## Same idea, but different approach. Faraday's Law

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

The magnitude of $$B$$ is increasing at $$100T/s$$ and the solenoids (the source of the induced EMF) is infinitely long. I want to find the current through the resistors

3. The attempt at a solution

I basically used the same idea, but I used Mesh's method and I decided to traverse both currents CCW and thus giving me the equations

$$6I_A + 3I_A - 3I_B = -\phi'_{i} = -\pi$$
$$-5I_B - 3I_B + 3I_A = -\phi'_{o} = -2.25\pi$$

Solving the equations http://www.wolframalpha.com/input/?i=RowReduce{{9%2C-3%2C-pi}%2C{3%2C-8%2C-2.25pi}}

I get $$I_A= -0.0623A$$ and $$I_B = 0.860A$$

So this tells me that $$I_A$$ actually travels clockwise? This contradicts the answer key. Also, if the solenoids were not infinitely long, would the flux be through the square circuit because the magnetic field wouldn't then be confined in the solenoid? Why do they only take positive emf in the key?

Thank you very much for reading
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 Recognitions: Homework Help Check the sign of the induced emf-s. The magnetic fields are of opposite sign, and the magnitude of both increasing in the same rate, so the fluxes are opposite, their derivative are of opposite sign and so are the induced emf-s. In the left loop, the change of flux starts counter-clockwise current, so the induced emf is positive. The circuit corresponds that in the attachment. ehild Attached Thumbnails

 Quote by ehild Check the sign of the induced emf-s. The magnetic fields are of opposite sign, and the magnitude of both increasing in the same rate, so the fluxes are opposite, their derivative are of opposite sign and so are the induced emf-s. In the left loop, the change of flux starts counter-clockwise current, so the induced emf is positive. The circuit corresponds that in the attachment. ehild
I think $$E_A$$'s polarity is wrong, it should be the other way because the current flows CCW, the electric field must also be in the same direction and the potential different must point from + to -

Recognitions:
Homework Help

## Same idea, but different approach. Faraday's Law

In case of a single loop with a single voltage source, the current flows out from the positive terminal of the source toward the negative terminal. EA polarity ensures CCW current in loop A, as it was stated in the problem, so it is correct in the figure, and EA=∏ (Volt).

ehild

 Quote by ehild In case of a single loop with a single voltage source, the current flows out from the positive terminal of the source toward the negative terminal. EA polarity ensures CCW current in loop A, as it was stated in the problem, so it is correct in the figure, and EA=∏ (Volt). ehild
I got confused with the electric field INSIDE the battery and the electric field in the wire. Could you answer my other problem regarding the field confined in the solenoid? Thanks
 One question I never asked is, why is the emf RHS taken for the absolute value? Isn't $$\oint \mathbf{E}\cdot d\mathbf{s} = -\phi'$$ And why does the key say Kirchoff's Loop rule when the RHS isn't 0? Makes no sense
 Recognitions: Homework Help The varying magnetic field makes the problem non-conservative, the loop integral will not be zero. Kirchhoff"s Loop Rule in the original form says that the sum of voltage drops is the sum of the emf-s for a closed loop. Imagine a loop of resistive wire connected between the terminals of a battery. The direction of the electric field inside the wire is parallel with it, as shown in the picture. If the distance between the terminals of the battery is very small compared to the length of the wire the integral from the positive terminal to the negative one along the wire is almost the total loop integral. You need to do the last step, integrating from the negative terminal of the battery to the positive one. If you choose the path inside the battery, you do not know the electric field there. The inside of the battery is beyond Electricity: it involves also Chemistry, with non-conservative forces, so the integral of the electric field along the loop which involves the inside of the battery can be different from zero. To get zero loop integral you need to choose the integration path outside the battery, between the terminals, which potential difference is equal to the emf. The electric field between the terminals points from the positive terminal to the negative one, so it is opposite to the direction of ds: EΔs=-ε. That makes the loop integral=(integral along the resistive wire) -ε=0 or ∮E⋅ds=ϵ The integral ∫Eds along a resistor is IR. So you can write the circuit integral in the form ∑IkRk-ε=0 or ∑IkRk=ε. The derivative of the flux correspond to a battery with emf=-Φ'. In the problem, the magnitude of B changes with the rate of 100 T/s. The directions of the B fields are opposite in the loops, so are the fluxes and their derivatives. For one loop, Φ'=100 and Φ'=-100 for the other one. ehild Attached Thumbnails

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