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Given is the voltage U and the Angular velocity $\omega$. The rotating cylinder is inside the other cylinder and U is measured between them. To calculate the magnetic field, I tried to solve

$\oint B ds = \mu_0 I$.

I chose to integrate along a rectangle (AB,BC,CD,DA) so that the two sides (BC,DA) of the rectangle are going to be 0 since B ⦝ BC/DA. The length AB goes through the middle of the rotating cylinder and CD is 0 because it's far away..

With $ I = Q*\nu = 2 \pi \omega Q$

I would get the result $B = \frac{2 \pi \omega Q}{AB}$.

Despite this could be the correct result, I'm not really happy with it because the charge isn't given. I also haven't used the voltage U.

Can somebody please help me findig the correct result by giving me a hint how to calculate Q?

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  • $\begingroup$ A diagram will be very helpful. To calculate Q, I can see that you will need to use U and the radius of the two cylinders, treating them as making a capacitor. Find out the capacitance and use U=Q/C. $\endgroup$ – verdelite Sep 11 '18 at 14:08
  • $\begingroup$ I have allready tried this but since the radius is not given, I don't know how to calculate the capacitance. C would be $ 2 \pi \epsilon_0 \frac{l}{ln\frac{R2}{R1}}$. $\endgroup$ – gamma Sep 11 '18 at 14:18
  • $\begingroup$ Is it possible to solve this if I say the cylinder is an inductor. So that would mean that $U_{ind} = -n * \partial_t\Phi$ and with $\Phi = 2 \pi r l B$ I would get $U_{ind} = -n \mu 4 \pi ^2 r Q \partial_t \omega$. I now could calculate Q but since $\omega$ is constant that would mean $Q = 0$. $\endgroup$ – gamma Sep 12 '18 at 8:13

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