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To calculate the current in a closed loop of resistance R subject to a changing magnetic field which generates an emf $ \epsilon= - \frac {d(\phi)} {dt} $, we use the relation ohm's law in the form.

$\epsilon=I*R$

My question is how do we use ohm's law here? Isn't ohm's law only valid in the form

$V=IR$ where V is the potential difference between two points? To that extent, how do we define potential difference between two points in a closed loop? Between which two points do we define the potential difference?

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  • $\begingroup$ No, That's where it gets interesting, potential differences Are defined for conservative electrostatic forces, The electric field generated during change in magnetic flux is not Conservative, so we need to use the root equation for Current analysis, i.e $\vec J=\sigma \vec E$ $\endgroup$ Apr 19 at 15:53
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    $\begingroup$ See my answer here physics.stackexchange.com/questions/667777/… . Might help you on establishing a separation between voltage and (scalar) potential difference. (The V in Ohm's law is voltage). $\endgroup$
    – Peltio
    Apr 19 at 19:23

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Ohm's law in the form $V=IR$ can be regarded as a special case of the arguably more fundamental local relationship $\vec J = \sigma \vec E$ in which $\vec J$ is the current density, $\sigma$ is the conductivity and $\vec E$ is the electric field strength.

The point I'm trying to make here is that $\vec E$ need not arise from separated charges, to which the notion of potential difference is applicable. Instead $\vec E$ might be the non-conservative field that arises at points around a loop through which the magnetic field is changing. If we integrate $\vec J = \sigma \vec E$ around the loop and across its cross-section we get $I=\tfrac 1R\mathscr E$.

[We shall assume that $\vec J$ and $\vec E$ are uniform across the wire's cross-section. Let $\vec A$ be the vector of magnitude $A$ normal to the wire's cross-section at some point, and in the same direction as $\vec J$ and $\vec E$ at that point. Taking the dot product with $\vec A$ of both sides of $\vec J = \sigma \vec E$, we get $\vec J.\vec A = \sigma \vec E.\vec A$. So we have $$I=\sigma EA\ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \frac 1{\sigma A}I=E$$ So for a small length, $dl$ of the wire we have $$\frac {dl}{\sigma A}I=E\ dl\ \ \ \ \ \ \text{that is}\ \ \ \ \ \ dR\ I=E\ dl$$ So integrating once round the loop, of resistance $R$, $$RI=\mathscr E\ ]$$

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  • $\begingroup$ I'm not sure how you're integrating $\vec{J}=\sigma \vec{E}$. Taking the surface integral one gets $I=\sigma \int_{S}\vec{E}\cdot d\vec{S}$, but the RHS is not the EMF. I guess if you first integrate around the loop and then do the surface integral, you can get something like $I = \sigma r \mathscr{E}/2$, and then redefine that as the resistance? $\endgroup$
    – agaminon
    Apr 19 at 16:08
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    $\begingroup$ Here's an informal treatment... If we assume that $\vec J$ is uniform across the wire's cross-section, we have $\vec J.A\hat n=\sigma \vec E.A\hat n$ in which $\hat n$ is the unit vector pointing along the loop at some point. So $I=\sigma AE$, that is $\frac 1{\sigma A}I= E$. So for a small length $dl$ of loop, $\frac{dl}{\sigma A}I=E dl$, that is $dR\ I=E dl$. Integrating once around the loop, $RI=\mathscr E$. $\endgroup$ Apr 20 at 7:50
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There's a little handwaving going on. All of it legal, but they did not specify all of the steps.

You have a loop of wire with a resistance in a changing magnetic field. As you noticed, by Faraday's Law, $\mathcal E=-\frac{d\Phi_B}{dt}$. His law is arguably for an ideal wire (no resistance), but it is easily modified. The resistance of the wire is distributed along its length and so is the EMF. We model this as a voltage source, whose voltage is $\mathcal E$ in series with a resistor $R$. We claim this simplified model is equivalent to the wire you started with.

If you are comfortable with that model, you can skip these italicized paragraphs. If its uncofmortable that somehow you can make this assumption, we can use a more general model, the "distributed element model," where its easier to disect the assumptions. We can create a distributed model where each little section of wire is modeled as a small resistance, $dR$ in series with a voltage source modeling the EMF, $d\mathcal E$. We're integrating along this wire. Of course, voltage sources and resistors in a series permit reordering. So we can lump all of the $dR$'s into one resistor of resistance $R$ and one ideal voltage source of voltage $\mathcal E$. This gets us to the above model.

(If one is still unconvinced by the distributed model, one can go all the way down to expressing Maxwell's equations, but that math gets way beyond the scope of a stack exchange answer)

Now that we have this model, we can see where $\mathcal E=IR$ comes from. You describe this as a closed loop. Thus this voltage source and resistor are all there is. Now we get to use Ohm's law because we can focus on the resistor. The entire magnetic effect has been captured in the voltage source, so it won't affect our study of the resistor. On any ideal resistor, $V=IR$. We know $R$ (presumably it was measured). We know V, because there's only one other element in this loop, our voltage source whose voltage is $\mathcal E$. This leads us to $\mathcal E=IR$.

One thing I find challenging is that there's not many uses for a closed loop with a given resistance on its own. It's a bit of an academic exercise, so its hard to check your work by asking whether you're making progress on something useful. It's almost simplified to the point of being difficult. However, once this closed loop makes sense, one can start to ask about more interesting circuits, like two loops on the same closed circuit. That gets a lot more interesting as one might convert energy from a magnetic field into current on the loop, and the other can act as an electromagnet to convert energy into a magnetic field elsewhere.

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