A potential difference of 600 V is applied across the plates of a parallel plate condenser. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of `2xx10^(6)m//s` moves undeflected between the plates. Find the magnitude of the magnetic field (in T) in the region between the condenser plates. (Neglect the edge effects) `square` .
(Charge of the electron `=1.6xx10^(-19)C`)

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.