# 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

• What is $\vec{w}$? Jan 17, 2015 at 11:58
• @toliveira $\vec{w}$ is the velocity (absolute) of charge carriers in the wire Jan 17, 2015 at 12:00
• is $\mathbf B$ inhomogeneous in space? otherwise, once the loop has entered in the region with the magnetic field the emf would be zero until it gets out of it Jan 17, 2015 at 12:38
• No B is variable Jan 17, 2015 at 12:56

$$\mathcal{E}_m=\oint_{\partial{S}} \vec{F_B}\cdot \vec{\mathrm{d}l}$$ $$\vec{F_B} = \vec{w}\times\vec{B}$$ $$\mathcal{E}_m=\oint_{\partial{S}} (\vec{w}\times\vec{B})\cdot \vec{\mathrm{d}l} = \oint_{\partial{S}} \vec{B}\cdot (\vec{\mathrm{d}l}\times\vec{w}) = 0$$ 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.

• That's an incorrect analysis. $\vec{w}$ isn't parallel to $\vec{dl}$ Jan 17, 2015 at 12:14
• Aren't both tangential to the wire? Jan 17, 2015 at 12:17
• No. $\vec{dl}$ is parallel to wire element. The element will have some velocity $\vec{v}$. The charge carrier has some velocity $\vec{u}$ w.r.t the element and $\vec{w}=\vec{u}+\vec{v}$ Jan 17, 2015 at 12:22
• Sure! I am sorry. Jan 17, 2015 at 12:24
• See another try. Jan 17, 2015 at 12:44