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

Answer

D

Explanation

The force on a particle is
So `F=q(E+V+B)` or `F=F_(e)+F_(m)`
`:.F_(e)=qE=-16xx10^(-18)xx10^(4)(-j)`
`=16xx10^(-14)k` and `F_(m)=-16xx10^(-18)(10ixxBj)`
`=-16xx10^(-17)xxB(+k) =-16xx10^(-17)Bk`

Since, particle will continue to move along + x-axis, so resultant force is equal to zero.
`F_(e)=F_(m)=0implies16xx10^(-14)=16xx10^(-17)BimpliesB=(16xx10^(-14))/(16xx10^(-17)=10^(3)`
or `b=10^(3)Wb//m^(2)`