Is Ampere's Law Affected by External Currents?

[ubergewehr273]

When we try to find magnetic field due to a set of current carrying wires in a region we draw an imaginary amperian loop and using ampere's law find the magnitude of the magnetic field.
$\oint \vec B \cdot d\vec l = \mu_{0}i_{enclosed}$
The RHS involves only the enclosed current inside the loop but however the net magnetic field is due to the current carrying wires both inside and outside the loop. I don't seem to understand this part. Because in another scenario if I have the same current carrying wires to be enclosed in the loop that were enclosed in the previous case but now I remove all the wires that are present outside the loop. Now the net magnetic field must change but when I apply ampere's law, I still get the same magnitude of magnetic field as the previous case. Isn't this contradicting the law ?

 

[ZapperZ]

When we try to find magnetic field due to a set of current carrying wires in a region we draw an imaginary amperian loop and using ampere's law find the magnitude of the magnetic field.
$\oint \vec B \cdot d\vec l = \mu_{0}i_{enclosed}$
The RHS involves only the enclosed current inside the loop but however the net magnetic field is due to the current carrying wires both inside and outside the loop. I don't seem to understand this part. Because in another scenario if I have the same current carrying wires to be enclosed in the loop that were enclosed in the previous case but now I remove all the wires that are present outside the loop. Now the net magnetic field must change but when I apply ampere's law, I still get the same magnitude of magnetic field as the previous case. Isn't this contradicting the law ?

But you HAVE seen something similar to this before, haven't you? Look at Gauss's law for a uniform spherical charge distribution, for instance. The E-field at points on the surface is due ONLY to charge enclosed INSIDE the gaussian surface, and not due to the charges outside of the surface. This is the E-field ONLY at points on the surface, not everywhere in space. You will have to do another gaussian surface to represent regions outside of the charge distribution.

The Ampere's circuital law is similar to such a concept. The B-field around the loop represented by the Ampere's loop is only due to the current enclosed within the loop. But as with Gauss's law, this is the B-field only on that loop. If you want to find the B-field outside of the current distribution, then you have to do another Amperian loop OUTSIDE of the current volume.

Zz.

 

[Dale]

The RHS involves only the enclosed current inside the loop but however the net magnetic field is due to the current carrying wires both inside and outside the loop.

It is important to note that the LHS is not just the magnetic field, but is the integral of B.dl. So, let's consider a uniform "external" field, such as a solenoid located outside of the loop. Because of the form of the integral the result is 0, regardless of the value of the uniform field and therefore regardless of the current in the solenoid outside of the loop. Although it is not obvious, the same thing happens for a non-uniform field also.

 

[ubergewehr273]

It is important to note that the LHS is not just the magnetic field, but is the integral of B.dl.

The whole point of Ampere's law is to use the power of symmetry to reduce the problem to simple cases so as to make it easy to find the magnitude of the B field. But then what's the point of finding integral of B.dl if the B field itself is different in cases where external magnetic field is present outside the loop and otherwise.

 

[Dale]

The whole point of Ampere's law is to use the power of symmetry to reduce the problem to simple cases so as to make it easy to find the magnitude of the B field. But then what's the point of finding integral of B.dl if the B field itself is different in cases where external magnetic field is present outside the loop and otherwise.

Well, it only works for that purpose in highly symmetric situations, such as close to long straight wires or inside a coaxial cable. It cannot be used for that purpose in general cases.

However, it can always be used as one of Maxwell's equations, and the differential form is more generally useful as well.

 

[ZapperZ]

Well, it only works for that purpose in highly symmetric situations, such as close to long straight wires or inside a coaxial cable. It cannot be used for that purpose in general cases.

However, it can always be used as one of Maxwell's equations, and the differential form is more generally useful as well.

Actually, just like Gauss's Law, it can be used in any general situations. It is just that it can't be solved easily or analytically for non-symmetric configurations. There's nothing to stop anyone from solving it numerically for non-symmetric situations.

Zz.

 

[Dale]

Actually, just like Gauss's Law, it can be used in any general situations. It is just that it can't be solved easily or analytically for non-symmetric configurations. There's nothing to stop anyone from solving it numerically for non-symmetric situations.

Yeah, good point. You are right.

 

[ubergewehr273]

I still haven't got my doubt clarified. How does the B field change whether I place an external current carrying conductor or not ?

 

[Dale]

I still haven't got my doubt clarified. How does the B field change whether I place an external current carrying conductor or not ?

Hmm, I thought I already explained above:

the LHS is not just the magnetic field, but is the integral of B.dl. So, let's consider a uniform "external" field, such as a solenoid located outside of the loop. Because of the form of the integral the result is 0, regardless of the value of the uniform field and therefore regardless of the current in the solenoid outside of the loop.

Was this not clear?

 

[ubergewehr273]

Was this not clear?

I didn't quite get your point.

It is important to note that the LHS is not just the magnetic field, but is the integral of B.dl. So, let's consider a uniform "external" field, such as a solenoid located outside of the loop. Because of the form of the integral the result is 0, regardless of the value of the uniform field and therefore regardless of the current in the solenoid outside of the loop. Although it is not obvious, the same thing happens for a non-uniform field also.

Then what is the point of finding integral B.dl if effects of external magentic field gets ignored?

 

[Dale]

Then what is the point of finding integral B.dl if effects of external magentic field gets ignored?

It tells you what is the effect of the sources on the field. By itself it does not completely and uniquely tell everything about the magnetic field, but it does tell something important.

 

[ubergewehr273]

It tells you what is the effect of the sources on the field. By itself it does not completely and uniquely tell everything about the magnetic field, but it does tell something important.

What do you mean by "sources"?

 

[Dale]

What do you mean by "sources"?

The currents are the sources. Ampere’s law describes how currents are related to the magnetic field.

 

[ubergewehr273]

The currents are the sources. Ampere’s law describes how currents are related to the magnetic field.

Ok thanks a lot.