Consider an infinitely long ideal solenoid with current $I$, radius $a$, turns per unit length $n$. Put a closed conducting loop around it (radius $b > a$), on a common axis through their centers.

I understand that there will be a flux $\Phi$ through the second loop, and if the solenoid's current is changing in time, the flux will also ($\frac{d\Phi}{dt} \not= 0)$.

This will induce a current in the circular loop, and I can find that current.

What about the force on the loop though? Assume the circular loop and the solenoid are held still. Lenz's law says the the induced current will act to oppose the change in flux. Then the magnetic fields repulse I suppose?

I'm having trouble pinning down what is causing what, and how one could quantify the force applied to the circular loop.


2 Answers 2


I think a source of difficulty lies in the fact that the solenoid is infinitely long, so strictly speaking there shouldn't be any field lines outside the solenoid. But imagine for a minute the field that the loop sets up due to the induced current. The solenoid's windings will intersect these field lines and so there will be a force on the solenoid. By Newton's third law, one would expect an equal and opposite force on the loop.

  • $\begingroup$ But the direction of the resultant force is perpendicular to the solenoid windings and to the $B$ field from the circular loop, so radially away from the loop, and by symmetry, the net force is zero. So in this case is there no net force at all? $\endgroup$
    – Henry
    Commented Dec 6, 2014 at 3:48
  • 1
    $\begingroup$ Yes, almost. The forces are balanced so there is no net force, but there exists an outward pressure. This is measurable and one of the reasons transformers buzz (the other is magnetic effects in their cores). It also limits the strength of high field magnets, as they much withstand these forces. $\endgroup$ Commented Dec 6, 2014 at 10:06
  • $\begingroup$ @lionelbrits If the current in the infinite solenoid weren't changing in time, there would be no field outside. But, as the current changes two things happen, for simplicity assume the current increases. One effect is that the outer loop sees a stronger current from the part nearby than from farther away, so it's like the field of a finite solenoid in addition to the field of the infinite solenoid. So there's a magnetic field. But that effect is small if you get close to the infinite solenoid. Secondly because of the changing current, there is an electric field outside the infinite solenoid $\endgroup$
    – Timaeus
    Commented Feb 12, 2015 at 0:15

The question of what causes what is a good one. It's best to think of fields as changing in time due to the spatial configurations of the fields. An example is:

$$ \frac{\partial \vec{B}}{\partial t}=-\vec{\nabla}\times \vec{E}.$$

The instantaneous values of $\vec{E}$ not only exert forces on charges (to contribute to their acceleration) but also completely and 100% determine the change of the $\vec{B}$ field. Similarly, the electric field changes from:

$$ \frac{\partial \vec{E}}{\partial t}=\frac{1}{\epsilon_0}\left(-\vec{J}+\frac{1}{\mu_0}\vec{\nabla}\times \vec{B}\right).$$

The electric and $\vec{B}$ field could evolve normally, as they do in free space, but that $\vec{J}$ terms changes everything. By adjusting $\vec{J}$, we can change the electric field in new non-vacuum fashions. In this case, we change the current by exerting forces on the mobile charges, this changes the $\vec{J}$, which changes the electric field, which changes the magnetic field. In every case, the change is caused by a formula with $\partial/\partial t$ and in every case it is the only fundamental (i.e. Maxwell-Lorentz) formula with a $\partial/\partial t$ so it is most definitely and unarguably the cause.

So I can tell you that if there were rigorously no $\vec{B}$ field outside your solenoid, and no current outside your solenoid, then the electric field is constant. And that if you have no circulation in your electric field outside your solenoid then your $\vec{B}$ is constant outside your solenoid. However, I don't think either of those is true (well, there is no $\vec{J}$ outside your solenoid, that part is true.) But if, for instance you have a current that is uniform throughout your wire, and is $I(t)=I_\tau t/\tau$ in your infinite solenoid, then this simply does not have a zero $\vec{B}$ field outside. Outside you see a field due to a superposition of a bunch of finite solenoids centered about the part closest to you, because you are seeing the parts closer to you from closer to the present, when $I$ is strongest. This is pretty easy to see directly from the retarded potentials.

There is also a completely different way to see causality, which is to use Jefimenko's Equations:

$$ \vec{E}(\vec{P},t)=\iiint \frac{\left(c^2\rho(\vec{r},t_r)+c|\vec{P}-\vec{r}|\dot{\rho}(\vec{r},t_r)\right)(\vec{P}-\vec{r})-|\vec{P}-\vec{r}|^2\dot{\vec{J}}(\vec{r},t_r)}{c^2|\vec{P}-\vec{r}|^34\pi\epsilon_0}d^3\tau$$ $$ \vec{B}(\vec{P},t)=\frac{\mu_0}{4\pi}\iiint \frac{\left(c\vec{J}(\vec{r},t_r)+|\vec{P}-\vec{r}|\dot{\vec{J}}(\vec{r},t_r)\right)\times (\vec{P}-\vec{r})}{c|\vec{P}-\vec{r}|^3}d^3\tau.$$


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