- #1

zenterix

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- Homework Statement
- A rectangular conducting coil of length ##l##, height ##h##, and resistance ##R## is moving parallel to the ##(x,y)## plane at a constant velocity ##\vec{v}=v_0\hat{i}## as shown below. It continues to move through a region containing a uniform magnetic field of magnitude ##B_0##. The magnetic field is in the ##-\hat{k}## direction and is confined in the region ##0\leq x\leq 3l##. Consider the unit normal to the surface enclosed by the loop to be in the same direction as the magnetic field, ##\hat{n}=-\hat{k}##.

- Relevant Equations
- Calculate ##\mathcal{E}## and ##I##, the induced emf and the current in the loop, at the instant when the loop is entering the region of magnetic field. A clockwise current is positive. A counterclockwise current is negative.

My issue is with the signs.

We have

$$\mathcal{E}=\oint \vec{v}\times\vec{B}\cdot d\vec{s}$$

$$=\oint v_0\hat{i}\times B_0(-\hat{k})\cdot dy\hat{j}$$

$$=\int_0^h v_0B_0\hat{j}\cdot dy \hat{j}$$

$$=v_0Bh$$

Now, apparently, the sign should be negative.

Intuitively, the magnetic force on any positive charge in the loop that is in the magnetic field points up. The charges on the right vertical portion of the loop move up and we have a counterclockwise current.

The current is ##I=\frac{\mathcal{E}}{R}##.

At what point in the equations above do I have an incorrect sign? In other words, how do I get a negative sign into the calculation of ##\mathcal{E}##?

One thing I just thought about is a law that I am just coming across. Faraday's law.

$$\mathcal{E}=-\frac{d\Phi_B}{dt}$$

Where ##\Phi_B## is the magnetic flux through the loop.

Let ##x(t)## be the x-coordinate of the right side of the loop relative to the start of the portion with the magnetic field on the left side of the picture above.

$$\Phi(t)=\iint_{loop}\vec{B}\cdot d\vec{A}$$

$$=B_0hx(t)$$

Then, ##\mathcal{E}=-B_0hv_0##.

This seems to work out. But how do I get the correct sign when I compute the emf directly?

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