Derive an expression for the force acting on a current carrying conductor placed in a uniform magnetic field. Name the rule which gives the direction of the force. Write the condition for which this force will have (i) maximum (ii) minimum value.

Explanation

Consider a current conductor of length l and cross-section area A. If n be the free electron density for the metal of conductor then total number of free electrons in n A l. For a steady current I in the conductor, let each free electron has an average drift `vecv_d` velocity in the presence of a magnetic field `vecB`
Hence, Lorentz magnetic force on a mobile electron is
`vecf = -e (vecv_d xx vecB)`
`:.` Force on all the mobile/free electrons of the conductor
`vecF = vecf cdot n A l = - n A l e (vecv_d xx vecB)`
`= I (vecl xx vecB) " " [ :. I = - n A e vecv_d]`
Direction of the force is given by Fleming.s left hand rule.
If magnetic field `vecB` makes an angle `theta` from the conductor then magnitude wise
`F = I l B sin theta`
(i) The force will have a maximum value `F_("max") = I l B` when `theta = 90^@ and sin theta = 1`
(ii) The force will have a zero value, when `theta = 0^@` or `180^@` and `sin theta = 0`.