Force on a current carrying wire in a magnetic field
$ \displaystyle \vec{F} = q(\vec{v}\times \vec{B}) $
We can say
$ \displaystyle d\vec{F} = dq(\vec{v}\times \vec{B} )$
$ \displaystyle d\vec{F} = dq(\frac{\vec{dl}}{dt} \times \vec{B} )$
$ \displaystyle d\vec{F} = I(\vec{dl}\times \vec{B} )$
Actually, this force gives the force on the charge carriers within the length dl→
However, this force is converted, by collisions, into a force on the wire as a whole, a force which, moreover, is capable of doing work on the wire. The net force on a wire is found by integrating along length.
A corollary of this is that there is no net force on a current carrying loop in a uniform magnetic field. In this case, l→ = 0
Fleming’s left-hand rule
The direction of the force F→ = l(L→ x B→) is given by the Fleming’s left hand rule.
Close your left fist and then, ” shoot your index finger in the direction of the magnetic field. Relax your middle finger in the direction of the current. The force on the conductor is shown by the direction of the erect thumb .
Example : A conductor of length 2.5 m with one end located at z = 0 , x = 4m carries a current of 12 A parallel to the negative y-axis. Find the magnetic field in the region if the force on the conductor is 1.2 × 10-2 N in the direction (−i^ + k^)/√2