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

  • $\begingroup$ What is $\vec{w}$? $\endgroup$ – toliveira Jan 17 '15 at 11:58
  • $\begingroup$ @toliveira $\vec{w}$ is the velocity (absolute) of charge carriers in the wire $\endgroup$ – SUper Jan 17 '15 at 12:00
  • $\begingroup$ 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 $\endgroup$ – Phoenix87 Jan 17 '15 at 12:38
  • $\begingroup$ No B is variable $\endgroup$ – SUper Jan 17 '15 at 12:56

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.

  • $\begingroup$ That's an incorrect analysis. $\vec{w}$ isn't parallel to $\vec{dl}$ $\endgroup$ – SUper Jan 17 '15 at 12:14
  • $\begingroup$ Aren't both tangential to the wire? $\endgroup$ – toliveira Jan 17 '15 at 12:17
  • $\begingroup$ 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 $\vec{w}=\vec{u}+\vec{v}$ $\endgroup$ – SUper Jan 17 '15 at 12:22
  • $\begingroup$ Sure! I am sorry. $\endgroup$ – toliveira Jan 17 '15 at 12:24
  • $\begingroup$ See another try. $\endgroup$ – toliveira Jan 17 '15 at 12:44

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