A square loop of wire is held in a uniform 0.34 T magnetic field directed perpendicular to the plane of the loop. The length of each side of the square is decreasing at a constant rate of 4.9 cm/s. What emf is induced in the loop when the length is 15 cm?

Question

z4k4r1y4 · Answer

I got:
$$\frac{ dA }{ dt }=2.4mm ^{2}$$
Then I don't know where to go from there.

z4k4r1y4 · Answer

how do I get to Faraday's Law from there?

michele_laino · Answer

Let's suppose that the initial length of each side of the square is $l_0$, then we can write:
$$l\left( t \right) = {l_0} - vt$$
where $v=4.9$, and the corresponding area $S(t)$, as a function of time $t$, is:
$$S\left( t \right) = {\left( {{l_0} - vt} \right)^2}$$
Next we can write, the magnetic flux $ \Phi(t)$ of field $B$, is:
$$\Phi \left( t \right) = BS = B{\left( {{l_0} - vt} \right)^2}$$
and the corresponding induced voltage $E$ ( $Farady$ law,s), is:
$$E =  - \frac{{d\Phi }}{{dt}} = 2Bv\left( {{l_0} - vt} \right)$$
of course at time $\tau$ the length is $l_1=l_0-v \tau=15$, so we can write:
$$E\left( \tau  \right) = 2Bv\left( {{l_0} - v\tau } \right) = 2Bv{l_1} = ...?$$
please continue

z4k4r1y4 · Answer

Thank you very much. I was struggling to derive an equation for the magnetic field with respect to time.