Motional EMF and Ohm's law Suppose a thin, closed loop of wire is in motion in a static magnetic field. 
Let's define a few things
$$\vec{F_B} = \vec{w}\times\vec{B}$$
$$\Phi = \int_S \vec{B}.\vec{dS}$$
$$\mathcal{E}_m=\oint_{\partial{S}} \vec{F_B}.\vec{dl}$$
$S$ is the surface enclosed by the wire loop. Wire loop itself becomes ${\partial{S}}$. Also suppose that material of wire loop follows ohm's law in microscopic form, $\vec{j} = \sigma\vec{f}$ where $\vec{f}$ is force per unit charge and $\sigma$ is it's conductivity.
Can anyone mathematically prove, Ohm's law in integral form for $\mathcal{E}_m$, i.e.
$$\mathcal{E}_m = -\frac{d\Phi}{dt} = IR$$
where $I$ is the current in the wire loop and $R$ is it's resistance or point me in the right direction.
I can prove ohm's law in the case of voltage applied at the ends of a resistor. In that case electrical field inside the resistor is parallel and proportional to conductivity. But motional emf defies such an analysis
 A: Wrong answer; further information on the comments:
$$\mathcal{E}_m=\oint_{\partial{S}} \vec{F_B}\cdot \vec{\mathrm{d}l}$$
$$\vec{F_B} = \vec{w}\times\vec{B}$$
$$\mathcal{E}_m=\oint_{\partial{S}} (\vec{w}\times\vec{B})\cdot \vec{\mathrm{d}l} = \oint_{\partial{S}} \vec{B}\cdot (\vec{\mathrm{d}l}\times\vec{w}) = 0$$ since $\vec{\mathrm{d}l}$ and $\vec{w}$ are parallel.
Another try:
$$\mathcal{E}_m=\oint_{\partial{S}} (\vec{w}\times\vec{B})\cdot \vec{\mathrm{d}l} = \oint_{\partial{S}} \vec{B}\cdot (\vec{\mathrm{d}l}\times\vec{w}) = \oint_{\partial{S}} \vec{B}\cdot(\vec{\mathrm{d}l}\times(\vec{u}+\vec{v}))$$
$$\mathcal{E}_m= \oint_{\partial{S}} \vec{B}\cdot (\vec{\mathrm{d}l}\times\vec{v})$$
since $\vec{\mathrm{d}l}$ and $\vec{u}$ are parallel.
$$\mathcal{E}_m= \oint_{\partial{S}} \vec{B}\cdot \left(\vec{\mathrm{d}l}\times\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\oint_{\partial{S}} \vec{B}\cdot (\vec{\mathrm{d}l}\times\vec{r})=\frac{\mathrm{d}}{\mathrm{d}t}\int_{S} \vec{B}\cdot \vec{\mathrm{d}S}=\frac{\mathrm{d}\Phi}{\mathrm{d}t}$$
A minus signal got lost somewhere.
