Flux between two current-carrying wires

  • Thread starter Gene Naden
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In summary, the correct expression for the total flux between the wires is given by: ##\Phi=\frac{\mu_0 i}{\pi} ln(\frac{s}{R})##. The fraction of this flux that lies inside the wires is then calculated by dividing ##\Phi_1## by ##\Phi##, giving a value of 17%. The textbook answer of 15.3% is incorrect.
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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?
 
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  • #2
Your expression for the ##\Phi_2## integral should be twice of what you have. Would that do it?
 
  • #3
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.
 
  • #4
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.
 
  • #5
Same result: ##\frac{\frac{1}{4}}{\frac{1}{4}+\frac{ln(\frac{s}{R})}{2}}=.153##
 
  • #6
Gene Naden said:
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.
 
  • #8
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?
 
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  • #9
Hello.
Gene Naden said:

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).
 

What is flux between two current-carrying wires?

Flux between two current-carrying wires, also known as mutual flux, is the measure of the magnetic field created by one wire that passes through the other wire. This phenomenon occurs due to the interaction of the magnetic fields produced by the currents flowing through the wires.

How is flux between two current-carrying wires calculated?

The flux between two current-carrying wires can be calculated using the formula Φ = μ0*I1*I2*l/2πd, where Φ is the flux, μ0 is the permeability of free space, I1 and I2 are the currents in the wires, l is the length of the wires, and d is the distance between the wires.

What factors affect the flux between two current-carrying wires?

The flux between two current-carrying wires is affected by the currents in the wires, the distance between the wires, and the permeability of the medium between the wires. The length of the wires also has an impact on the flux, as a longer wire will produce a stronger magnetic field.

What is the direction of the flux between two current-carrying wires?

The direction of the flux between two current-carrying wires is perpendicular to the plane formed by the two wires. It follows the right-hand rule, where the thumb points in the direction of the current in one wire, and the fingers curl in the direction of the flux.

What are the applications of flux between two current-carrying wires?

The flux between two current-carrying wires is used in various applications, such as in transformers, generators, and motors. It is also utilized in measuring devices, such as current clamps, and is an important concept in the study of electromagnetism.

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