Current from induced emf

If the induced emf in a circuit is negative, and current from this emf is the emf over the resistance, what happens to the negative sign in the induced emf when solving for the current? Surely there's no such thing as "negative" current?

For example, if we have a constant magnetic field that is pointing down, and the area of a loop is increasing (so that the flux is positive), then the emf will be negative. But if the magnetic field is pointing down, then since the emf opposes a change, it will point up and the current will go counterclockwise. So is this what the negative current is accounting for?

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If the plus and minus signs are confusing, you can just use Lenz's Law. en.wikipedia.org/wiki/Lenz's_law –  Mark Eichenlaub Dec 10 '10 at 8:16
@Mark yes but if I use it does it mean I should ignore the signs? –  wrongusername Dec 10 '10 at 8:18
Why the flux is positive? –  KennyTM Dec 10 '10 at 8:22
@Kenny because area is increasing in a constant B field –  wrongusername Dec 10 '10 at 8:24
@wrongusername Personally, I would use Lenz's law to work out the direction of the current, then use that to double-check the work I did with the signs. You don't have to ignore the signs, since everything will technically work out fine, but you can check your answer that way. If you do rely on Lenz's law and ignore the signs, that should actually work out fine, too. It might be considered a little sloppy by a TA, though. –  Mark Eichenlaub Dec 10 '10 at 8:30

Negative current just means it flows opposite to the chosen direction.

Take your example, you have a loop with increasing area and downward pointing B-field

x   x  x  x  x  x  x  x  x
+-------------|-----------  ^
x | x  x  x  x  |  x  x  x    |     y
|             | --->        h     ^
x | x  x  x  x  |  x  x  x    |     |
+-------------|-----------  v     o---> x
x   x  x  x  x  x  x  x  x         z

<--- w(t) ---->


Therefore, the flux is $\Phi = hw(t)\mathbf B\cdot \hat{\mathbf z} = -Bhw(t)$ which is negative. Since the area is increasing, $dw/dt > 0$, so $d\Phi/dt = -Bh\,dw/dt < 0$ is also negative. The emf $\mathcal E = -d\Phi/dt > 0$ is thus positive, and the induced current flows counterclockwise as expected.

The sign of flux doesn't just depend on the area, but also the direction of the B-field relative to the surface normal.

As suggested by Mark in the comments, it would be easier to determine the sign using Lenz's law.

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I updated my question with a reply :) –  wrongusername Dec 10 '10 at 8:14