Time-dependent potential difference between two ends of a loop

In summary: Thus, there is no change in flux and no induced EMF. Therefore, the potential difference between the two ends of the loop is also zero. In summary, the potential difference (V) between the two ends of the loop is zero due to the fact that the magnetic flux through the loop is always zero, resulting in no change in flux and no induced EMF.
  • #1
TheBigDig
65
2

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


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

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
[tex] 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
\bigg) [/tex]
[tex] = -\frac{1}{50}\pi \frac{\mu_0}{4\pi} \frac{d}{dt} (\cos(100\pi t)) [/tex]
[tex] = \frac{1}{200}100\pi \mu_0sin(100\pi t) [/tex]
[tex] = \frac{\pi \mu_0}{2}sin(100\pi t) [/tex]

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.
 
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  • #2
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) :smile:
 
  • #3
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
 
  • #4
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...
 
  • #5
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.
 
  • #6
TheBigDig said:

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
[tex] 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
\bigg) [/tex]
[tex] = -\frac{1}{50}\pi \frac{\mu_0}{4\pi} \frac{d}{dt} (\cos(100\pi t)) [/tex]
[tex] = \frac{1}{200}100\pi \mu_0sin(100\pi t) [/tex]
[tex] = \frac{\pi \mu_0}{2}sin(100\pi t) [/tex]

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 ...
 
  • #7
If this is the setup, then the magnetic flux through the ring is always zero.
upload_2018-3-8_9-23-49.png
 

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Related to Time-dependent potential difference between two ends of a loop

1. What is a time-dependent potential difference?

A time-dependent potential difference refers to a changing or varying potential difference between two points in an electric circuit over a period of time. This can be caused by a changing magnetic field or varying current in the circuit.

2. How is a time-dependent potential difference measured?

A time-dependent potential difference can be measured using a voltmeter, which is a device that measures the potential difference between two points in a circuit. The voltmeter should be connected in parallel to the circuit to accurately measure the potential difference.

3. What is the significance of a time-dependent potential difference?

A time-dependent potential difference is significant in understanding the behavior of electric circuits and the flow of electricity. It can also be used to measure the rate of change of current and to study the effects of changing magnetic fields on a circuit.

4. How does a time-dependent potential difference affect the flow of electricity?

A time-dependent potential difference can cause a changing current in a circuit, as the potential difference is what drives the flow of electricity. A varying potential difference can also cause changes in the direction and magnitude of the current in the circuit.

5. How does a time-dependent potential difference relate to Faraday's Law?

A time-dependent potential difference is a key concept in Faraday's Law, which states that a changing magnetic field can induce an electromotive force (EMF) and a resulting potential difference in a conductor. This potential difference is time-dependent and can be measured using a voltmeter.

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