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 0.93 V and a current of 2.7 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?

Question

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 0.93 V and a current of 2.7 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?

Answer 1

If the area of the coil remains same, then emf and current remain same. I am uncertain of the question here,I am thinking it is trying you to take a length of wire l, and make a circle with it, then a square, and compare the areas.

I wonder if that is the question.

Answer 2

Yes, that's the question, but I'm confused as to how I should find the length.

Answer 3

To find the emf and current induced in the square coil, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced emf is directly proportional to the rate of change of magnetic flux.

First, let's find the initial magnetic flux Φ1 through the circular coil. The formula for magnetic flux through a coil is given by Φ = B*A, where B is the magnetic field and A is the area of the coil.

Since the magnetic field is perpendicular to the plane of the coil, and the coil is a single turn circular coil, the area of the coil is just the area of the circle formed by the coil.

Let's assume the radius of the circular coil is r. Therefore, A = πr^2.

Now, let's consider the square coil. The square coil is formed by reforming the wire, so the number of turns in the square coil is still 1. The side length of the square coil will be equal to the diameter of the circular coil, which is 2r.

Therefore, the area of the square coil is A' = (2r)^2 = 4r^2.

Now, let's find the final magnetic flux Φ2 through the square coil using the formula Φ = B*A'.

Since the magnetic field and the rate of change of the magnetic field are the same for both the circular and square coils, the change in magnetic flux (ΔΦ) is the same for both coils.

According to Faraday's law, the induced emf (E) is given by E = -ΔΦ/Δt, where Δt represents the time interval over which the change in magnetic flux occurs.

Since the rate of change of the magnetic field is the same for both coils, Δt is also the same for both coils.

Now, to find the emf and current induced in the square coil, we can use the ratios of the areas A' and A:

E' (emf induced in the square coil) = E (emf induced in the circular coil) * (A' / A)

And since the number of turns is the same for both coils, the induced current (I') in the square coil will be the same as the current (I) in the circular coil.

Therefore, the emf and current induced in the square coil will be:

E' = 0.93 V * (4r^2) / (πr^2)
I' = 2.7 A

Simplifying the equation, we get:

E' = 3.72 V
I' = 2.7 A

So, the emf induced in the square coil is 3.72 V and the current induced in the square coil is 2.7 A.