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Importance of Ampere's Law

by CAF123
Tags: ampere, importance
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CAF123
#1
Apr30-13, 05:09 AM
PF Gold
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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.
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dev70
#2
Apr30-13, 06:11 AM
P: 58
well...i believe you should consider Biot-Savart law to get into this..
CAF123
#3
Apr30-13, 06:18 AM
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P: 2,308
Quote Quote by dev70 View Post
well...i believe you should consider Biot-Savart law to get into this..
Could you expand what you mean?

dev70
#4
May1-13, 06:47 AM
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..
CAF123
#5
May1-13, 04:51 PM
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,
WannabeNewton
#6
May1-13, 06:10 PM
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Quote Quote by CAF123 View Post

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.
CAF123
#7
May2-13, 01:42 AM
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?
WannabeNewton
#8
May2-13, 12:26 PM
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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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