Derive the expression for magnetic force on a conductor carrying current kept in a magnetic field.

Explanation

Consider a conducting rod of length l and uniform area of cross-section A in an external magnetic field `vecB` as shown in fig.
When a steady current .I. flows in the rod, the total charge of mobile charge carriers is Q=n A l a
The magnetic force on these charge carriers is
`vecF = Q (vecV_d xx vecB) " " V_d to ` drift velocity
`vecF = nAl q(vecV_d xx vecB) ` .....(1)
But `I = n q A V_d` ......(2)
Equation (2) in (1) we get
`vecF = I(vecl xx vecB)`