A force `vecF` acting on an electric charge q, in presence of an electromagnetic field, moves the charge parallel to the magnetic field with velocity `vecv`. The `vecF` is equal to (where `vecE` and `vecB` are electric field and magnetic field respectively)

Options

A. `qvecE`
B. `q(vecvxxvecB)`
C. `q(vecvxxvecE)`
D. `qvecB`

Answer

A

Explanation

Force on the charge q due to electric field `vecE, vecF_(E)=qvecE`
and force on the charge q due to magnetic

`vecF_(B)=q(vecvxxvecB)rArrvecF_(B)=qvB sin theta`
In the given case `vecv||vecB`
`rArrtheta=0^(@)rArrvecF_(B)=0`
Hence, the force `vecF` acting on the charge q is equal to `vecF_(E)=qvecE`