 Motional electromotive force

Motional electromotive force is the contribution ${\oint }_C(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$ from an electrical conductor moving with local velocity $\mathbf{v}$ through a magnetic field. The cross product comes from the magnetic part of the Lorentz force. For a straight rod of length $\ell$, perpendicular to both its velocity and a uniform magnetic field, its magnitude is $B\ell v$. The sign depends on the direction of integration along the rod. ^[ref].

For a computed example, take $\ell =0.50\mathrm{m}$, velocity $\mathbf{v}=2.0\hat{\mathbf{x}}m/s$ to the right, and magnetic field $\mathbf{B}=0.40\hat{\mathbf{z}}\mathrm{T}$ out of the page. The rod lies along $y$. ThenPositive electric charge is driven downward. For an isolated rod after charge separation has equilibrated, $\mathbf{E}=-\mathbf{v}\times \mathbf{B}$ inside it, and the electric potential difference is $V_{\mathrm{bottom}}-V_{\mathrm{top}}=+0.40\mathrm{V}$; the bottom is positive. This open-circuit statement does not imply a sustained electric current.

Close the rod's path with stationary U-shaped rails and an electrical resistance $R>0$ at the left. With out-of-page unit normal, the positive loop direction is counterclockwise, hence upward along the rod. If the rod's horizontal position is $x(t)$, the enclosed area is $A=\ell x$ and the external magnetic flux is ${\Phi }_{\mathrm{ext}}=B\ell x$. ThusHere the resistive approximation neglects self-inductance. The negative electric current is clockwise, downward along the rod, consistent with the charge-separation calculation and Lenz's law.

The closed-loop result illustrates why the moving-circuit form of Faraday's law of induction includes motional electromotive force: the magnetic field is constant in time, so the electric field circulation alone would miss this drive. Its mechanical consequence is electromagnetic braking of a sliding rod.