A copper has `8*0xx10^28` electrons per cubic metre. A copper wire of length `1m` and cross-sectional area `8*0xx10^-6m^2` carrying a current and lying at right angle to a magnetic field of strength `5xx10^-3T` experiences a force of `8*0xx10^-2N`. Calculate the drift velocity of free electrons in wire.

Explanation

Here, `n=8*0xx10^(28)m^-3`, `l=1m`,
`A=8xx10^-6m^2`, `e=1*6xx10^-19C`,
`B=5xx10^-3T`, `F=8*0xx10^-2N`, `theta=90^@`
Total charge contained in the wire is
`q=`Volume of wire `xx "ne" =`Alne
`=(8xx10^-6)xx1xx(8xx10^(28))xx(1*6xx10^(-19))`
`=1*024xx10^5C`
If `v_d` is the drift velocity of electron, then
`F=qv_dBsintheta`
or `v_d=(F)/(qBsintheta)`
`=((8*0xx10^-2))/((1*024xx10^5)xx(5xx10^-3)xxsin 90^@)`
`=1*56xx10^-4ms^-1`