Question

A length of copper wire carries a current of 11 A, uniformly distributed through its cross section. Calculate the energy density of (a) the magnetic field and (b) the electric field at the surface of the wire. The wire diameter is 2.8 mm, and its resistance per unit length is 3.7 Ω/km.

Answer 1

Answer:

a)$0.983 \frac{J}{m^3}$

b)$u_E =7.329x10^-3 \frac{J}{m^3}$

Explanation:

The energy density is "the energy per unit volume, in the electric field.  The energy stored between the plates of the capacitor equals the energy per unit volume stored in the electric field times the volume between the plates".

A magnetic field is a "vector field that describes the magnetic influence of electric charges in relative motion and magnetized materials".

Part a

For this case we can assume use the equation for the magnetic field in terms of the energy per unit of volume.

$B=\sqrt{2\mu_o u}$

Where μ0 represent the permeability constant, also known as the magnetic constant. If we solve for u we got:

$u=\frac{B^2}{2\mu_o}$

We also know that the magnetic field can be expressed in terms of the current and the radius of action R like this:

$B=\frac{\mu_o i}{2\pi R}$

Replacing this on the formula for u we have:

$u=\frac{1}{2\mu_o}(\frac{\mu_o i}{2\pi R})^2$

And simplyfing we got:

$u=\frac{\mu_o i^2}{8\pi^2 R^2}$

Replacing the values given we have:

$u=\frac{(4\pix10^{-7} \frac{H}{m} (11A)^2}{8\pi^2 (0.0014m)^2} =0.983 \frac{J}{m^3}$

Part b

The density current is given by this formula $J=i/A$ and the resistance by $R=\frac{\rho l}{A}$

If we use the equation for the energy density we have this:

$u_E =\frac{1}{2}\varepsilon_o E^2 =\frac{\varepsilon}{2}(\rho J)^2=\frac{\varepsilon}{2}(\frac{iR}{l})^2$

And replacing the values given we have:

$u_E =\frac{8.85x10^{-12}\frac{F}{m}}{2}(\frac{11A(3700\frac{\Omega}{m})}{l})^2 =7.329x10^-3 \frac{J}{m^3}$