Challenge question: A rectangular circuit ABCD with resistance R moves away from a wire with current I at a speed v, while the side AB remains parallel to the wire. The lenght of the sides AB and BC are respectively 2a 2b. How much current flows through the circuit when the distance between the wire and the circuit is r? (Denote the magnetic permeability by μ0). The magnetic flux created by the circuit is negligible.

- ChillOut

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- ChillOut

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- ChillOut

Note: I can do this with calculus but I'm supposed to get an answer WITHOUT calculus.

- ChillOut

Be back in 5.

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## More answers

- ChillOut

Just for reference: The answer is \[\frac{2\mu_{0}abvI}{piR(r^{2}-b^{2)}}\]

- Michele_Laino

what is R?

- Michele_Laino

sorry I know

- ChillOut

If you are interested in the answer using calculus i started by doing this \(dB=\frac{\mu_{0}I}{4\pi}\frac{\vec{ds}×\vec{r}}{r^{2}}\)

- IrishBoy123

if you know the calculus, there is no harm in my having a go. and i get this solution.
with \(\hat r\) to right, \(\hat z\) up and down page, & \(\hat x\) into and out of page
\(\large \mathcal E = - \frac{d \phi}{dt}\)
\(\large \vec B(r) = \frac{\mu_o I}{2 \pi \ r} \hat x\)
\(\large \Phi = \int \int \vec B . \hat n \ dA \\ \large = \int_{z=-a}^{a} \int_{r=r-b}^{r+b} \vec B . \hat n \ dr \ dz \\ \large = 2a \int \vec B . \hat x \ dr \\ \large = \large 2a \int_{r-b}^{r+b} \frac{\mu_o I}{2 \pi \ r} \ dr = \frac{a \mu_o I}{ \pi } \int \frac{1}{r} \ dr\)
\(\large \frac{d \Phi}{dt} = \frac{d \Phi}{dr}.\frac{dr}{dt} = \frac{a \mu_o I}{ \pi \ r }.v\)
\( |I_{induced} (r)| = \large \frac{\mathcal E}{R}(r) = \frac{a \ \mu_o \ I \ v}{ \pi \ R \ r} \ \)
this looks right-ish to me so maybe the question needs to be more specific and that might resolve the differences in our solutions
to be clear, i used r as the centre of the coil: whereas the qu asks about "when the distance between the wire and the circuit is r" which is ambiguous...
of course, i used calculus but i thought that i may as well know the answer before i start doing it without calculus. but now i do not think it is possible, even having a sensible answer.

- ChillOut

Yes, r is the distance between the center of the circuit and the wire. I forgot to add that.

- ChillOut

And yes, I do know calculus albeit this is is (you might doubt it) a high school question.

- ChillOut

I might as well finish what I started

- IrishBoy123

pretty darned cool for high school!!!

- IrishBoy123

finish it and show me i am wrong :-))
i am fairly sure that i am very close to right track and just shifting the axis a little doesn't seem to get me to you.

- IrishBoy123

geometric solutions only!!

- Michele_Laino

another possible reasoning is:
let's consider the position of the circuit as in your drawing, namely:
|dw:1438889983018:dw|
where f1 an f2 are the electromagnetic forces acting on the parallel side of the circuit with respect to the wire

- ChillOut

\[\epsilon= Blv \rightarrow i=\frac{Blv}{R}\]\[i=\frac{\int\limits_{0}^{\pi}\int\limits_{r-b}^{r+b}\frac{\mu_{0}I}{4\pi r^{2}}dr(\sin(\theta)d\theta)lv}{R}\]\[i=\frac{2\int\limits_{r-b}^{r+b}\frac{\mu_{0}I}{4\pi r^{2}}drlv}{R}\]
Hold on...

- Michele_Laino

now we define the induce voltage as the subsequent formula:
\[\Large E = \frac{1}{e}\int {{\mathbf{f}} \cdot d{\mathbf{s}}} \]

- ChillOut

As I thought, the only way to approach is using calculus, right?

- Michele_Laino

where e is the electron charge.
Solving that integral we get:
\[\Large \begin{gathered}
E = 2avB\left( {r - b} \right) - 2avB\left( {r + b} \right) = \hfill \\
\hfill \\
= 2av\frac{{{\mu _0}I}}{{2\pi }}\frac{1}{{r - b}} - 2av\frac{{{\mu _0}I}}{{2\pi }}\frac{1}{{r + b}} = \hfill \\
\hfill \\
= 2av\frac{{{\mu _0}I}}{{2\pi }}\left( {\frac{1}{{r - b}} - \frac{1}{{r + b}}} \right) = 2av\frac{{{\mu _0}I}}{{2\pi }}\frac{{2b}}{{{r^2} - {b^2}}} = \hfill \\
\hfill \\
= \frac{{2abv{\mu _0}I}}{{\pi \left( {{r^2} - {b^2}} \right)}} \hfill \\
\end{gathered} \]

- Michele_Laino

I have supposed that the charge travels inside the circuit in a very short time with respect to the time needed to the same circuit to change the position

- ChillOut

Well, I guess that's it. Now to explain that for a high school student will be somehow hard :).
Thanks, guys. You gave me 2 more approaches to this question.

- Michele_Laino

since the forces f1 and f2 are constant, then my integral above becoems a simple multiplication force times distance, namely:
\[\Large E = \frac{1}{e}\left( {{f_1}2a - {f_2}2a} \right)\]
where f1 and f2 are the lorentz force acting on the sides of the circuit like in my drawing

- Michele_Laino

becomes*

- ChillOut

Well, We could get straight to the answer just making E=RI

- Michele_Laino

yes!

- Michele_Laino

\[\Large current = \frac{E}{R} = \frac{{2abv{\mu _0}I}}{{\pi R\left( {{r^2} - {b^2}} \right)}}\]

- ChillOut

I've tried searching in two old textbooks of mine and... Yeah, no high school approach for this one. This problem can be found here: http://www.studyjapan.go.jp/pdf/questions/10/ga-phy.pdf . The last question

- IrishBoy123

thank you @Michele_Laino for your input
i cut a corner and paid the price
\(\large \Phi = \frac{a \ \mu_o \ I}{\pi} \int_{r-b}^{r+b} \frac{dr}{r} = \\ \large \frac{a \ \mu_o \ I}{\pi} ln (\frac{r+b}{r-b})\)
so, in fact:
\(\large \frac{d \Phi}{dt} = \frac{d \Phi}{dr}.\frac{d r}{dt} = \\ \large \frac{a \ \mu_o \ I}{\pi} (\frac{1}{r+b} - \frac{1}{r-b}) . v \\ \large = - \frac{a \ \mu_o \ I \ v}{\pi} (\frac{2b}{r^2 - b^2})\)
and the answers tally
cool!
@ChillOut
i agree with you, hard to see a way to do this, beyond calculus....

- Michele_Laino

thank you! @IrishBoy123 @ChillOut

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