# Induced emf in a wire hard mathematically

1. Jan 5, 2010

### fluidistic

1. The problem statement, all variables and given/known data
See picture for clarity.
There are 2 infinite wires carrying a current i, one wire does it in the upward direction and the other wire in the downward direction. They are separated by a distance L+2d.
Between them lies a circuit such that the part that moves with a velocity v upward has a resistance R and the rest of the circuit has a negligible resistance.
1)Calculate the induced current through the rod moving at constant velocity v.
2)Calculate the force needed to maintain the constant velocity. (applied by an external agent)
3)The work done by the external agent by unit of time.

2. Relevant equations
None given.

3. The attempt at a solution
First I calculated the magnetic field between the 2 wires. I used Ampere's law. I reached $$B=\frac{\mu _0 I}{2\pi r}+\frac{\mu _0 I}{2\pi (L+2d-r)}$$, pointing into the sheet.
Now $$V=RI\Rightarrow I=\frac{V}{R}=-\frac{d\Phi _B}{dt}\cdot \frac{1}{R}$$.
So I must find $$\frac{d \Phi _B}{dt}$$.
First, I'm looking for $$\Phi _B$$.
I believe that $$\Phi _B = A\cdot B_{\text{enclosed}}$$.
So I integrated the expression I had found of B from d to L-d, which gave me $$B_{\text{enclosed}}=\frac{\mu _0 I}{2\pi r}\left [ \ln \left ( \frac{L-d}{d} \right) + \ln \left ( \frac{d+L}{3d} \right ) \right ]$$.
Then $$\frac{d \Phi _B}{dt}=LvB_{\text{enclosed}}$$, thus $$I=-\frac{Lv B_{\text{enclosed}}}{R}$$.

It was so messy to calculate, I'm sure at 99.9999999999% I've made at least an error...
Could someone confirm it? Or at least tell me if my reasoning was right.

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2. Jan 5, 2010

### Redbelly98

Staff Emeritus
Hi there ,

You're vaguely on the right track, but here are a few comments:

"Benclosed" does not have any real meaning. Instead, you are actually trying to find
ΦB = B·dA​
for the loop.

After writing B·dA in terms of r, the distance from a wire, the integration limits will be from d to L+d These are the distances from the vertical parts of the loop to either of the wires.

The final expression for ΦB should not depend on r (an integration variable), and should depend on time t somehow.

3. Jan 5, 2010

### fluidistic

Thank you so much Redbelly. I'm going to try tomorrow (a bit late for now).
Wow, so when I will calculate $$\frac{d \Phi _B}{dt}$$, it will go even messier since ∫B·dA depends on time!
I'll probably ask for further help tomorrow... but as of now, I say a big thank you.

4. Jan 5, 2010

### Redbelly98

Staff Emeritus
Okay, good luck. By the way, it will be a fairly simple time-dependence, and a simple derivative.

5. Jan 6, 2010

### fluidistic

Here's my attempt: ΦB=∫B·dA=$$B\cdot A$$ because the 2 vectors are orthogonals. This gives me ΦB$$=\frac{\mu _0 I}{2\pi} \left[ \frac{1}{r}+\frac{1}{L+2d-r} \right] Lvt$$.
Integrating, $$\int_d^{L+d} \Phi _B dr=\frac{\mu _0 I Lvt}{2\pi} \left[ \ln \left ( \frac{L+d}{d} \right) +\ln \left ( \frac{L+3d}{d} \right ) \right ]$$.
Derivating with respect to time, $$\frac{d \Phi _B}{dt}=\frac{\mu _0 I Lv}{2\pi} \left[ \ln \left ( \frac{L+d}{d} \right) +\ln \left ( \frac{L+3d}{d} \right ) \right ]$$, thus $$I=-\frac{\mu _0 I Lv}{2\pi} \left[ \ln \left ( \frac{L+d}{d} \right) +\ln \left ( \frac{L+3d}{d} \right ) \right ] \cdot \frac{1}{R}$$. Unfortunately I cancels out... What am I doing wrong?

6. Jan 7, 2010

### rl.bhat

I didn't understand how did you get expression for ΦB.
Inside the loop field due to two current carrying conductors is identical.
So net ΦB= 2*Lvt*μo/2π*Intg[1/r]*dr between the limits d to (d+L).

7. Jan 7, 2010

### Redbelly98

Staff Emeritus
Use rl.bhat's hint, just do the integral for one wire and multiply the result by 2.

Also, dΦB/dt (1/R) gives you the current in the loop, not the current in the outside wires. They are different currents.

8. Jan 8, 2010

### fluidistic

Ok I made an error. Without using the trick of multiplying by 2 the integral of the first wire, I do it the hard way.
Here's my new attempt: From Ampere's law, the total electric field is given by $$B(r)=\frac{\mu _0I}{2\pi} \left[ \frac{1}{r}+\frac{1}{r-2d-L} \right]$$.
Do you buy that? If so, then you should also buy that $$B\cdot dA=\frac{\mu _0I}{2\pi} \left[ \frac{1}{r}+\frac{1}{r-2d-L} \right] \codt Lvdt$$.
Integrating from d to L+d, I reached $$\Phi _B= \frac{Lvdt \mu _0 I}{2\pi}\int _d ^{L+d} \frac{1}{r}+\frac{1}{r-2d-L} dr=\frac{Lvdt \mu_0 I}{2\pi} \cdot \left [ \ln \left ( \frac{L+d}{d} \right ) + \ln \left ( \frac{d}{L+d} \right) \right ]$$. It seems to me it's worth 0...
Where did I go wrong?

9. Jan 8, 2010

### rl.bhat

The distance of a point from the second wire in terms of r is (2d + L - r). Now you will get the required result. When you find magnetic field due to current carrying conductor, distance should be always positive. In your expression for second wire it is negative. That is why you got zero result.

10. Jan 9, 2010

### fluidistic

Ok thank you very much. So I'm the same case as before. I get $$\Phi _B =\frac{\mu _0 I vLdt}{2\pi} \left [ \ln \left ( \frac{L+d}{d} \right ) + \ln \left ( \frac{L+3d}{d}\right ) \right ]$$.
I don't know what I'm missing... I'm sure it's wrong. Now I have to differentiate with respect to time...

11. Jan 9, 2010

### Redbelly98

Staff Emeritus
The term in question is a result of doing this integral:

$$\int_{d}^{L+d}\frac{1}{L+2d-r}dr =-ln(L+2d-r)|_{d}^{L+d}$$

L+2d-r is the distance to the right-hand wire here.