Suppose I have a constant magnetic field, and I switch it off (abruptly or slowly) , a ring of uniform charge would acquire same angular velocity if it was present in a plane perpendicular to B and assumed to be on a desk. Who supplied this kinetic energy? A more important case is a simply constantly increasing magnetic field, the ring's angular velocity would keep on increasing, is there any difference in field structure or in something else if the ring was not present, if not then again, from where did this energy come?
5
-
$\begingroup$ A magnetic field contains energy. Switching it off (or increasing it) means that work has to be performed on or by something. That's why we use inductors in electronics and electrical engineering, they are fabulous short term energy storage devices from the second to the ps range. $\endgroup$– CuriousOneCommented Mar 4, 2016 at 8:55
-
$\begingroup$ Ok, increasing or decreasing magnetic field supposedly by a line charge would require battery, accepted. problem is that the energy acquired by ring ,where was this when there was no ring, where was it dissipated? $\endgroup$– MrigankCommented Mar 4, 2016 at 9:03
-
$\begingroup$ Capacitors and inductors store energy. Capacitors in an electric field and inductors in a magnetic field. For many it just seems harder to visualise the storage in the magnetic case than in the electric case possibly because the opposite charges producing an electric field are easier to picture? $\endgroup$– FarcherCommented Mar 4, 2016 at 9:04
-
$\begingroup$ It's either dissipated in the switch or as an electromagnetic wave or both. You can design experiments that optimize either. A typical application is a car's ignition coil, by the way. The spark is generated by the voltage buildup on a coil that has its current flow rapidly disrupted. $\endgroup$– CuriousOneCommented Mar 4, 2016 at 9:10
-
$\begingroup$ So, energy dissipated in switch or EM wave decreases when we put ring there? $\endgroup$– MrigankCommented Mar 4, 2016 at 9:45
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Think about how you would decrease a magnetic field.
One option is that you could be inside a super large solenoid with a steady current and you build a smaller solenoid and run current the opposite direction. In order to run the current you need to supply an electric field that circulates around the wires. And as that current ramps up, then Jefimenko's equations:
$$\vec E(\vec r,t)=\frac{1}{4\pi\epsilon_0}\int\left[\frac{\rho(\vec r',t_r)}{|\vec r -\vec r'|}+\frac{\partial \rho(\vec r',t_r)}{c\partial t}\right]\frac{\vec r -\vec r'}{|\vec r -\vec r'|^2}\; \mathrm{d}^3\vec{r}' -\frac{1}{4\pi\epsilon_0c^2}\int\frac{1}{|\vec r-\vec r'|}\frac{\partial \vec J(\vec r',t_r)}{\partial t}\mathbb{d}^3\vec r'$$ and $$\vec B(\vec r,t)=\frac{\mu_0}{4\pi}\int\left[\frac{\vec J(\vec r',t_r)}{|\vec r -\vec r'|^3}+\frac{1}{|\vec r -\vec r'|^2}\frac{\partial \vec J(\vec r',t_r)}{c\partial t}\right]\times(\vec r -\vec r')\mathbb{d}^3\vec r'$$ (where $t_r$ is actually a function of $\vec r'$, specifically $t_r=t-\frac{|\vec r-\vec r'|}{c}$) show an electric and a magnetic field tracing itself back to that changing current.
Or you can even just look at the fact that electric fields and magnetic fields have energy and since charged particles gain $\vec J\cdot \vec E$ then the fields lose $\vec J\cdot \vec E.$ And we also know how magnetic fields change:
$$\frac{\partial \vec B}{\partial t}=-\vec \nabla \times \vec E.$$
So magnetic fields at a point only change in time when there is a circulating electric field at that same point. Every single one of these arguments point to the same fact. You cannot change a magnetic field over time without electric fields.
When you put energy into magnetic fields or take it out of magentic fields, both cases require electric fields. Those electric fields might be in empty space. Or they might be where wires are located and thus a current might start to flow.
