Derive an expression for the maximum force experienced by a straight conductor of length I, carrying current I and kept in a uniform magnetic field, B.

Consider a straight conductor PQ of length I, area of cross section A carrying current I placed in a uniform magnetic field . Suppose the conductor is placed along x-axis and magnetic field acts along y-axis. Current I flows from end P to Q and electrons drift from Q to P.
Let 
       - e = charge on each electron

Magnetic Lorentz force on an electron is given by
                                  

If n is the number of free electrons per unit volume of the conductor, then total number of free electrons in the conductor will be
N = n (AI) = nAl
Total force on the conductor is

But the current through a conductor is related with drift velocity be the relation.
                                 
                     Il = n A e v_dl
We represent I as current element vector. It acts in the direction of flow of current i.e., along OX. Then have I and  opposite directions. So
                                     ...(ii)
From (i) and (ii), we have

Magnitude of
F = Il B sin θ
When   = IlB.
Fleming’s left hand rule: This rule gives the direction of force on current carring conductor placed in magnetic field perpendicularly. If we stretch the fore finger, central finger and the thumb of our left hand mutually perpendicular to each other such that the fore finger points in the direction of magnetic field, central finger in the direction of current, then the thumb gives the direction of force experienced by the conductor.