A piece of insulated wire is shaped into a figure eight as shown in the figure b

Question

A piece of insulated wire is shaped into a figure eight as shown in the figure below. For simplicity, model the two halves of the figure eight as circles. the radius of the upper circle is 3.00 cm and that of the lower circle is 7.00 cm. the wire has a uniform resistance per unit length of 10.00 Ohm/m. A uniform magnetic field is applied perpendicular to the plane of the two circles, in the direction shown. the magnetic field is increasing at a constant rate of 1.80 T/s. Find the magnitude of the induced current in the wire. A Find the direction of the induced current in the wire. (Select all that apply.) clockwise in the upper loop clockwise in the lower loop counterclockwise in the upper loop counterclockwise in the lower loop

Explanation / Answer

This question has to do with Faraday's Law, which states that a changing magnetic field creates an electric field:
EMF = -d(Flux)/dt
where EMF is the induced EMF,
and Flux is the magnetic flux through a given surface: Flux = B*A,
where B is the magnetic field and A is the area of the surface

(usually, the Flux is an integral [Flux = Integral(B.dA)] but everything is perpendicular here so you don't need the integral)

Nature hates changed in flux.
So the indecued current will create an induced magnetic field that opposes the changing external field.

From looking at the geometry, we can see that the induced current will want to circulate in the same direction in both circles (wither BOTH clockwise or BOTH counterclockwise)

If you trace the path of the current, you can see that these induced currents will oppose one another.

So, the one with the bigger EMF wins!!!!
This winner is the one with the bigger area.

So now that you know which one wins, you know which way the current will flow, but what about the magnitude of the current?

Well, since the induced EMF's oppose each other the total EMF will be given by the difference between the two, which is :

EMF = (dB/dt)(A_1 - A_2) = 1.8 * Pi * (.07^2 - .03^2) = 2.262 V

now, the induced current is related to EMF by Ohm's Law:

V = IR
or in this case
EMF = IR

So, I = EMF / R.

Now you have to figure out what R is.
Since you're given a resistance per length, and you know the total length, you can see that
R = 10.00 * 2 * Pi * (.03 + .07) = 6.283 ohm

I = 2.262 / 6.283 = 0.36 A

Clockwise in lower loop

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