Question

A conducting loop in the form of a circle placed perpendicular to a magnetic field of 0.40 T. If the area of the loop increases at m^2/s, what is the induced emf in the loop?

Answer 1

We must use Faraday's Law of Induction for this one.

We start with the magnetic flux, \(\Phi_B\).
$$\large{
\Phi_B = \iint\limits_{\Sigma(t)} \mathbf{B}(\mathbf{r}, t) \cdot d \mathbf{A}\\
}
$$

Where \(\Sigma(t)\) means take the integral about the imaginary surface (which is to be integrated) attached to the loop. Here we have a double integral because the surface can be ANY surface, even something that looks like a bag, but it must be attached to the loop.

Faraday says that it is the CHANGE in flux that induces an EMF:
$$
\Large{
\mathcal{E} = -{{d\Phi_B} \over dt} \\
\mathcal{E} = - \frac{d}{dt} \int_{\Sigma} \mathbf{B} \cdot d\mathbf{A}.
}
$$
I used the single integral because I choose my surface to be a flat surface.

B is the magnetic field and A is the area. B and A are perpendicular, so when we multiply B times A in the dot product the cosine term equals 1.

The only thing changing is the area, A.

So we have:
$$
\large
\mathcal{E} = - \frac{d}{dt} B\times \pi r(t)^2
$$
Note that Area (\(\pi r^2(t)\)) is a function of time because it is changing. Let's denote this dependence of area on t by \(A(t)\).

I can't tell by how much it is changing because in your problem statement, this was left out. But let's say it was changing by \(M~\cfrac{m^2}{s} \).

So, because B is constant:
$$
\large{
\frac{d}{dt} A(t)= M\\
\mathcal{E} = - \frac{d}{dt} B\times M(t)\\
=- B\times \frac{d}{dt} A(t)\\
=- B\times M~\text{Volts}
}
$$
Now just plug in B and M to get your answer.

Hope this helps.

Answer 2

*Correction - I had in my last set of equations M as a function of t. This should have been simply M NOT \(M(t)\). Sorry about that.

Answer 3

is it 0.00112 ??

Answer 4

@ybarrap

Answer 5

Did you see my solution?

Answer 6

yes;s

Answer 7

So what is the solution missing. Once you get that you'll have the answer.

Hint: M.

Answer 8

ok:s