Time-dependent potential difference between two ends of a loop

[TheBigDig]

Homework Statement

A straight copper wire that carries a sinusoidal current with an alternating frequency of 50 Hz and a maximum amplitude of 0.5 A passes through the centre of a circular ring of a second copper wire, with the two wires orientated perpendicularly to each other. The radius of the ring is 1 cm. The ring is cut in one place, to form a loop. What is the time-dependent potential difference V(t) between the two ends of the loop?

Homework Equations

I = Acos(\omega t)= 0.5cos(100\pi t)
B = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 cos(100\pi t)}{4\pi r}
V = \oint_{\delta s} \textbf{E}\cdot d\textbf{l} = -\frac{d}{dt} \int \int_S \textbf{B}\cdot d\textbf{a}

The Attempt at a Solution

I assume we have to treat this as a coil of 1 turn as opposed to a loop (in which case the answer would be zero as the flux would be zero because they're perpendicular). I replaced da with rdrdθ to get
V = \oint_{\delta s} \textbf{E}\cdot d\textbf{l} = -\frac{d}{dt} \bigg(\frac{\mu_0 cos(100\pi t)}{4\pi} \int_{0}^{2\pi} \int_{0}^{0.01} \frac{1}{r} r dr d\theta <br /> \bigg)
= -\frac{1}{50}\pi \frac{\mu_0}{4\pi} \frac{d}{dt} (\cos(100\pi t))
= \frac{1}{200}100\pi \mu_0sin(100\pi t)
= \frac{\pi \mu_0}{2}sin(100\pi t)

However this answer is different from some of my classmates but I'm not really sure where I went wrong. We haven't reached a general consensus on the answer and a few of us are getting different things so any help would be appreciated.

 

[gneill]

I think that you need to go with your first instinct, wherein the magnetic flux from the wire does not cut the plane of the loop so that no change takes place for Faraday or Lenz to apply, and hence no potential is induced. Of course, I would be perfectly happy to have my thoughts on the matter corrected (with an appropriate reference or demonstration)

 

[TheBigDig]

I discussed it with the professor who set the problem and he informed me that regarding it as a closed loop or a coil of one turn would make no difference. While he didn't confirm that the answer was 0, I think that's probably good enough to go on. I won't mark it solved as of yet in case someone has some last minute ingenious solution that I've yet to see :D

 

[gneill]

Fair enough.

If you're interested you might investigate the construction of those current clamp meters that are used to measure the current passing through a wire. They enclose the wire without touching the conductor. If it was simply a matter of forming a loop around the wire to have a potential difference induced, they wouldn't need the more complex arrangement that they employ...

 

[rude man]

Assume very thin loop wire. Compute the total flux inside the loop, then rate of change of flux = your voltage.

The answer is definitely not zero.

 

[rude man]

The Attempt at a Solution

I assume we have to treat this as a coil of 1 turn as opposed to a loop (in which case the answer would be zero as the flux would be zero because they're perpendicular). I replaced da with rdrdθ to get
V = \oint_{\delta s} \textbf{E}\cdot d\textbf{l} = -\frac{d}{dt} \bigg(\frac{\mu_0 cos(100\pi t)}{4\pi} \int_{0}^{2\pi} \int_{0}^{0.01} \frac{1}{r} r dr d\theta<br /> \bigg)
= -\frac{1}{50}\pi \frac{\mu_0}{4\pi} \frac{d}{dt} (\cos(100\pi t))
= \frac{1}{200}100\pi \mu_0sin(100\pi t)
= \frac{\pi \mu_0}{2}sin(100\pi t)

However this answer is different from some of my classmates but I'm not really sure where I went wrong. We haven't reached a general consensus on the answer and a few of us are getting different things so any help would be appreciated.

1. Don't use numbers until the end. Use symbols like B, I, r etc.
2. Use the differential area of a differential annulus of constant radius r, then integrate from 0 to the ring's radius to get total flux. Reason is that B is constant over such an annulus. Using the differential area r dr dθ makes life more tedious (forces a double integration when only one is needed).

Otherwise, looks like you're on the right track. Can't imagine why you stated at one point that the answer is zero ...

 

[TSny]

If this is the setup, then the magnetic flux through the ring is always zero.