# Magnetic field created by a current carrying wire

Hi,

I studied the Maxwell laws and the Biot Savart law and I found something I cannot answer.

If you have a finite wire carrying current (let say 5m long) and you want to determine the magnetic induction vector due to it at some point that has distance r from the wire, you have 2 options I think.

If you use the Biot Savart law and you integrate over the lenght of the wire you get an answer which is definitely less than the answer you would get if the wire would be infinite.(B=µ0*I/(2*r* ), where B is the magnetic induction vector, I is the current of the wire and r is the distance from the wire)

However if you use Ampere's law for a circle around the wire (the wire is at the center of the circle and the plane of the circle is perpendicular to the line of the wire) you get that the magnetic field is B=µ0*I/(2*r* ) since there is only the current I crossing the circle and there is no change in flux. It is the same as the magnetic field generated by an infinite long wire. So that is definitely bigger than u get from the Biot Savart law.

I can't figure out why the 2 methods don't give the same result so if you could help me out please do so. It would be really appreciated.

Thanks,
Istvan

Dale
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a finite wire carrying current
A finite wire carrying current violates the conservation of charge.

I can't figure out why the 2 methods don't give the same result
The conservation of charge is built in to Maxwell’s equations, so a scenario that violates it will cause problems when you use them to analyze it.

BvU
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As far as I know, Ampere is for an infinitely long wire (namely where there is no ##z## component of ##\vec B## anywhere).

A finite wire carrying current violates the conservation of charge.

The conservation of charge is built in to Maxwell’s equations, so a scenario that violates it will cause problems when you use them to analyze it.
Under finite wire I mean like a circuit that is big enough that the wire closing the circuit is far away so its effect to the magnetic field is negligible.

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vela
Staff Emeritus
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As far as I know, Ampere is for an infinitely long wire (namely where there is no ##z## component of ##\vec B## anywhere).
Nah, Ampere's Law always works. It's just that an infinitely long straight wire is a situation where it's straightforward to apply and find B.

• BvU
Dale
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Under finite wire I mean like a circuit that is big enough that the wire closing the circuit is far away so its effect to the magnetic field is negligible.
OK, that doesn’t change my comment. The Biot Savart law can be applied to situations that violate the conservation of charge. Maxwell’s equations cannot. If you choose to apply both to such a scenario then you are guaranteed to run into problems.

vela
Staff Emeritus
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Under finite wire I mean like a circuit that is big enough that the wire closing the circuit is far away so its effect to the magnetic field is negligible.
"Big enough" is another way of saying "infinitely far away." When you apply Ampere's Law to an infinitely long wire, you're taking advantage of the cylindrical symmetry of the system to argue that the magnitude of the field is constant on a circle centered on the wire. If the rest of the circuit is far enough away so that approximation holds true, you're saying it's infinitely far away.

If you use the Biot-Savart law on an finite length of wire to find the field at a point, you have to assume that the contributions from the rest of the circuit cancel at that point. Even though they cancel at that one point, they won't necessarily cancel at other points on a circle which goes through the point in question, so the result derived using Ampere's Law assuming the field strength is constant on the loop doesn't apply.

• Istvan01
OK, that doesn’t change my comment. The Biot Savart law can be applied to situations that violate the conservation of charge. Maxwell’s equations cannot. If you choose to apply both to such a scenario then you are guaranteed to run into problems.
Then I don't see how does it violates the conservation of charge..... I'm sorry if it's very obvious but I don't understand

Dale
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Then I don't see how does it violates the conservation of charge..... I'm sorry if it's very obvious but I don't understand
Sure, no problem. It is the ends. At one end you have current flowing out of nowhere and at the other end you have current flowing into nowhere. That appearing and disappearing current is the problem since current needs to either travel in loops or start/end in a place where charge is decreasing/increasing.

BvU
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I was a bit hasty in #3, so perhaps I can try to fix things a bit by referring to amperes law in a higly unrealistic case where you manage to get a linear section of current with charge conservation: the charge just piles up at the top end. Would the time derivative of the E-flux make the difference between infinitely long wire and a finite length section aclculated with Biot-Savart ?

Dale
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a higly unrealistic case where you manage to get a linear section of current with charge conservation: the charge just piles up at the top end. Would the time derivative of the E-flux make the difference between infinitely long wire and a finite length section aclculated with Biot-Savart ?
Mostly, yes. Having the charge pile up like that will make Maxwell’s equations work, but at the same time it violates one of the assumptions of Biot Savart. In the end it is more important to not violate Maxwell’s equations, so overall I think it is an improvement even if it doesn’t quite reconcile the two

• vsv86 and BvU
I agree with Dale. When applying integral form of Ampere's law remember that the requirement on the surface bounded by the imaginary loop you draw around the current line, the loop that you use to get your magnetic field, is only that this surface is bounded by the loop. So I could deform this surface, without moving bounds until finite current line no longer pierces it. The magnetic field would then actually be zero! Thus you get contradictions even with Ampere's law alone.

More specifically the problem is that in magnetostatics ##\vec{\nabla}\times\vec{B}=\mu_0 \vec{J}## (##\vec{B}## magnetic field, ##\vec{J}## current density), but this implies ##\vec{\nabla}.\vec{J}=0##, yet if your current line is finite, then at the ends the divergence of the current density will not be zero, so you are violating the very law you are trying to use.

• Dale
Dale
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@vsv86 that is very well said. The issue with deforming the surface is quite important and very easily forgotten.

@Dale sorry, I did not tag you, still getting used to forum :-)

BvU
It would if there wasn't a term $${1\over c^2 }{\partial \over \partial t}\int \vec E\cdot\vec {dA}\ ,$$but there is.
It would if there wasn't a term $${1\over c^2 }{\partial \over \partial t}\int \vec E\cdot\vec {dA}\ ,$$but there is.