It's obvious that a metallic ring rotating in a magnetic field will have an emf $\epsilon$ induced across it. using a combination of the centripetal force and the Lorentz force, this makes sesne. BUt I'm unable to explain it using faraday's alw. The only idea I can think of is drawing an imaginary radius vector from the centre to the ring, and saying that the area swept out by that changes with time.
$\begingroup$
$\endgroup$
4
asked Oct 18, 2022 at 13:16
-
$\begingroup$ Can you explain what axis the ring is rotating around, and what direction the magnetic field points in? Is the magnetic field static? $\endgroup$– Brian BiCommented Oct 18, 2022 at 22:23
-
$\begingroup$ My bad. The field is static and into the screen, the ring rotates clockwise on the plane of teh screen. $\endgroup$– math and physics foreverCommented Oct 19, 2022 at 13:58
-
$\begingroup$ And the ring lies in the plane of the page? $\endgroup$– Brian BiCommented Oct 19, 2022 at 15:46
-
$\begingroup$ yes, it does... $\endgroup$– math and physics foreverCommented Oct 19, 2022 at 16:05
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
In the OP's setup, with the ring in the plane of the page and rotating clockwise, and the magnetic field pointing into the page, a unit positive test charge in the ring will experience a magnetic force given by the product of its velocity, which is tangential to the ring, and the magnetic field. This magnetic force will point radially outward. For a negative charge, the force will be radially inward.
There will be no emf. The emf around the ring is given by
\begin{equation} \mathcal{E} = \int (E + v \times B) \cdot \mathrm{d}s \end{equation}
where the integral is taken along a given path (e.g. a loop that coincides with the ring) and $v$ is the velocity of the wire at the point where the integrand is evaluated. Since $v \times B$ is perpendicular to $\mathrm{d}s$, the integrand vanishes.
This result, $\mathcal{E} = 0$, coincides with that given by Faraday's law.
Note that the "centripetal force" mentioned by the OP is irrelevant; the charges in the ring must accelerate in the radial direction in order to stay in the ring, but the Lorentz force only depends on velocity.
$\begingroup$
$\endgroup$
in Faraday's law the area is a vector, so the area changes fom parallel to B to perpendicular to B.
