# Flux between two current-carrying wires

## Homework Statement

Two long, parallel copper wires of diameter 2.5 mm carry currents of 10 A in opposite directions. Assuming that their central axes are 20 mm apart, calculate the magnetic flux per meter of wire that exists between those axes. What fraction of this flux lies inside the wires?

## Homework Equations

Inside a wire, ##B=\frac{\mu_0 i}{i R^2}r##.
Outside a wire, ##B=\frac{\mu_0i}{2\pi r}## where R is the radius of the wire and r is the distance from the center of the wire.

## The Attempt at a Solution

##\Phi_1=2\int _{0}^{R} \frac{\mu_0 i x}{2\pi R^2} dx=\frac{\mu_0 i}{2\pi}##
where x is the vertical distance downward from the center of the wire.
##\Phi_2=2\int_{R}^{s}\frac{mu_0 i}{2\pi r} dr=\frac{\mu_0 i}{2\pi} ln(\frac{s}{R})##
where r is the vertical distance downward from the center of the wire and s is the separation.
##\Phi=\Phi_1+\Phi_2=13.09\mu W/m##
##Frac=\frac{\Phi_1}{\Phi_1+\Phi_2}=15.3 \:percent##

For the flux, I agree with the textbook (Halliday, Resnick & Walker Fundamentals of Physics 5th edition, chapter 31, problem 23). But I disagree for the fraction of flux inside the wire. They get 17%. Can anyone verify my answer or point out where I went wrong?

kuruman
Homework Helper
Gold Member
2021 Award
Your expression for the ##\Phi_2## integral should be twice of what you have. Would that do it?

Thanks for looking at this. But actually, I mistyped the expression when I posted the problem. I actually used the correct integral, ##\frac{\mu_0 i}{\pi} ln(\frac{s}{R})##. So that is not the problem.

kuruman
Homework Helper
Gold Member
2021 Award
If you don't substitute numbers, what do you get for the fraction ##\frac{\Phi_1}{\Phi_1+\Phi_2}##? Note that the common factor ##\frac{\mu_0i}{\pi}## cancels out.

Same result: ##\frac{\frac{1}{4}}{\frac{1}{4}+\frac{ln(\frac{s}{R})}{2}}=.153##

kuruman
Homework Helper
Gold Member
2021 Award
Same result: ##\frac{\frac{1}{4}}{\frac{1}{4}+\frac{ln(\frac{s}{R})}{2}}=.153##
OK, then that's the answer. If your answer to part (a) agrees with the answer in the book and doing part (b) without relying on the numbers from (a) gives the same answer, then the answer in the book for part (b) must be incorrect.

Thank you

haruspex
Homework Helper
Gold Member
Not an area I know much about, but I believe flux lines in the same direction effectively repel each other. Is it possible the presence of the current in one wire would displace some into the other wire?
What happens if you sum the two fields before performing the integral?

Last edited:
TSny
Homework Helper
Gold Member
Hello.

## The Attempt at a Solution

##\Phi_1=2\int _{0}^{R} \frac{\mu_0 i x}{2\pi R^2} dx=\frac{\mu_0 i}{2\pi}##
If ##\Phi_1## denotes the total flux inside the two wires, then this is not quite correct. It does not take into account the flux produced inside wire 1 by the field of wire 2, and vice versa.

##\Phi_2=2\int_{R}^{s}\frac{mu_0 i}{2\pi r} dr=\frac{\mu_0 i}{2\pi} ln(\frac{s}{R})##
If ##\Phi_2## is the total flux outside the wires, then the upper limit of your integral is not correct.

The sum of your expressions for ##\Phi_1## and ##\Phi_2## does, however, give the correct value for the total flux (inside and outside the wires).