Retarded potential of a wire loop

[delyle]

Homework Statement

This is the problem 10.10 in Griffiths' Introduction to Electrodynamics, 3e

(See attached image)

I am only concerned about finding the retarded vector potential [itex] \textbf{A} [/itex]

Homework Equations

[itex] \textbf{A}(r',t) = \frac{\mu_{0}}{4\pi} \int \frac{ \textbf{I}(t_{r}) dl}{r'}[/itex]
[itex] t_{r} = t - r'/c [/itex]

The Attempt at a Solution

I got this far:

[itex] \textbf{A}(r',t) = \frac{\mu_{0}k}{4\pi} \left[ t \int \frac{\textbf{dl}}{r'} + \int \frac{\textbf{dl}}{c} \right][/itex]

Which matches the given solution. My next step would be to take the first integral and divide it into components, that is, take the line integral over the inner semi circle, the outer semi circle, and the two linear components on the x-axis seperately. For the line integral over the inner circle, I would take,

[itex] \int_{inner} \frac{\textbf{dl}}{r'} = -\hat{\theta} \int_{0}^{\pi} \frac{a d\theta}{a}[/itex]

which should be equal to [itex] -\pi \hat{\theta} [/itex] (r' is a in this case because we want to find the potential at the centre). However, the solution says that

[itex] \int_{inner} \frac{\textbf{dl}}{r'} = 2a\hat{x} [/itex]

The solution given for the outer integral is similar.

Either a) I've forgotten how to do line integrals (a distinct possibility), b) There is some physical reason for the solution given or c) the solution is incorrect.

I'd be very much obliged if someone could let me know what is going on for the solution given, and why my approach is incorrect.

 

[TSny]

Hello delyle. Welcome to PF!

In the integrals, dl is a vector. So the integration corresponds to vector addition. You should consider 2 integrals, one for adding the x-components of dl to find the x-component of A and another integral for the y-component.Note that [itex] \int_{inner} \frac{\textbf{dl}}{r'} = 2a\hat{x} [/itex] is dimensionally inconsistent. The left side is dimensionless while the right side has dimension of length.

 

[delyle]

Hello TSny,

Thank you for your reply. I think I understand now. As far as the dimensional inconsistency, I had made a mistake in my explanation; the solution said that [itex] \int_{inner} d\textbf{l} =2a\hat{x} [/itex], which makes more sense, dimensionally.

So, let's see if I follow you. Explicitly, [itex] -d\textbf{l} = a\sin(\theta)d\theta\hat{x} - a\cos(\theta)d\theta\hat{y} [/itex], and so the integral [itex] \int_{inner} d\textbf{l} [/itex] can be rewritten as [itex] a\int_{0}^{\pi} \left[\sin(\theta)\hat{x} - \cos(\theta)\hat{y} \right]d\theta [/itex]. The integral over cos is 0, so we're left with [itex] \int_{inner} d\textbf{l} = 2a\hat{x} [/itex], which is consistent with the solution. Thanks a ton!

 

[TSny]

That's essentially it. I'm not sure I see your choice of signs, but that might have to do with how you're defining the positive direction of θ. dl points in the same direction as the current. So along the arc of radius a, dl should have a positive x-component, but you wrote -dl as having a positive x-component (I think). Anyway, your result looks good.

 

[delyle]

Yes, you're right. That minus sign in front of [itex] d\textbf{l} [/itex] should be removed