# Force on a current carrying conductor in a uniform magnetic field

Current through a conductor is due to the motion of free electrons in it in a definite direction. If such a conductor be placed in a magnetic field, each electron experiences a force and thus the conductor experiences a force when placed in the magnetic field.

Let us consider a straight conductor of length I, area of cross-section A and carrying a current i be placed in a uniform magnetic field of induction \vec B [in the figure given below]. Also let the conductor be placed along X-axis and the magnetic field be acting in XY plane making an angle \theta with X-axis. The current i is flowing parallel to X-axis as shown in the figure, so the electrons must be moving in opposite direction.

If V_d be the drift velocity of the electrons and – e be the charge on each electron, then magnetic Lorentz force acting on an electron is given by,

\vec f= -e(\vec v_d\times\vec B)

If n be the number of free electrons per unit volume, then total number N of free electrons in the conductor will be

N=nAl

\therefore Total force \vec F on the conductor is given by,

\vec F=N\vec f=-nAle(\vec v_d\times \vec B)

Now, current i through the conductor is

i= nAev_d

\therefore il=nAlev_d

Let i\vec l be representedas current element vector. It acts in the direction of flow of the current (i.e., along OX). As i\vec l and \vec v_d are oppositely directed so,

i\vec l= -nAle\vec v_d

\therefore \vec F = i\vec l \times \vec B

i.e., F= ilB\sin\theta

where \theta is the smaller angle between i\vec l and \vec B