- #1

crick

- 43

- 4

## Homework Statement

A cylindrical conductor of radius ##a## has a hole of radius ##b## and there is a current ##i## uniformly spread inside the conductor.

Calculate the magnetic energy of the conductor.

## Homework Equations

$$U_{m}=\int_{Volume} \frac{B^2}{2 \mu_0} dV$$

## The Attempt at a Solution

I know that this system is equivalent to a integer cylindrical conductor with uniformly spread current ##i_{+}##, plus another conductor in the place of the hole, with uniformly spread current ##i_{-}##, where

$$i_{+}=\frac{i \, \pi a^2}{\pi (a^2-b^2)} \,\,\,\, \mathrm{and} \,\,\,\, i_{-}\frac{i \, \pi b^2}{\pi (a^2-b^2)}$$

I see that on my textbook this exercise is solved by saying that the total energy would be the sum of the energy due to ##i_{+}## and the energy due to ##i_{-}##, that is (doing the integral above):

$$U_{m}=\frac{\mu_0}{16 \pi} (i_{+}^2-i_{-}^2)$$

But I do not agree with that. The

**superposition principle**is valid in the calculation of the

**magnetic field**but not in the calculation of the total energy.

Roughly said, the sum of squares is not the square of the sum, so shouldn't I instead

**firstly calculate the total magnetic field il all the conductor**(summing vectorially the magnetic field due to ##i_{+}## and the one due to ##i_{-}## in every point)

**and then calculate**

$$U_{m}=\int \int \int_{Volume} \frac{B_{total}(x,y,z)^2}{2 \mu_0} dx dy dz$$

?

This process looks more correct because the superposition principle is allowed while calculating fields.

Did I get this right is the process done on my textbook valid to calculate the total energy?