Obtain an expression for the magnetic force on a current carrying conductor.

Explanation

Consider a conducting rod of length l, and uniform sectional area A in an external magnetic field `vecB`. Let the number density of the mobile charge carriers (electrons) be n.
`therefore` Number density (n) =`"no. of electrons"/"Volume"`
no. of electrons = n x volume
no. of electrons = n x Al
If q is the charge of each charge carrier
Total charge = q x nAl
When a steady current I flows in this conducting rod , each mobile charge carrier has an average drift velocity `vecV_d` in the presence of an external magnetic field `vecB`.
`therefore` Force on these charge carriers is `vecF` =Total Charge `(vecVxxvecB)`
`vecF=qnAl(vecV_d xx vecB)`
`vecF=Al(vecjxxvecB)`
`vecF=I(veclxxvecB)`
Magnitude `F=BI l sin theta`
Where `vecl` is a vector of magnitude l and vector sign is transferred from j to `vecl` direction of `vecl` is same as j (Current ) .