Posts Tagged ‘Electric Current’

Physics Friday 128

July 23, 2010

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, .

Physics Friday 125

June 25, 2010

Part 5: Work Done on Currents

Consider a point charge of charge q moving from position r0 at time t0 to position r1 at later time t1, and let the curve C be the path it takes between these points. Further, let their be an external electric field E(r). What, then, is the total work done on the charge by this field in the time interval from t0 to t1?

The work done by a force field F on a particle traveling on a curve C is the line integral of F on C:
.
For an electric force, we have F=qE; and with position as a function of time r(t), then the line element is , where v(t) is the velocity. Thus, our work becomes:

Thus, the work done per unit of time, the power transferred to the particle by the field, for a given moment of time is thus

(by the fundamental theorem of calculus).

Now, let us have a current I flowing along the curve C, with external field E. Consider a line element ds of the current. Let the time for a bit of charge in the current to traverse this length be dt. Since the current is the amount of charge flowing per unit time, in this (infinitesimal) time interval, we have a charge of  passing through. And since the velocity of this charge is , we have , and so the work done per unit time on this piece of the current is
, and so the total power delivered to the current by the field is
.
Now, if the electric field corresponds to an electric potential φ, then , and so we see
,
and so by the gradient theorem (fundamental theorem of calculus for line integrals):
.
Thus, if we have a drop in potential, a voltage drop, of , then the power delivered to the current by the source of this voltage difference (and the corresponding electric field) is given by the above ; power is current times voltage. This law is a key tool in the analysis of circuits.

Lastly, let us replace our current in a loop with a general current density distribution . Consider within this distribution a small segment ds parallel in direction to the local current density vector J. So further, let us consider a small path C, with a small cross section Δσ perpendicular to this curve. Then the current in this elemental “tube” is , with , and so the power imparted to this element by the field is

But since ds is in the direction of J, then , and so we have
.
Now, considering the entire current density distribution as broken up into such elements, we see , and so our total power is given by the volume integral
.

For magnetic fields, the force exerted on a point charge q with velocity v is the Lorentz force
.
Thus, the magnetic force is always perpendicular to the path of the particle, , and so the magnetic field does no work.

Physics Friday 65

March 27, 2009

Consider a loop of wire, forming a simple closed plane curve. We let the plane of the loop be the xy plane. Then let us parametrize the curve via (x(u),y(u)), 0≤uumaxwith counterclockwise orientation (so that the normal vector to the loop via the right-hand rule is in the positive z direction).

Suppose we then have a uniform magnetic field of magnitude B directed at an angle α from the positive z axis; we can choose our x axis so that  is in the xz plane with positive x component.

Now, let us have a current I in the loop (positive I indicates counterclockwise current). What then, is the force on an element du at
(x(u),y(u)), and what is the net effect of this force on the loop?

The magnetic force due to field  on a length  of wire carrying a current I is . Thus, for an element du, the length vector is . We also have . Thus, we have force
.
To find the total force, we integrate over u:

(here we have used the fact that , as our curve is closed, and similarly for ).

So there is no net force on the loop. However, let us pick a point (xc,yc,0) in the plane of the loop, and find the torque about that point. For an element of the curve, the torque element is
.

Integrating this, and using again the fact that , we get
.

Now,
,
and similarly, , so we see that the x and z components of the torque are zero, and
. Note that the torque is independent of the plane point (xc,yc,0) chosen, so we can choose it to be the center of mass of the loop.

For the remaining integral in our torque, we note that
, and to this integral over our closed curve (c), we apply Green’s Theorem (see also here), which says
, where D is the plane region bound by the simple closed curve ∂D. Here, f(x,y)=0, g(x,y)=x, so
, and so
,
where A is the area enclosed by our loop. Our torque is thus:
.
Now, using the area vector , we see that , so
.
Noting that the magnetic moment of an object can be defined by the torque it experiences from an external magnetic field via
, we see that the magnetic moment of any planar current loop is
. Lastly, notice that the direction of our torque is such that the loop rotates to align the magnetic moment (and thus the area vector) with the magnetic field.