= Motional electromotive force
{title2=$\mathcal E_{\rm mot}=\oint_C(\mathbf v\times\mathbf B)\cdot d\mathbf l$}
<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. https://openstax.org/books/university-physics-volume-2/pages/13-3-motional-emf[OpenStax University Physics 2, §13.3: motional emf]{ref}.
For a computed example, take $\ell=0.50\,\mathrm m$, <velocity> $\mathbf v=2.0\widehat{\mathbf x}\,\mathrm{m/s}$ to the right, and <magnetic field> $\mathbf B=0.40\widehat{\mathbf z}\,\mathrm T$ out of the page. The rod lies along $y$. Then
$$
\mathbf v\times\mathbf B=-0.80\widehat{\mathbf y}\,\mathrm{V/m},\qquad
\int_{\mathrm{top}}^{\mathrm{bottom}}(\mathbf v\times\mathbf B)\cdot d\mathbf l
=B\ell v=0.40\,\mathrm V.
$$
Positive <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 $\Phi_{\mathrm{ext}}=B\ell x$. Thus
$$
\mathcal E_{\mathrm{CCW}}=-B\ell\dot x=-0.40\,\mathrm V,\qquad
I_{\mathrm{CCW}}=-\frac{0.40\,\mathrm V}{R}.
$$
Here 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>.
\Image[/motional-emf-sliding-rod.png]
{title=<Motional electromotive force> from a rod sliding right through an out-of-page <magnetic field>}
{description=The U-shaped return path closes the <electrical circuit> to the left of the rod. Increasing <area> increases positive <magnetic flux>; orange arrows show the clockwise induced <electric current>. The rod's bottom end is positive in the isolated-rod limit. The <electromotive force> has magnitude $0.40\,\mathrm V$, or signed value $-0.40\,\mathrm V$ for counterclockwise orientation.}
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>.
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