}^{-1}\right)(1.0\mathrm{T})=50{\mathrm{Vm}}^{-1}$
This field will be in a direction which is radially outward at P