A ciruclar coil of 20 turns and radius `10cm` is placed in a uniform magnetic field of `0.1T` normal to the plane of the coil . If the current in the coil is `5.0A` what is the average force on each electron in the coil due to the magnetic field `(` The coil is made of copper wire of cross`-` sectional area `10^(-5)m^(2)` and the free electron density in copper is given to be about `10^(29)m^(-3))`.

Options

A. `2.5xx10^(-25)N`
B. `7.5xx10^(-25)N`
C. `5xx10^(-25)N`
D. `10^(-25)N`

Answer

3

Explanation

Force acting on each electron, `i.e.,`
Lorentz force, `F_(m)=ev_(d)B`
`or F_(m)=e((1)/(An e))B=(IB)/(An) (asI=n e A v_(d))`
or `F_(m)=(5xx0.10)/(10^(-5)xx10^(29))N=5xx10^(-25)N`
`(` as `A=` cross `-` sectional area of the wire `=10^(-5)m^(2),n=` free electron density `=10^(-29)m^(-3))`