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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?

wrongusername
  • 1,937

1 Answers1

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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.

kennytm
  • 2,593