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

An electron experiences electric force `vec(F_(e))` from the negative to the positive plate. Therefore, if the path should be undeflected, the magnetic force `vec F_(mag)` must be equal and opposite to `vecF_(e)`
i.e. `vec F_(mag)`must be directed from the positive plate to the negative plate. Then the magnetic field must be perpendicular to the plane of `vec F_(e)` and v.
`qE=q vB, B=E/v` `rArr E=V/d=600/3xx10^(-3)=2xx10^(5)N` `rArr v=2xx10^(6)m//s`
So, `B=2xx10^(5)/2xx10^(6)=10^(-1)`tesla.