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)

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.