1.

In given region electric and magnetic field are prpendicular to each other.Electron moving in this region do not get deflected,charge on cathode will be….. [V=potential between anode and cathode]

Answer»

`(B^(2))/(2VE^(2))`
`(2VB^(2))/(E^(2))`
`(2VE^(2))/(B^(2))`
`(E^(2))/(2VB^(2))`

Solution :f beam do not GET deflected then eE=evB
`impliesv(E )/(B)` ALSO `(1)/(2)mv^(2)=eV`
`therefore (e )/(m)=(v^(2))/(2V)=(E^(2))/(2VB^(2))`


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