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I just want to validate something I inferred from studying Griffiths (1999).

The instantaneous magnetic field $\vec{b}(t)$ at a distance $r$ from a long infinite conductor carrying a current $i(t)$ is

$$\vec{b}(t) = \frac{\mu}{2\pi r} \hat{\psi} \cdot i(t)$$

The energy density per unit volume is defined in terms of $\vec{b}(t)$ as

$$w_V(t) = \frac{b^2(t)}{2\mu_0}$$

Question 1: Does it immediately follow that the power density available per unit volume is $\frac{dw_V(t)}{dt}$?

Question 2: If I place an inductor in that time-varying magnetic field and connect it to some electrical load, I'll be able to capture that power. Assuming the surface normals $\vec{A}$ of the turns of the inductor coils are perfectly collinear with $\hat{\psi}$, what might be some energy loss mechanisms that prevent me from getting all that power, aside from electrical series resistance in the inductor?

This is a self-study question, not a homework.

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2 Answers 2

A time-varying magnetic field produces an electric field. In your example $c$ $\rightarrow$ $\infty$ -- I'll work in this limit so that we can get rid of second-order effects (i.e. the quasi-static approximation.) The relevant Maxwell equation is

$$\nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}.$$

We know that the magnetic field will curl around the wire by the right-hand rule. Let's take, as an example, the current $I(t)=I_0t$ so that ${\bf B}({\bf r},t)=B(r)t{\bf \hat{\phi}}$ where $B(r)=\frac{B_0}{r}$. Then

$$\nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}=-\frac{B_0}{r}{\bf \hat{\phi}}$$ which is constant in time. We can use Stoke's theorem to solve for ${\bf E}$: $$\int \nabla \times {\bf E} \cdot d{\bf A}=\int {\bf E} \cdot d{\bf l}=-B_0\int \frac{1}{r}\hat{\phi}\cdot d{\bf A}$$

where $d{\bf A}=drdz\hat{\phi}$ and $d{\bf l}=dz\hat{z}$. Integrating from $r=a$ to $r=b$:

$$E(a)-E(b)=-B_0\ln(b/a)$$

so that in general

$${\bf E}(r) \approx B_0\ln(r)\hat{z}.$$

This is the electric field which does work on charges. The magnetic field doesn't do work on the charges because of the Lorentz force law. Even though the magnetic field points along the coils, the charges are directed into little circles, causing no net flow of current. More simply, ${\bf B} \cdot {\bf v} = 0.$

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Okay, so I can solve for $E(t)$ from $B(t)$. Therefore, should I consider the energy density in the electric field $\frac{\epsilon_0e^2(t)}{2}$, then differentiate it over time so I can get the power density available? Given that $B(t)$ does no work on charges, I can just safely leave it out? –  Kit Mar 26 '13 at 4:46
    
I'm not sure if I know how to interpret "power density available." If by power you mean $dE/dt$ then you must include the magnetic field, for although it can't do work it still contains energy. On the other hand it won't be doing work on charges. So I suppose if you could somehow transform all of the ${\bf E}$ field energy into the motion of charges (excluding second order effects) then you could think about this as the power density available. But I'm not actually sure, maybe someone else can chime in. –  santa claus Mar 27 '13 at 21:55

First of all, your formula for the energy density is missing a term for the electric field:

$$ w(t)=\frac{\varepsilon_0 e^2(t)}{2}+\frac{b^2(t)}{2\mu_0 } $$

You will always have an electric field if the magnetic field is time-varying.

However, this is not that important for the more conceptional question you seem to be struggling with. I think a good analogy to help you here is to think of this energy as a resorvoir of water with an inflow and an outflow. The change in the amount of water in the reservoir is the inflow minus the outflow. But the flow through the reservoir can be much larger than the change in water in the reservoir. For instance, if the inflow is equal to the outflow, there is no change in water in the reservoir even though there is a flow through the reservoir.

Similarly, the electomagnetic energy flowing through a unit volume can be much larger than $dw(t)/dt$. What it is in your example and how that energy flow could be "captured" is a question I'll leave to someone else.

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