A charged particle with a velocity `2 xx 10^3 ms^-1` passes undeflected through electric field and magnetic fields in mutually perpendicular directions. The magnetic field is 1.5 T. The magnitude of electric field will be

Options

A. `1.5xx10^(3)NC^(-1)`
B. `2xx10^(3)NC^(-1)`
C. `3xx10^(3)NC^(-1)`
D. `1.33xx10^(3)NC^(-1)`

Answer

C

Explanation

The charged particle goes undeflected through both the fields, therefore, force experienced by charged particle due magnetic field must be equal to the fore experienced by the charge particle due to electric field, i.e, `F_(m)=F_(e)`
or `ev B sin theta =eE`
Given `v=2xx10^(3) ms^(-1)`
`B=1.5T`
and `theta=90^(@)`
Hence `E=v B sin theta`
`=2xx10^(3)xx1.5 sin 90^(@)`
`=3xx10^(3)V//m`