Ampere's law on solenoid, using a circular loop

Question

Suppose we take a long and tightly wound solenoid with current I going in it. We can find the field inside, by ampere's law, taking a rectangular loop and assuming the magnetic field just outside the solenoid is 0.

However if we take a circular loop around the solenoid, then:

$\int \mathbf{B}\cdot d\mathbf{r} = 0$ as the field outside the solenoid is 0, but current passing through the loop is not 0. The current is entering at a near-90 degree angle, but it passes through the loop(as charge comes from left side, it must exit through the loop through the right side).

Since the angle at which the current goes through the loop does not matter in Ampere's law, we will still count that "I" current passes through the loop. This means that Ampere's law is not working, as current is passing through the loop, but $\int\mathbf{B}\cdot d\mathbf{r} = 0$.

Why isn't Ampere's law working in the above case?

Answer 1

You've discovered an important, real feature of coiled wire solenoids. Ampere's law is working correctly, and it shows that the magnetic field outside is indeed not zero. Instead, even for an infinite solenoid, it approximately obeys $$B_{\mathrm{out}} (2 \pi r) = \mu_0 I.$$ We don't mention this in introductory textbooks because it's a relatively small contribution. Right outside the solenoid, we will have $B_{\mathrm{out}} = \mu_0 I / (2 \pi R)$, but the field inside the solenoid is $$B_{\mathrm{in}} = \mu_0 n I = \frac{\mu_0 I}{d}$$ where $d$ is the spacing between wires. So we have $B_{\mathrm{out}} / B_{\mathrm{in}} = d / (2 \pi R)$ which is very small, if the solenoid is wound tightly. The more tightly you wind it, the smaller the ratio gets.

But if you're designing a real solenoid, and you need the field outside to be very precisely zero, then you'd need to account for this effect. For example, you can "counterwind" the solenoid, meaning you wind it clockwise going down the axis, and clockwise again going back up the axis. Then the current enclosed in your Amperian loop will be zero.

Answer 2

So, you're just doing:

$$ \epsilon \rightarrow 0 $$

but never getting there, and then asking:

"Hey, why is $\epsilon \ne 0$"

The field outside the solenoid is not zero, it is only zero if:

1. The 'noid is infinitely long

2. the screw-rate of the coils is 0

So imagine an infinite stack of a ring of disconnected super conducting rings of current.

Expensive to make, but should make it all work out.

Answer 3

The previous answers explain what is happening, I will just throw in some equations for completeness. You can model your solenoid as a cylindrical sheet of current. For an idealized solenoid, the current density is purely orthoradial, but in general it will have an longitudinal component as well.

Mathematically, for current $I$ and $n$ loops per unit length: $$ |j|=nI $$ and at an angle $\alpha$ with respect to the radial plane, in cylindrical coordinates: $$ j=nI\cos\alpha e_\phi+nI\sin\alpha e_z $$ The corresponding magnetic field outside the solenoid is: $$ B=\frac{\mu_0}{2\pi r}nI\sin\alpha e_\phi $$ which vanishes for fixed $I,n$ but shallow angle $\alpha\to0$. For completeness, inside it is: $$ B=\frac{\mu_0}{2\pi R}nI\sin\alpha e_\phi+\mu_0 nI\cos\alpha e_z $$ with $R$ the radius of the solenoid. In the same limit, only the logitudinal component survives.