For an inductor, the electromotive force comes from Faraday's law, as I understand it. The governing equation includes the number of turns, the magnetic flux (field and area), and a time differential.

$$ \text{emf} = n \frac{ d \phi }{ dt } = n \frac{ d ( B A) }{ dt } $$

The idea is that $B$ changes, because that comes directly from the current value.

So does $A$ want to change too? Since increasing area would decrease the flux, I picture a natural tendency to close or open the area in the same way as the inductor does for the current, because they're contained in the same derivative.

Does that imply that the wire experiences an outward or inward force? Inductors you can buy tend to be fairly well secured, but if they weren't, is there are combination of $I$ and $dI/dt$ that would make the loop balloon out, and also contract in?

I ask because I'm curious if this is used as artificial muscle fibers. Make a connected series of loops like a chain, and then alter the current to contract and expand it. Other examples of such artificial muscle fibers I've seen use less direct material properties. But I've not seen the idea mentioned before, and in fact, I've never seen it mentioned that inductors experience wire forces. I'm not really sure how to formalize that in the first place.


For a wire, the force on a current element $\boldsymbol{dl}$ in a magnetic flux density field $\boldsymbol{B}$ is just the Lorentz force $I \boldsymbol{dl \times B}$.

In some situations, you can usefully think of lines of magnetic flux density $\boldsymbol{B}$ as:

  • repelling each other laterally
  • being under tension.

Specifically, for a current sheet (think solenoid) that has flux density $B$ on one side and 0 on the other, the current sheet feels a pressure $B^2/(2 \mu_0)$ in the direction away from the field. For a solenoid, this pressure tries to expand the coil cross-section.


This is not a complete answer. It's more of a, hopefully, helpful observation:

I'm not convinced that for a loop that is not fixed in time, the equation you wrote down is correct. Here's why.

Edit 1. I recant my position on this, please read on and see Edit 2 below!

The differential form of Faraday's law is $$ \nabla\times\mathbf E = -\frac{\partial\mathbf B}{\partial t} $$ Now let $C_t$ denote a closed curve (topologically a circle) at any given time $t$, and let $\Sigma_t$ be any surface (topologically a disk) whose boundary is $C_t$. Then integrating both sides over $\Sigma_t$ and using Stokes' theorem gives $$ \int_{C_t} \mathbf E\cdot d\boldsymbol\ell = -\int_{\Sigma_t}\frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a $$ The expression you wrote down follows from taking the time derivative outside of the integral and noting that what remains is the flux. However, if the surface over which the flux is being computed is changing, then you can't take the derivative outside of the integral without incurring an extra term. In fact, one has $$ \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a = \int_{\Sigma_t}\frac{\partial \mathbf B}{\partial t}\cdot d\mathbf a + \eta(t) $$ where the expression for $\eta$ is given in the Appendix below. It follows from this that for a time-varying loop, Faraday's law is $$ \int_{C_t} \mathbf E\cdot d\boldsymbol\ell = -\frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a + \eta(t) $$ If the loop lies in a plane, and if $\mathbf B$ points perpendicular to the surface $\Sigma_t$ lying in the plane whose boundary is the loop, and if the magnetic field is spatially constant on this surface, then this reduces to $$ \int_{C_t} \mathbf E\cdot d\boldsymbol\ell = -\frac{d}{dt}\Big(B(t)A(t)\Big) + \eta(t) $$ and it seems to me like your expression is therefore missing the extra $\eta(t)$ term.

Edit 2. Prompted by Art Brown's comments, upon examining the Leibniz integral formula it becomes clear that the expression for $\eta(t)$ can be written in a vastly simpler (and manifestly gauge-invariant which is nice) form than in my appendix; $$ \eta(t) = -\int_{C_t} \mathbf v\times\mathbf B\cdot d\boldsymbol\ell, \qquad \mathbf v = \frac{\partial\gamma}{\partial t} $$ If you'd like to see the proof of this, take a look at the question I asked on math.SE when I got confused about the proof: https://math.stackexchange.com/questions/453297/leibniz-integral-rule-confusion/453316?noredirect=1#453316

What's interesting about this is that the integral form of Faraday's Law can then be written as follows: $$ \int_{C_t}(\mathbf E +\mathbf v\times\mathbf B)\cdot d\boldsymbol\ell = -\frac{d}{dt}\int_{\Sigma_t} \mathbf B\cdot d\mathbf a $$ but notice that the left-hand integrand is simply the Lorentz force per unit charge, and it is precisely this quantity that when integrated around the loop at a given time gives the emf $\mathcal E$, the emf is not just the line integral of the electric field around the loop, this is only true when the loop is stationary! The punchline is therefore that the expression $$ \mathcal E=-\frac{d\Phi}{dt} $$ is actually valid for a loop with arbitrary movement and deformation, you just have to be careful about the definition of EMF.

Appendix. Let $\boldsymbol\gamma(t,\lambda)$ denote a function that for each $t$ traverses the curve $C_t$ once as $\lambda$ goes from parameter value $0$ to $1$, then $$ \eta(t) = \int_0^1d\lambda \left(\frac{\partial\gamma_i}{\partial \lambda}(t,\lambda)\frac{\partial\boldsymbol\gamma}{\partial t}(t,\lambda)\cdot \nabla A_i(t,\boldsymbol\gamma(t,\lambda) + A_i(t,\boldsymbol\gamma(t,\lambda)\frac{\partial^2\gamma_i}{\partial \lambda\partial t}(t,\lambda)\right) $$ where $A_i$ are the components of the magnetic vector potential. We derive this expression as follows. First, recall that the magnetic field can be written as the curl of a vector potential; $$ \mathbf B = \nabla\times\mathbf A $$ This tells us that \begin{align} \int_{\Sigma_t}\mathbf B\cdot d\mathbf a &= \int_{\Sigma_t}\nabla\times\mathbf A\cdot d\mathbf a = \int_{C_t}\mathbf A\cdot d\boldsymbol \ell = \int_0^1 A_i(t,\boldsymbol\gamma(t,\lambda))\cdot\frac{\partial\gamma_i}{\partial\lambda}(t,\lambda)d\lambda \end{align} where is the last equality I restored some suppressed arguments of functions for clarity in the next step, and I rewrote the dot product in components. Now we take the time derivative of both sides. We need to use the product rule and the chain rule, and this results in the following expression: \begin{align} \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a &= \int_0^1 \left(\frac{\partial A_i}{\partial t}\cdot\frac{\partial\gamma_i}{\partial\lambda}+\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda\\ &= \int_0^1 \frac{\partial A_i}{\partial t}\cdot\frac{\partial\gamma_i}{\partial\lambda}d\lambda + \int_0^1 \left(\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda \\ &=\int_{C_t} \frac{\partial\mathbf A}{\partial t}\cdot d\boldsymbol\ell + \underbrace{\int_0^1 \left(\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda}_{\equiv \eta(t)}\\ &= \int_{\Sigma_t}\frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a + \eta(t) \end{align}

  • $\begingroup$ "I'm not convinced that ..., the equation you wrote down is correct" such is often a possibility, now isn't it? This is very helpful. $\endgroup$ – Alan Rominger Jul 25 '13 at 22:13
  • $\begingroup$ @AlanSE Glad it wasn't useless. Incidentally, I added a derivation of the extra term $\eta(t)$; I'd appreciate a look through to make sure I didn't make an error if you have time. $\endgroup$ – joshphysics Jul 26 '13 at 20:53
  • $\begingroup$ I think the formula in the question is correct as is, at least for a one-turn loop. For a constant B field, as the wire is deformed with velocity $v$ the emf around the loop is the line integral of the Lorentz force $q\boldsymbol{v \times B}$ around the circuit. If $B$ is changing as well you also get the usual time rate of change of flux density. For a uniform (but time-varying) B the emf is just $-d(BA)/dt = -A dB/dt - B dA/dt$; for a non-uniform B replace the 1st term with an area integral and the 2nd with a line integral. $\endgroup$ – Art Brown Jul 27 '13 at 3:16
  • $\begingroup$ For $n$ turns, the enclosed flux may be different for each turn, so multiplying the one-turn result by $n$ may not be accurate. $\endgroup$ – Art Brown Jul 27 '13 at 3:18
  • $\begingroup$ Reference: physics.stackexchange.com/questions/61851/… $\endgroup$ – Art Brown Jul 27 '13 at 3:18

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