Calculating Magnetic Field in Center of Twisted Conductor - Circuit Question

[Icheb]

This is the circuit in question:
http://www.atnetzwerk.de/temp/windung.gif

There is an infinitely long isolated conductor and at one point it has a twist in the shape of a circle. Now I am supposed to find the strength of the magnetic field in the center of the twist when there's a current I applied to the conductor.

I know that H = \frac{I}{2r} for the magnetic field at the center of a ring, but how does the infinitely long conductor come into place here? If I just take the conductor without the ring I could use H = \frac{I}{2 \pi r}, but that doesn't really help either.

My question is, how do I incorporate the infinitely long conductor into the equation?

 

[Ahmes]

It seems to me adding up the vector fields will do the trick. According to the right hand rule (if I'm not mistaken) the two fields will be anti-parallel, where the field of the wire "comes out of the page towards you".

 

[Icheb]

I'm not sure what you mean with "adding up the vector fields"? How do I use the fact that the conductor is infinitely long?

 

[Icheb]

I thought about it more and I'm wondering whether my solution is correct, because it seems overly simple:

H = \frac{I}{2r} + \frac{I}{2\pi r}

And then just simplify and done?

 

[Hootenanny]

Not quite. Firstly for some corrections. The magentic fields are given by

Infitely Long Straight Wire

B = \frac{\mu_{0}I}{2\pi R}

At The centre of a Loop of wire

B = \frac{\mu_{0}I}{R}

Now, you are partially right when you said this;

I thought about it more and I'm wondering whether my solution is correct, because it seems overly simple:

H = \frac{I}{2r} + \frac{I}{2\pi r}

And then just simplify and done?

However, as Ahmes said, you need to consider the vector sum of the fields, which means taking into account the direction of the field lines. Now, if you analyse the direction of the current you will find that the field of the loop is directed perpendicular to the plain of your moniter acting towards you. And the field of the wire is also acting in the same plane.

Can you go from here?

~H

 

[Icheb]

If I understand you correctly the vector of the magnetic field in the loop is
\left(<br /> \begin{array}{c}<br /> 0\\<br /> 0\\<br /> \frac{\mu_0 I}{R}<br /> \end{array}<br /> \right)<br />
and the vector of the magnetic field in the wire is
\left(<br /> \begin{array}{c}<br /> 0\\<br /> -\frac{\mu_0 I}{2\pi R}\\<br /> 0<br /> \end{array}<br /> \right)<br />
Then I just add those vectors and I have the solution?

 

[Ahmes]

Icheb, both fields are the same component (you can randomly choose it to be z). Only the signs matter here. So it should be (I didn't check the numbers, I don't remember the wire/ring field formulae by heart):
\left(<br /> \begin{array}{c}<br /> 0\\<br /> 0\\<br /> \frac{\mu_0 I}{R}-\frac{\mu_0 I}{2\pi R}<br /> \end{array}<br /> \right)<br />

 

[Icheb]

Whoops, of course! Thank you. :)

 

[dimensionless]

I have B = -\frac{\mu_{0}I}{2 R} for the magnetic field produced by the current loop. This is probably arbitrary but I would write the vector as

\left(<br /> \begin{array}{c}<br /> 0\\<br /> 0\\<br /> \frac{\mu_0 I}{2\pi R}-\frac{\mu_0 I}{2R}<br /> \end{array}<br /> \right)<br />

instead.

 

[Hootenanny]

I have B = -\frac{\mu_{0}I}{2 R} for the magnetic field produced by the current loop. This is probably arbitrary but I would write the vector as

\left(<br /> \begin{array}{c}<br /> 0\\<br /> 0\\<br /> \frac{\mu_0 I}{2\pi R}-\frac{\mu_0 I}{2R}<br /> \end{array}<br /> \right)<br />

instead.

Your vector is identical to Ahmes'