# Ampere's law and external currents

In Ampere's law, the current outside the curve taken is not included in the expression. Does it mean that the magnetic field calculated by using the law gives the contribution of only the currents crossing the area bounded by the curve?

The value of the line integral $\oint_C \vec B \cdot d\vec l$ really does only depend upon the current bounded by the closed path $C$. That's a consequence of Ampere's law. However, the value of the magnetic field $\vec B$ at any point along the path $C$ depends on every current, even those outside. Knowing the value of the line integral is not the same thing as knowing the value of the magnetic field.

Here's an example that illustrates the difference:

Imagine just a single wire carrying current $I_o$. Take the Amperian to be a circle that encloses the wire but is not centered on it. By Ampere's law, $\oint_C \vec B \cdot d \vec l=\mu_o I_o$. Done and done. However, you can't use this Ampere's law to "factor out" $\vec B$ from the integral, so you can't use it to solve for $\vec B$. Why? Because the magnetic field isn't constant over the path, and since it's not constant, it can't be pulled out of the integral in the manner you normally would. In order to rewrite $\oint \vec B \cdot d\vec l$ as $B \oint dl$, one requirement is that the magnitude of the magnetic is constant over the path.

One more example.

Imagine two wires that each carry current $I_o$, but the Amperian loop is centered around only one of them; the other wire is not bounded by the loop. Alright, here again, $\oint \vec B \cdot d\vec l = \mu_o I_o$. Done. But the magnetic field $\vec B$ in the integral is really the net magnetic field $\vec B_\text{net}$. As mentioned earlier, all currents contribute to the magnetic field $\vec B$ in Ampere's law. But since the net magnetic field isn't constant, you can't factor it out of the integral and solve for it.

So the Ampere's law as you know it is mainly useful when you can factor out the magnetic field and solve for it. In these cases, you typically have highly symmetric situations that allow you to do this. But even in non-symmetric cases, Ampere's law is true, but doesn't allow you to solve for the magnetic field.

• thank you so can i say that we will get magnetic field of inside if we can take B outside i.e when magnetic field is constant Aug 18, 2014 at 5:38
• You might be able to say that, but I don't recommend memorizing it as a rule. Instead, write it out the law completely every time and make sure you know what $\vec B$ does and doesn't represent at each step.
– BMS
Aug 18, 2014 at 6:12

It would be nice to provide a picture or diagram to make the question a little more clear.

The answer is that the curve of integration (to find the associated magnetic field) should cover the area or loop into which the current flows, as such there is no current outside which is not taken into account. Of course one can divide the calculation in areas and then sum these together (as in integral form).

Since the equations are linear (with respect to both $B$ and $E$, i.e magnetic and electric fields), does not make a difference.

• OK will try to add diagram next time Aug 17, 2014 at 13:10
• @rjmessibarca, no problem, the question is clear, but in general helps to make your point across :) Aug 17, 2014 at 13:11

No problem - I think this one is clear, rj. The answer is No: Amperes law is correct and exact irrespective of whatever currents are flowing outside the loop without crossing it. Of course these currents still affect the total field at any give point - it's just that these contributions sum to zero.

A good example is the field near a long straight wire. If you draw a loop that doesn't contain the wire, the line integral of the field is zero. You can choose a loop that is like a slice of pie with a bite taken out of the pointy bit - two circular arcs and two straight radial segments. The integral along the straight parts is zero by symmetry, and that along the curved parts must cancel - which shows the field $~1/R$.

Amperes law has lots of interesting consequences when symmetry is present. For example, the field inside a uniform axisymmetric current sheet with toroidal topology is $~I/R$, where $I$ is the current flowing up the centre of the toroid and R the distance from the symmetry axis, independent of z, no matter what the shape of the cross-section. Best of luck.

• The contribution of external currents to the field does not necessarily sum to zero. May 15, 2015 at 15:01