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`)

Answer

0.1

Explanation

Electron pass undeviated. Therefore,
`|F_(e)|=|F_(e)|` or `eE=eBv`
or `B=E/V=(V//d)/v`(V= potential difference between the plates ) or substituting the values, we have
`B=600/(3xx10^(-3)xx2xx10^(6))=.1T`
Further, direction of `F_(e)` should be opposite of `F_(m)`
or `eE uarr darre(vxxB) " "Euarr darrvxxB`
Here E is in positive x-direction
Therefore, `vxxB` should be in negative x-direction or B should be in negative z-direction or perpendicular to paper inwards, because velocity of electron is in positive y-direction.