# Importance of Ampere's Law

by CAF123
Tags: ampere, importance
 PF Gold P: 2,308 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.
 P: 58 well...i believe you should consider Biot-Savart law to get into this..
PF Gold
P: 2,308
 Quote by dev70 well...i believe you should consider Biot-Savart law to get into this..
Could you expand what you mean?

 P: 58 Importance of Ampere's Law 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..
 PF Gold P: 2,308 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,
C. Spirit
Thanks
P: 5,635
 Quote by CAF123 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}##.
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.
 PF Gold P: 2,308 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?
 C. Spirit Sci Advisor Thanks P: 5,635 That is one way to do it yes. You can use superposition; see this tutorial: http://einstein1.byu.edu/~masong/ems...Q50/S3Q50.html

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