An electron in a television picture tube travels at `3xx10^(7)` m/s. It is subjected to a transverse magnetic field of `2xx10^(-3)Wb//m^(2).` What is the magnitude of the lateral foce acting on the electron, due to the action of the magnetic field ?
Charge on the electron `=-1.6 xx10^(-19) C.`

Options

A. `4.8xx10^(-15)N`
B. `9.6xx10^(-15)N`
C. `7.2xx10^(-15)N`
D. `2.4xx10^(-15)N`

Answer

B

Explanation

`F =qvB sin theta `
where `q=1.6 xx10^(-19) C` in magnitude and `theta =90^(@)`
`therefore F =1.6 xx10^(-19) xx3xx10^(7) xx2xx10^(-3)xx sin 90^(@)=9.6 xx10^(-15)N`