The Unapologetic Mathematician

Mathematics for the interested outsider

Energy and the Magnetic Field

Last time we calculated the energy of the electric field. Now let’s repeat with the magnetic field, and let’s try to be a little more careful about it since magnetic fields can be slippery.

Let’s consider a static magnetic field B generated by a collection of circuits C_i, each carrying a current I_i. Recall that Gauss’ law for magnetism tells us that \nabla\cdot B=0; since space is contractible, we know that its homology is trivial, and thus B must be the curl of some other vector field A, which we call the “magnetic potential” or “vector potential”. Now we can write down the flux of the magnetic field through each circuit:

\displaystyle\Phi_i=\int\limits_{S_i}B\cdot dS_i=\int\limits_{C_i}A\cdot dr_i

Now Faraday’s law tells us about the electromotive force induced on the circuit:

\displaystyle V_i=\frac{d\Phi_i}{dt}

This electromotive force must be counterbalanced by a battery maintaining the current or else the magnetic field wouldn’t be static.

We can determine how much power the battery must expend to maintain the current; a charge q moving around the circuit goes down by qV_i in potential energy, which the battery must replace to send it around again. If n such charges pass around in unit time, this is a work of nqV_i per unit time; since nq=I — the current — we find that the power expenditure is P_i=I_iV_i, or.

\displaystyle P_i=I_i\frac{d\Phi_i}{dt}

Thus if we want to ramp the currents — and the field — up from a cold start in a time T it takes a total work of

\displaystyle W=\sum\limits_{i=1}^N\int\limits_0^TI_i\frac{d\Phi_i}{dt}\,dt

which is then the energy stored in the magnetic field.

This expression doesn’t depend on exactly how the field turns on, so let’s say the currents ramp up linearly:

\displaystyle I_i(t)=I_i(T)\frac{t}{T}

and since the fluxes are proportional to the currents they must also ramp up linearly:

\displaystyle\Phi_i(t)=\Phi_i(T)\frac{t}{T}

Plugging these in above, we find:

\displaystyle W=\sum\limits_{i=1}^N\int\limits_0^TI_i(T)\Phi_i(T)\frac{t}{T^2}\,dt=\frac{1}{2}\sum\limits_{i=1}^NI_i(T)\Phi_i(T)

Now we can plug in our original expression for the flux:

\displaystyle W=\frac{1}{2}\sum\limits_{i=1}^NI_i\int\limits_{C_i}A\cdot dr_i

This is great. But to be more general, let’s replace our currents with a current distribution:

\displaystyle W=\frac{1}{2}\int\limits_{\mathbb{R}^3}A\cdot J\,dV

Now we can use Ampère’s law to write

\displaystyle\begin{aligned}W&=\frac{1}{2\mu_0}\int\limits_{\mathbb{R}^3}A\cdot(\nabla\times B)\,dV\\&=\frac{1}{2\mu_0}\int\limits_{\mathbb{R}^3}B\cdot(\nabla\times A)-\nabla\cdot(A\times B)\,dV\\&=\frac{1}{2\mu_0}\int\limits_{\mathbb{R}^3}B\cdot B\,dV-\frac{1}{2\mu_0}\int\limits_{\mathbb{R}^3}\nabla\cdot(A\times B)\,dV\end{aligned}

We can pull the same sort of trick last time to make the second integral go away; use the divergence theorem to convert to

\displaystyle\frac{1}{2\mu_0}\lim\limits_{R\to\infty}\int\limits_{S_R}(A\times B)\cdot dA

and take the surface far enough away that the integral becomes negligible. We handwave that A\times B falls off roughly as the inverse fifth power of R, while the area of S_R only grows as the second power, and say that the term goes to zero.

So now we have a similar expression as last time for a magnetic energy density:

\displaystyle u_B=\frac{1}{2\mu_0}\lvert B\rvert^2

Again, we can check the units; the magnetic field has units of force per unit charge per unit velocity:

\displaystyle\frac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}

while the magnetic constant has units of henries per meter:

\displaystyle\frac{\mathrm{m}\cdot\mathrm{kg}}{\mathrm{C}^2}

Putting together an inverse factor of the magnetic constant and two of the magnetic field and we get:

\displaystyle\frac{\mathrm{C}^2}{\mathrm{m}\cdot\mathrm{kg}}\frac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}\frac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}=\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{m}^3\cdot\mathrm{s}^2}=\frac{\mathrm{J}}{\mathrm{m}^3}

or, units of energy per unit volume, just like we expect for an energy density.

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February 14, 2012 - Posted by | Electromagnetism, Mathematical Physics

4 Comments »

  1. […] calculating the potential energy of the magnetic field, we calculated the power needed to run a certain current around a certain circuit. Let’s look […]

    Pingback by Ohm’s Law « The Unapologetic Mathematician | February 16, 2012 | Reply

  2. […] On the left, we’re taking the derivative of the combined energy densities of the electric and magnetic […]

    Pingback by Conservation of Electromagnetic Energy « The Unapologetic Mathematician | February 17, 2012 | Reply

  3. […] contractible spaces, but we must make mention of the fact that our space is contractible! In fact, I did exactly that when I needed to get ahold of the magnetic potential […]

    Pingback by A Short Rant about Electromagnetism Texts « The Unapologetic Mathematician | February 18, 2012 | Reply

  4. […] Energy and the Magnetic Field (unapologetic.wordpress.com) […]

    Pingback by Magnetic fields, again | cartesian product | February 26, 2012 | Reply


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