# Net Force on a circular current carrying wire, from an infinite wire.

1. Apr 26, 2014

1. The problem statement, all variables and given/known data

-I've attached a picture of the problem-

An inﬁnitely long straight wire of steady current I1 is placed to the left of a circular wire of current I2 and radius a as shown. The center of the circular wire is distance d(≥ a) away from the straight wire. Let’s ﬁnd the net magnetic (Lorentz) force acting on the entire circular wire as follows:

(A) Express the magnetic ﬁeld Bp at point P (due to the
current I1) in given quantities (including its direction).

(B) The Lorentz force due to the magnetic ﬁeld Bp acting on a small current segment I2dℓ at P is given by

dF = I2dℓ × Bp

Express dF = (dFx, dFy, dFz) in component representation in given quantities.

(C) By integrating your results from (B) show explicitly that the net Lorentz force for the
entire circular wire is given by

Fnet = µoI1I2$\left(1-[itex]\frac{d}{\sqrt{d2-a2}}$\right)$[itex]\hat{y}$​

2. Relevant equations

Most given in question.

Bp=$\frac{-μ0I1}{2π(d+x)}$​

3. The attempt at a solution

The first two parts I got through pretty easily. Part (A) was just giving the equation of the B-field for an infinite wire. For part (B) I ended up with:

dF=$\left(\frac{-μ0I1I2acosødø}{2(d+acosø)}\right)$$\hat{x}$+$\left(\frac{-μ0I1I2asinødø}{2(d+acosø)}\right)$$\hat{y}$​

which I got from taking the cross product of I2dl and Bp, and am fairly certain is correct.

Now, for part (C) I keep getting stuck, I see what I'm supposed to get for Fnet, but my answer always has a natural log, or becomes zero, and I don't know how else to approach the problem.

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2. Apr 26, 2014

### TSny

Yes, I think that's correct. (I tried to fix the latex, hope I got it the way you intended.)

The integration of the y-component is straightforward. (The answer for this component should also be evident by symmetry.) The x-integration seems more difficult. I threw it into Mathematica and it popped out the answer. I also managed to do it by converting it to a contour integral in the complex plane. I'm not seeing an elementary way to do it, but I'm probably just not seeing it. I spent a few minutes searching integral tables on the internet but did not find it.

The answer for the net force that you stated in part (C) appears to have the wrong direction.

3. Apr 26, 2014