Force on a current carrying conductor in a magnetic field.

Consider a conductor PQ of length l and area of cross section A.

The magnetic field B acts along +ve Z direction.

The drift velocity of electrons is v_d

Each electron experience a magnetic Lorentz force ,f = -e( v_d×B)

If n is the number of free electrons per unit volume,then total number of electrons in the conductor is-

N=n×volume  = n×Al

Total force on the conductor is, F= Nf = nAl[-e( v_d×B)]

=enA[-l v_d×B]

Now, if Il represents a current element vector in the direction of current, the vectors I and v_d will have opposite directions and we can take,

-lv_d=v_dl

So, F= enAv_d (l×B) = IlB sinΘ

(Θ is the angle between the direction of the magnetic field and the direction of flow of current).

Direction of force- Direction of force is given by Flemings left hand rule.

According to this, Stretch the thumb and the first two fingers of the left hand in mutually perpendicular direction.If the forefinger points in direction of magnetic field,central finger in the direction of current then the thumb gives the direction of force on the conductor.