A particle of charge `-16xx10^(-18) coulomb` moving with velocity `10 ms^(-1)` along the ` x- axis `, and an electric field of magnitude `(10^(4))//(m)` is along the negative ` z- ais`. If the charged particle continues moving along the ` x`- axis , the magnitude of `B` is

Options

A. `10^(-3) Wb//m^(2)`
B. `10^(3) Wb//m^(2)`
C. `10^(5) Wb//m^(2)`
D. `10^(16) Wb//m^(2)`

Answer

B

Explanation

Since particle is moving underflected.
So `qE = qvB implies B = E//v = (10^(4))/(10) = 10^(3) Wb//m^(2)`