# Homework Help: Electromagnetic Induction - Concentric Circles

1. May 10, 2015

### PhysixChick

1. The problem statement, all variables and given/known data
A small circular metal ring of radius r is concentric with a large circular metal ring of radius 10r. Current in the outer ring flows counterclockwise due to an unpictured power supply. By adjusting the power supply, you can adjust I, the current in the large ring. The graph below shows I(t). Notice that I increases linearly from I0 to 2I0 from time t = 0 to time t = T

a. Sketch a rough, qualitative graph of the current in the small ring as a function of time. On your graph, let 'positive' current correspond to current flowing CW in the ring.

b.The small ring is made of a copper wire of cross-sectional area a. What is the resistance of the small ring? Let ρ Cube the resistivity of copper. Call your answer R.

c. Using the Biot-Savart rule, find the magnetic field, B (magnitude and direction) due to the large ring at the center of the small ring when the current in the large ring is 1.5I0. The direction can be described by words such as “into the page” or “out of the page.”

d. When the current in the large ring is 1.5I0, what is the current in the small ring? Neglect the small ring’s self-inductance. Answer in terms of R, r, I0, T, and any universal constants. The B field can be taken as uniform over the small ring.

2. Relevant equations
ρL/A = R

∫ B ⋅ ds = μ0I

and not too sure which other ones I would use.

3. The attempt at a solution
So I attempted a, b, and c.
For:
A. I understand that from 0 - T and from T on, it will be zero because the flux isn't changing (current is constant) and , it will be -I0 in between the T interval.

B. I got ρ(2πr)/a = R

C. I don't know if this is right, but I came up with B = μ0Ilarge/(2πr+2π(10r)) - μ0Ismall/(2πr)

D. For this part, I'm not too sure how to approach this.

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2. May 10, 2015

### Hesch

In your answer you have involved current in the small ring. It is somehow absolutely right what you are doing, but it is just not an answer to the question.

The induced voltage in the ring: V = dΨ/dt, Ψ is the flux passing through the small ring.

3. May 10, 2015

### PhysixChick

Just curious though, what do you mean? I'm not trying to be sarcastic (I know tone of voice and emotions don't translate well through text), but how is that not the answer to the question? Because it's asking for B (magnetic field) and that's what I solved for.

And for the current, I would find the rate of change of the flux, which equals the potential difference, and then I would divide that by R (found in previous part, b) and divide to find current? Is that what I'm doing here?

4. May 10, 2015

### Hesch

In the question you are asked to find the magnetic field due to the large ring. Well, english is not my language, but I read it as you must only regard current in the large ring, not in the small ring.

Tell me if I'm wrong: It's your language, I assume.

5. May 10, 2015

### Hesch

Yes.

( There is no potential difference in the ring, but there is an emf ).

6. May 11, 2015

### PhysixChick

LOL Taha, is that you? hahahaha. If not, ignore that lol.