- #1

TheBigDig

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## 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 d

**a**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.