Importance of Ampere's Law

1. CAF123

2,433
Consider a hollow cylinder carrying a current I and a wire outside the cylinder carrying a current I'.
Let's say the cylinder is symmetrical with even current distribution etc.. so the B field at any point (due to current in cylinder) within the cylinder is zero by Amperes Law. However, this doesn't mean the B field is zero within the cylinder entirely - there is a B field contribution from the wire. So my question is: What is the usefullness of Amperes Law?

Does Ampere's Law only tell me something about the B field from a particular source?

Also say we have a solid cylinder inside a hollow cylinder with radii a and b respectively. They have opposite current directions. Then by Ampere, the B field at some point P where a < P < b is given as ##B = \frac{\mu I}{2\pi r}, I ##the current in the solid cylinder. Is it really? The B field from the hollow cylinder will be in the opposite direction at P and so acts to cancel the B field at P from the solid cylinder thus resulting in zero net B field, no? Yet the B field at P is in fact nonzero?

I understand how the non zero B field was obtained using Ampere's Law, but the Amperian loop which coincides with P does not simply shield the B field from the hollow cylinder. So I am struggling to see why the B field would be nonzero.

Many thanks.

2. dev70

58
well...i believe you should consider Biot-Savart law to get into this..

3. CAF123

2,433
Could you expand what you mean?

4. dev70

58
well, ampere law does explain it..but if you want to find the way it is done..you should use Biot-Savart law to find out the magnetic field. take a infinitesimal element, find the conditions, integrate it..add the vectors..you wont find them to be zero.. although ampere's law deals with the magnetic field due to closed loop carrying a current. but still it is feasible..

5. CAF123

2,433
Actually given that the B field inside the hollow cylinder from the cylinder is zero, I don't really have a problem since the B field is exclusively from the solid cylinder inside.

However, now consider the following rearrangement: A solid cylinder enclosed in a hollow cylinder and a wire outside the hollow cylinder. Take some point ##a< P < b##. Then by Ampere's Law, the B field is ##\oint \underline{B} \cdot d\underline{s} = \mu_0 I_{enc} \Rightarrow B = \frac{\mu_o I_{enc}}{2\pi R}##.

Now set up the wire such that the magnetic field it produces is in the opposite direction to the B field produced from the solid cylinder at P. (I.e so that the net B field is zero at B and the two B fields 'cancel'). We now know that the B field is zero (at P)but an application of Ampere's Law (as given before) gives a non zero field. Where is the flaw in this argument?

The Amperian loop I choose to measure the B field from the cylinder does not 'shield' the B field from the wire and yet by Ampere's Law, provided we have a closed loop enclosing the current from the solid cylinder we attain a non zero B field.

Thanks for any clarity,

6. WannabeNewton

5,827
I'm assuming ##a,b## are the radii of the solid and hollow cylinders respectively. Why should the implication you wrote down follow i.e. why does ##\oint B\cdot dl = \mu_{0}I_{\text{enc}}##, which is Ampere's law, imply ##B_{net} = \frac{\mu_o I_{enc}}{2\pi R}## for the set-up you described? You seem to be thinking that you can just pull the ##B_{net}## out of the integral like you would if you only had the two cylinders and no wire but in the latter case you have circular symmetry of ##B_{net}## around the cylinders whereas there is no such symmetry in the case where there is a wire next to the hollow cylinder. You have to use superposition on the cylinders + wire setup in order to get ##B_{net}##. What you have written down isn't ##B_{net}##, it is only the magnetic field at ##P## if we ignored the magnetic field of the wire.

7. CAF123

2,433
Hi WannabeNewton,
I see - so by including the wire this changes the magnetic field locally at point P (I.e it is no longer circular) and so ##\underline{B} \cdot d\underline{s} = B(2 \pi r)## no longer holds.

So we have to consider the field due to all possible sources at point P before we consider what the dot product of B and ds looks like?

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