# Importance of Ampere's Law

1. Apr 30, 2013

### CAF123

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. Apr 30, 2013

### dev70

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

3. Apr 30, 2013

### CAF123

Could you expand what you mean?

4. May 1, 2013

### dev70

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. May 1, 2013

### CAF123

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. May 1, 2013

### WannabeNewton

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. May 2, 2013

### CAF123

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?

8. May 2, 2013