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

Options

A. `1.5 xx 10^(3) N C^(-1)`
B. `2 xx 10^(3) N C^(-1)`
C. `3 xx 10^(3) N C^(-1)`
D. `1.33 xx 10^(3) N C^(-1)`

Answer

C

Explanation

As the charged particle passes undeflected through cross electric and magnetic fields
`:. qE= qvB, v= (E )/(B) or E= vB`
Here, `v=2 xx 10^(3) ms^(-1), B= 1.5T`
`:. E= (2 xx 10^(3) ms^(-1)) (1.5T) = 3 xx 10^(3) NC^(-1)`