# Magnetic field energy density per electrical power

1. Jan 20, 2008

### kmarinas86

http://scienceworld.wolfram.com/physics/MagneticFieldEnergyDensity.html

The magnetic field energy density is $U=\frac{B^2}{2\mu_0}$.

http://scienceworld.wolfram.com/physics/MagneticField.html

The derivative of the magnetic field is equal to $d\mathbf{B}=\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{r^2}$.

Therefore, the magnetic field is proportional to $\frac{Il}{r^2}$, and the magnetic field density is proportional to $(\frac{Il}{r^2})^2$. Electrical power is equal to $P=IV=RI^2$. Resistivity is equal to $\rho=R\frac{A}{l}$. Therefore, the central magnetic field energy density produced for a given electrical power is proportional to:

$\frac{(\frac{Il}{r^2})^2}{RI^2} =\frac{(\frac{l}{r^2})^2}{R} =\frac{(\frac{l}{r^2})^2}{\rho\frac{l}{A}} =\frac{l}{\frac{\rho}{A}r^4} =\frac{l*A}{\rho*r^4}$

.... The volume of the conductor divided by the product of resistivity and radius to the fourth power.

Is this valid?