# Archived Current of Coil, and Emf of Coil.

1. Feb 20, 2008

### rcmango

1. The problem statement, all variables and given/known data

A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of 0.91 V and a current of 3.2 A are induced in the coil. The wire is the re-formed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate).

What emf and current are induced in the square coil?

2. Relevant equations

L = 2*pi*r, A = L/4*pi, A = L^2/16

3. The attempt at a solution

Okay i know the lengths are the same for the circular coil and the square coil. Oh and the form has changed.

So A1/A2 = (L^2/4*pi) / (L^2/16) = 16/4*pi

and for the current i could use I2 = V2/R

and i'm not sure but i think R = V1/I1 ?

I have alot of help so far, not sure what the final answer is, anyone?

2. Feb 7, 2016

### Staff: Mentor

A complete solution is offered.

Faraday's law tells us that an emf is induced in a loop when the magnetic flux passing through the loop changes. In this case we have a changing magnetic field that is the same in both cases, and a change of loop area due to a change in the geometry of the loop. Only the magnetic field changes; the area of each loop is fixed.

When area is fixed, Faraday's law is:

$emf = A \frac{ΔB}{Δt}$

where $A$ is the loop area and $B$ is the magnetic field strength. $\frac{ΔB}{Δt}$ is the same for both scenarios. What is different is the loop area. Forming ratio for the two cases the changing flux cancels:

$\frac{emf_2}{emf_1} = \frac{A_2}{A_1}$

Apparently we need to look into the ratio of the loop areas.

The length of wire used to form the loop is the same in each case (which means its electrical resistance will be the same, too, which will be important later). So we want the areas in terms of the perimeter, L.

For a circular loop the perimeter is $L = 2 \pi r$ and the area is $A = \pi r^2$ Combining them gives:

$A_1 = \frac{L^2}{4 \pi}$

And for the square loop each side will have length L/4, so

$A_2 = \frac{L^2}{16}$

Giving us the ratio of areas:

$\frac{A_2}{A_1} = \frac{\pi}{4}$

Now returning to Faraday, our emf for the square loop will be:

$emf_2 = emf_1\frac{A_2}{A_1} = (0.91~V)\frac{\pi}{4} = 0.72~V$

With the same resistance the current will be proportional to the emf. So the current will be:

$I_2 = I_1 \frac{emf_2}{emf_1} = I_1 \frac{A_2}{A_1} = (3.2~A) \frac{\pi}{4} = 2.5~A$