Physics Friday 128

Part 8: Work, Induction, and Magnetic Energy Density

Previously, we derived the energy density of the electric field by considering the work done in assembling a charge distribution from infinite separation, with the assembly done slowly enough as to be a quasi-static system. While we can find the energy density of the magnetic field by considering the work done in forming a particular current density, we cannot do so using magnetostatics; we must use Faraday’s Law of induction.

First, consider a circuit with a constant current I. If the flux Φ through the circuit changes, then an electromotive force will be generated around the circuit. This will change the current in the circuit; to oppose this, and keep the current constant, the current source must do work. As we found here, the power delivered to a current (work per unit time) by a voltage V is P=IV. Thus, the emf does work per unit time of , and so the work per unit time needed to oppose this and keep the current constant is . Now, Faraday’s Law tells us that (using SI units) . Thus, our current source delivers power ; or, in terms of differentials, a small change in flux δΦ is countered by work .

Next, consider a system of n circuits, with respective currents Ii, i=1,2,…,n. Then the flux in the ith circuit is

where dSi is the vector surface element for a surface bounded by the ith circuit (I’ve used dS rather than the usual dA to avoid confusion with the magnetic vector potential A).
Now, as we noted here, the definition of the vector potential combined with the Kelvin-Stokes theorem tell us that
,
where dsi is a vector line element of the ith circuit.
Thus, by the above single-circuit case, when there is a change in the magnetic field, and thus the fluxes, the current source of current Ii to maintain the current must deliver power equal to
.

So, then the total work necessary to take these n circuits from zero current to some final values a time T later, is
.
Now, the result should be independent of the particular “path” through intermediate values, so to simplify, we ramp up the currents proportionally, so that there is some increasing function of time f(t), with f(0)=0 and , with some constants of proportionality ci, for all i=1,2,…,n. The key then, is to note that the magnetic field generated by a current has magnitude proportional to that current, and so B will be linearly proportional to f(t), and thus all the Φi will be proportional to f(t). Thus, dubbing the constants of this latter proportionality by ki,

and so
. Since the final values of the current and flux are , and , respectively, this says that for proportional ramp-up,
, and so the work in setting up these currents is
,
and thus, for n circuits with currents Ii and fluxes Φi, the energy stored is
,
and using our expression for flux in terms of the line integral of the vector potential,
.

Now, let us instead consider a continuous current distribution, with current density J. As we did in our argument here, we break up the distribution into elemental current loops. An elemental loop will have a path C with line element ds parallel to the local current density, so that Jds=Jds; and we have a small perpendicular cross-section Δσ, so that the current in the loop is I=JΔσ. Thus, the contribution to the total stored energy by this element is
;
but, as we noted in our argument here,
JΔσds=JΔσds=Jd3r, and so the sum over all of these elemental loops becomes a volume integral:
.
Now, recall that Ampère’s Law states that
, and so, using this to replace the current density in the above, we see
.
Now, the product rule for the divergence of a cross product states that for two vector fields v and w,
.
Letting v be B and w be A, and solving for the first term on the right-hand side, we see
,
and so
.
This second term is the volume integral of a divergence; thus, by the divergence theorem,
, where S is the surface bounding our volume of integration. Now, a realistic current distribution can be expected to be of finite spatial extent; thus, as we expand our volume, the surface will eventually come to be far from the current distribution. As we noted here, for distant fields, the dipole term dominates, and the vector potential goes as r-2, and the field goes as r-3; thus, their cross product will have a magnitude that goes as r-5, while the surface of integration has area that goes as r2; thus, as the volume is expanded to all space, this surface integral will go to zero, and we get
. But from the definition of the vector potential, , so the above is
, and we identify the quantity being integrated over all space as the energy density of the magnetic field: .
Compare this to the energy density of the electric field, .

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2 Responses to “Physics Friday 128”

  1. Physics Friday 131 « Twisted One 151's Weblog Says:

    […] with the curl, we use the product rule for divergence of a cross product (similar to our usage here, albeit with E instead of A): . Now, Faraday’s Law gives us the curl of the electric field: ; […]

  2. Physics Friday 136 « Twisted One 151's Weblog Says:

    […] case in each of those limits. Also, since the energy density for electric fields is , and the energy density for magnetic fields is . Applying these to our wave, and taking the time average over a cycle, the average energy […]

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