Suppose a wire of length L carrying a current I is kept in a uniform magnetic field B perpendicular to the current. The force on the wire will be IBL and work done by magnetic force when wire moves a distance d along the force will be IBLd. But magnetic force cannot do any work on a moving charged particle and hence total work done on all particles by magnetic force should be zero. Where does the work IBLd come from?
Answer
The work comes from the battery that is driving the current through the wire.
Even if the wire were stationary, the battery would be supplying work at a rate $I^{2}R$. But with the wire moving, the battery would need to be supplying extra work at a rate $\mathscr{E}I$ in order to overcome the emf generated by the moving wire.
Now, $\mathscr{E}$ is equal to the rate at which the wire cuts magnetic flux so $\mathscr{E}=BLv$ (in which $v=\frac{d}{t}$), so the extra rate of doing work has to be $\mathscr{E} I=BLvI=BLdI / t $. And this is equal to the rate of mechanical work done on the wire!
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