Question

A uniform electric field of magnitude $E=100kV/m$ is directed upward. Perpendicular to E and directed into the page, there exists a uniform magnetic field of magnitude $B=0.5T.$ A beam of particles of charge +q enters this region. What should be the chosen speed of the particles for which the particles will not be deflected by the crossed electric and magnetic field? Assume that the particles are moving in +ve direction.

Answer 1

Force acting on the charged particle.

Due to electric field=$qE$(upward)

Due to magnetic field=$qVB$(downwards from right hand screw rule)

As the two forces are equal and opposite

$qE=qVB$

$V=E/B$

Hence $V=\frac{100\times 1000}{0.5}=2\times {10}^{5}m/s$