Wikipedia:Reference desk/Archives/Science/2024 February 10

= February 10 =

Rotating a magnetic field
I've been thinking about a superconducting electric coil with some current, creating a stable magnetic field. Now assume an axis through the middle of that coil. My intuition tells me that rotating the coil around this axis should be inconsequential. But assume an axis perpendicular to the first one. If I rotate the coil around this axis (and, I assume, the magnetic field with it), a distant observer would see a regularly fluctuating magnetic field - and, I again assume, because auf the unity of electric and magnetic fields, this would appear as an electromagnetic wave. Is this correct? If so, where is the energy carried by the wave coming from? Is the magnetic field dissipating? Or do I need to put energy into the system to keep rotating the coil? Thanks! --Stephan Schulz (talk) 13:53, 10 February 2024 (UTC)


 * Assuming that the setup indeed creates a changing magnetic field, by Lenz's law this induces an opposing current which (as in an induction brake) should counteract the rotation. --Lambiam 15:00, 10 February 2024 (UTC)


 * An ordinary permanent magnet is a cheaper and more convenient alternative to a super-conducting coil. Spinning the magnet about its transverse axis will induce eddy currents in any nearby metal objects, which will oppose the rotation of the magnet and remove energy by resistive heating of the metal - this is induction braking.  In the absence of any nearby metal objects, then there will indeed be braking associated with the energy and angular momentum carried away by the radiated field.  However it is necessary to spin the magnet very fast or be very far from metal for the radiation braking to be significant compared with the induction braking, requiring something of the order of ω = c / d where ω is the angular frequency of the rotation and d is the distance from metal. catslash (talk) 18:09, 10 February 2024 (UTC)
 * Thank you - and of course. I should have thought of that! --Stephan Schulz (talk) 05:07, 11 February 2024 (UTC)
 * An example of such a strong, rapidly spinning permanent magnet with no metal anywhere near would be a pulsar. There is, however, some interesting plasma physics happening in the pretty good vacuum of the near field, so although there are no metals, there are electric currents. And pulsars do spin down. PiusImpavidus (talk) 11:57, 11 February 2024 (UTC)
 * A perhaps overly complicated analysis:


 * If an electrically small loop carrying $$I$$ amp-turns d.c. around an area $$A$$, spins with an angular frequency $$\omega$$ about its diameter on (say) the z-axis, then the resulting electromagnetic field is


 * $$\mathbf{E} = {Z}_{0} I A \left( \left( \frac{1}{i k r} + \frac{1}{(i k r)^{2}}\right) \frac{y \mathbf{\hat{z}} - z \mathbf{\hat{y}}}{r} + i \left( \frac{1}{i k r} + \frac{1}{(i k r)^{2}}\right) \frac{- x \mathbf{\hat{z}} + z \mathbf{\hat{x}}}{r}\right) i k^{3} \frac{e^{- i k r}}{4 \pi}$$


 * $$\mathbf{H} = I A \left( \left( \frac{1}{i k r} + \frac{1}{(i k r)^{2}} + \frac{1}{(i k r)^{3}}\right) (\mathbf{\hat{x}} + i \mathbf{\hat{y}}) - \left( \frac{1}{i k r} + \frac{3}{(i k r)^{2}} + \frac{3}{(i k r)^{3}}\right) \frac{x + i y}{r^{2}} \mathbf{r}\right) i k^{3} \frac{e^{- i k r}}{4 \pi}$$


 * where $$k = \frac{\omega}{c}$$ is the wavenumber, $$\mathbf{r} = x \mathbf{\hat{x}} + y \mathbf{\hat{y}} + z \mathbf{\hat{z}}$$ and $$r = \|\mathbf{r}\|$$ is the distance from the loop. Then the time-averaged Poynting vector (giving the electromagnetic power flow) is


 * $$\mathbf{P} = \frac{1}{2} \mathrm{Re}(\mathbf{E}\times \mathbf{H}^*) = \frac{{Z}_{0} I^{2} A^{2} k^{4}}{32 \pi^{2} r^{3}} \left( \frac{r^{2} + z^{2}}{r^{2}} \mathbf{r} + 2 \left( \frac{1}{r k} + \frac{1}{(k r)^{3}}\right) \mathbf{r}\times \mathbf{\hat{z}}\right)$$


 * and integrating the radial component of this over the sphere of all directions gives the total radiated power


 * $$P = \int _{0}^{\pi}\int _{0}^{2 \pi}(\mathbf{P}\cdot \mathbf{\hat{r}}) r \sin(\theta) \partial \phi r \partial \theta = \frac{{Z}_{0}}{6 \pi} \left( I A \left( \frac{\omega}{c}\right)^{2}\right)^{2}$$


 * Similarly, the time-averaged Maxwell stress tensor (giving the momentum flow) is


 * $$\mathbf{T} = \frac{1}{2} \mathrm{Re}\left( \varepsilon_{0} \mathbf{E} \mathbf{E}^* + \mu_{0} \mathbf{H} \mathbf{H}^* - \frac{1}{2} \mathbf{I} (\varepsilon_{0} (\mathbf{E}\cdot \mathbf{E}^*) + \mu_{0} (\mathbf{H}\cdot \mathbf{H}^*))\right) = \mathrm{complicated}$$


 * and integrating this to get the total radiated angular momentum gives


 * $$\mathbf{\tau} = \int _{0}^{\pi}\int _{0}^{2 \pi}\mathbf{r}\times (\mathbf{T}\cdot \mathbf{\hat{r}}) r \sin(\theta) \partial \phi r \partial \theta = \frac{{Z}_{0}}{6 \pi \omega} \left( I A \left( \frac{\omega}{c}\right)^{2}\right)^{2} \mathbf{\hat{z}}$$


 * But conservation of angular momentum demands a torque of $$- \mathbf{\tau}$$ on the source, and that will absorb a power of $$\omega (\mathbf{\tau}\cdot \mathbf{\hat{z}})$$ from the rotation, exactly accounting for the radiated power $$P$$, meaning that no power is derived from the current (the radiation resistance is zero) - as expected.


 * The $$c^{4}$$ in the denominator means that $$P$$ is really small for feasible $$I A$$ and $$\omega$$.


 * catslash (talk) 01:34, 19 February 2024 (UTC)