A potential difference of `600 volts` is applied across the plates of a parallel plate consenser . The separation between the plates is `3mm`. An electron projected vertically, parallel to the plates , with a velocity of `2 xx 10^(6) m//sec` moves underflected between the plates. Find the magnitude and direction of the magnetic field in the region between the condenser plates. ( Neglect the edge effects). ( Charge of the electron `= -1.6xx10^(-19) coulomb)

Explanation

The force on electron will be toward the left plane due to electric field and will be equal to `F_e=eE`.
For the electron to move undeflected between the plates, there should be a force (magnetic) which is equal to the electric force and opposite in direction to electric force. So the force should be directed toward the right as the electric force is toward the left. On applying Fleming's left hand rule, we get that the magnetic field should be directed perpendicular to the plane of paper and inwards. Therefore,
Force due to electric field=Force due to magnetic field
`eE=evB [:'E=V/f]`
`B=E/v=(V//d)/v`

where V=potential difference between the plates, and
d=distance between the plates
`B=(600//3xx10^-3)/(2xx10^6)=600/(3xx10^-3xx2xx10^6) implies B=0.1T`