Question

g What is the magnetic potential energy stored in a cylindrical volume of height hcylinhcylin = 50 mmmm and radius RcylinRcylin = 24 mmmm that symmetrically surrounds an infinitely long wire that has radius RwireRwire = 2.1 mmmm and carries current III = 4.9 AA ? The volume in which the energy should be calculated does not contain the wire.

Answer 1

Answer:

$2.9\cdot 10^{-7} J$

Explanation:

The energy density associated to a magnetic field is:

$u=\frac{B^2}{2\mu_0}$ (1)

where

B is the strength of the magnetic field

$\mu_0$ is the vacuum permeability

The magnetic field produced by a current-carrying wire is

$B=\frac{\mu_0 I}{2\pi r}$

where

I is the current in the wire

r is the distance from the wire at which the field is calculated

Substituting into (1),

$u=\frac{\mu_0 I^2}{8\pi^2 r^2}$ (2)

Since this is the energy density, the total energy stored in a certain element of volume $dV$ will be

$U=u\cdot dV=\frac{\mu_0I^2}{8\pi^2 r^2}dV$ (3)

Here the field strength changes as we move farther from the wire radially, so we can write dV as

$dV=2\pi h r dr$

where

h is the height of the cylinder

r is the distance from the wire

So eq(3) becomes:

$dU=\frac{\mu_0I^2}{8\pi^2 r^2} \cdot 2 \pi h r dr = \frac{\mu_0 I^2 h}{4\pi}\frac{1}{r}dr$

Now we have to integrate this expression to find the total energy stored in the cylindrical volume. We have:

h = 50 mm = 0.050 m is the height of the cylinder

I = 4.9 A is the current in the wire

$b=2.1 mm = 0.0021 m$ is the internal radius of the cylinder (the radius of the wire)

$a=24 mm=0.024 m$ is the external radius of the cylinder

So,

$U=\int\limits^a_b {dU} =\frac{\mu_0 I^2 h}{4\pi} \int\limits^{0.024}_{0.0021} \frac{1}{r}dr = \frac{\mu_0 I^2 h}{4\pi} [ln(a)-ln(b)]=\\=\frac{(4\pi \cdot 10^{-7})(4.9)^2(0.050)}{4\pi}[ln(0.024)-ln(0.0021)]=2.9\cdot 10^{-7} J$