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Magnetic energy stored in a cylindrical conductor

  • #1
158
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Homework Statement


So I came across with following problem:

> Consider a cylindrical conductor of infinite length and circular section of radius a and that is traversed by a stationary current I. What is the magnetic energy stored in the conductor.

Homework Equations


3. The Attempt at a Solution [/B]
So my question is more of a conceptual one.
I proceeded to apply Ampere's law to calculate the B field which is


\begin{cases} \frac{\mu_0Ir}{2\pi a^2} \,,\, r<a\\ \\ \\ \frac{\mu_0I}{2\pi r} \,,\, r>a\end{cases}

I checked the resolution of the problem and they seem to only calculate the magnetic energy on the conductor. But according to

$$\iiint_{all\,space} 0.5 B^2 \,dV$$

Shouldn't we take it all space? Because B isn't zero outside of the conductor. I'm so confused on why they just considered the conductor, am i misunderstanding something.

Also let me add that the integral in the cylinder gave us

$$\frac{\mu_0 I^2}{16 \pi}$$

I also have know idea on how to compute the integral outside the conductor. What limits of integration should I take?
I'm really confused, can someone help me?
 

Answers and Replies

  • #2
801
173
I believe the formula is E = 0.5 (L * I2). Then you would only need to find the inductance
 
  • #3
158
7
I believe the formula is E = 0.5 (L * I2). Then you would only need to find the inductance
The goal of the exercise is to do the opposite e.g. determine E and then use that formula to get L. I didn't post that last bit because it was not relevant for the question I guess.
 
  • #4
rude man
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$$\frac{\mu_0 I^2}{16 \pi}$$

I also have know idea on how to compute the integral outside the conductor. What limits of integration should I take?
I'm really confused, can someone help me?
First, I agree with you that the B field outside the conductor must be added to that of the inside. I also agree with the given answer for just the inside mag field.

Do you know how you got the energy (per unit length BTW) for the inside energy? If so, proceed exactly the same way for the outside:
find energy in a cylindrical shell of thickness dr, length l and radius r. Compute the energy density at distance r from the center of the cylinder using the formula for B(r), r > a, which you already have. So differential energy = energy density times differential volume. The it should be obvious what the limits of integration over r have to be.

Keep the length l throughout your computations so you can check dimensions as you go, then at the end you can make l=1.
 
Last edited:
  • #5
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energy density magnetic or electric, both are proportional to square of the module of either B or E

so , that says theres a lot of differents ways to express the energy density of a field , since both use their own distinct constants, e0 and u0 but there is a equivalency you can do by substitution of the energy density which is part of the potential energy of a system equation, and , link that formula , the the induction formula generated by a point in space .

hope this helps, i can post the formulas if you want
 

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