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Use compatible orientations for the loop and spanning surface. The relevant Maxwell equations areThe magnetic flux transport for a moving circuit separates the change of the field from the motion of its boundary. To see the boundary term, freeze $\mathbf{B}$ at time $t$. Orient the closed swept surface as $S(t+\delta t)-S(t)+S^{\prime }$. Its side element isto first order. Hence its flux is $-\delta t{\oint }_C\mathbf{B}\cdot (\mathbf{v}\times d\mathbf{r})$. By the divergence theorem and $\nabla \cdot \mathbf{B}=0$, the sum of the three fluxes is zero, giving the purely geometric changeAdding the temporal field change and applying Stokes theorem yieldsThusThe boundary contribution is the motional electromotive force; the field contribution is Faraday's law.

For a moving closed circuit $C(t)$ bounding an oriented surface $S(t)$, Faraday's law isHere $\mathbf{E}$ is the electric field, $\mathbf{B}$ the magnetic field, $\mathbf{v}$ the local velocity of a circuit element, and ${\mathcal{E}}_{\mathrm{em}}$ the electromagnetic electromotive force per unit charge around the circuit. The line orientation and normal $\mathbf{n}$ obey the right-hand rule. The flux derivative is a total derivative and includes both changing field and changing surface. The term $\mathbf{v}\times \mathbf{B}$ is the motional electromotive force. A battery contributes an additional non-electromagnetic source electromotive force.

Take positive current around $D\to E\to F\to G\to D$, so it runs upward through the bar. With $\mathbf{B}=B_0{\mathbf{e}}_z$, the oriented magnetic flux is $B_{0}Ls(t)$ and the motional electromotive force is $-B_{0}L\dot{s}$. Ohm's law therefore givesIf $I>0$, the bar current is $E\to F$; if $I<0$, it is $F\to E$. The induced part opposes the battery for a bar moving to the right when $B_0>0$.