What is the Magnetic Field at a Point 10cm from the Center of Two Finite Wires?

[rude man]

Oh he didn't tell us the numerical answer, he just showed what we had to do and the formula he showed was similar to the one above.. in post #22,

What did you do different from that to get the answer you have now?

In post 22 I solved the problem (looked it up, actually!) for a circular hoop, just for comparison. The circular loop had the same perimeter (circumference) of 2m and the observation point was of course the same 10 cm from the hoop's center on its axis.

In post 27 I did the Biot-Savart integration for your problem (square loop, 50 cm on a side).
If you're interested I can give you a few leads. I did not invoke any angles, just cartesian vectors.

 

[rude man]

To apply Ampere's law, you need a setup that does not violate the Maxwell equations - you need a current loop or an infinite wire. While you have a current loop here, you cannot use this as it does not have the needed symmetry.

With a finite wire, it is easy to see that Ampere's law has to fail: you can choose the surface such that you avoid the wire...

I have perfect symmetry if I have an isolated, current-carrying wire if I choose a circular path of integration about its middle. What's not symmetric? I don't deny that any other path would not work due to lack of symmetry. And I don't see any violation of Maxwell. Ampere is based on Maxwell: del x B = j + Stokes.

I still think the resolution of the paradox is that you simply cannot have an isolated, current-carrying wire.

BTW would you care to answer my question to you in my post 27? We got the same answer so I'm quite curious how you got it without integration.

 

[mfb]

I have perfect symmetry if I have an isolated, current-carrying wire if I choose a circular path of integration about its middle. What's not symmetric?

Then you have symmetry, but violation of charge conservation: your currents begin and end in the middle of nowhere. Maxwell's equations imply charge conservation, so they have to be violated.

You cannot have the required symmetry, the proper wires and a physical setup at the same time. You can have two, but not all three.

I still think the resolution of the paradox is that you simply cannot have an isolated, current-carrying wire.

... as it violates the Maxwell equations. That's what I wrote.

BTW would you care to answer my question to you in my post 27? We got the same answer so I'm quite curious how you got it without integration.

If the wire segment appears under an angle of +-θ around the central point, the field is sin(θ) times the value for an infinite wire. (This also allows to calculate non-symmetric setups, if you consider both parts separately).

 

[rude man]

@mfb, OK. Clever on the sin(theta) argument! Thanks.

@freshcoat, we have agreed on this and we got the same answer. If you understand what mfb wrote re: sin(theta) you can get the answer without integration. Or you can use Biot-Savart & integrate. I would go with the latter approach since it's not obvious at first blush that you can handle a finite-length wire the way mfb describes.

Let me know if you want hints as to doing the Biot-Savart integration.

 

[rude man]

No Calculus is needed for this problem.
Since the point where you are calculating the magnetic field is at equal distances from the two ends of each wire, the magnetic field from each wire at that point must be perpendicular to the wire. This means you can use the equation for the magnetic field of an infinite wire, although these wires are finite.

Since the point is 0.01m from the plane and 0.025m from each edge, the distance to each wire is r=\sqrt{0.01^{2}+0.025^{2}}=0.0269m.

The field from each wire is simply \textbf{B}_{w}= \frac{\mu_{o} I}{2 \pi r}

The field from each wire makes an angle with the vertical axis of \Theta = arctan \left(\frac{0.01}{0.025}\right)

The horizontal components of the field will cancel but the vertical components will add.

The vertical component of each field is \textbf{B}_{w} cos \left(\Theta \right)

The total field is then \textbf{B}=4 \left( \textbf{B}_{w} cos \left(\Theta \right) \right)

From this I get B = 2.77e-5 T wheras there seems to be general concurrence by now (2 separate computations of differing approaches) that B = 1.88e-6.