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Rotating and tilting charged disk induces a voltage inside a ring

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Homework Statement
Approximately calculate the voltage U(t) inside a conductive ring.
Homework Equations
none were given
Screenshot from 2019-10-09 15-16-02.png


As I`` m learning for an upcoming exam I found an electrodynamics problem I struggle with.
In the first task I need to calculate the magnetic dipole moment of a uniformly charged,thin disk with the Radius R and a total charge Q which rotates with a angular speed omega round its symmetry axis. Which i did in zylinder coordinates:
$$\vec m = \frac{1}{2} \int \vec r \times \vec j(\vec r) d^3 r= \frac{Q}{2 \pi} \int \vec r \times (\vec \omega \times \vec r) d^3 r = \frac{Q \omega R^2}{4} \vec e_z$$
so far so good.
The next task is to take this exact disk and rotate it slowly by the angle theta(t) around the x-axis. There is a conducting ring with the radius r (r<<L) in the xy-plane on the y-axis. It's distance to the disk is L and L>>R. The second task is to approximately calculate the voltage U(t) in the ring.
I thought of calculating the B-field in the z-direction first without an angle. $$\vec B (\vec r) = \frac{\mu_0}{4 \pi} \frac{3 \vec r (\vec m \cdot \vec r ) - \vec m |r|^2}{|r|^5}=\frac{-\mu_0 Q \omega R^2}{16 \pi L^3} \vec e_z$$
The problem begins here as I don't know if I can just add a cos(theta(t)) to the equation to get the right magnetic field that acts on the conductive ring. And then go on with calculating the magnetic flux with $$\Phi = \int_A \vec B \cdot d\vec A$$ and the voltage afterwards with $$U=\frac{d\Phi}{dt}$$ or if I got something wrong by thinking it's that easy. I would be very pleased if someone could help me in that regard.
 
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Hi. Say
[tex]\mathbf{m}=m\mathbf{e_z}[/tex]
[tex]3\mathbf{r}(\mathbf{m}\cdot\mathbf{r})-\mathbf{m}|r|^2=3mz\mathbf{r}-m\mathbf{e_z}|r|^2[/tex]
[tex]=3mzx\mathbf{e_x}+3mzy\mathbf{e_y}-m(x^2+y^2+2z^2)\mathbf{e_z}[/tex]
where position vector r is originated from the disk center.
At the posion of the ring
[tex]x=0,y=L\ cos(-\theta), z=L \ sin(-\theta)[/tex]
and area vector is
[tex] S(0,cos(\pi/2-\theta),sin(\pi/2-\theta))[/tex]
Here I take the disk is at still but the ring rotates for easiness for me.
Do such expressions may help you ?
 

vanhees71

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Hm, at the first glance this looks like a tough time-dependent problem. For not too large ##L## you can use the quasistatic approximation though.
 
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Thanks for your answers. I will think of your tips and try to work on with the problem.
 

TSny

Homework Helper
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##\vec B(\vec r) ## can be expressed nicely in spherical coordinates with unit vectors ##\hat r## and ##\hat \theta##. Decide if you need the ##\hat r## component or the ##\hat \theta## component.

 
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[tex]=3mzx\mathbf{e_x}+3mzy\mathbf{e_y}-m(x^2+y^2+2z^2)\mathbf{e_z}[/tex]
Correction of a sign
[tex]=3mzx\mathbf{e_x}+3mzy\mathbf{e_y}-m(x^2+y^2-2z^2)\mathbf{e_z}[/tex]
 

Delta2

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Hm, at the first glance this looks like a tough time-dependent problem. For not too large ##L## you can use the quasistatic approximation though.
I believe in order for the quasistatic approximation to be valid we also need that the tilting is done with small angular velocity/acceleration, otherwise the radiation term will not be neglicible.
 

vanhees71

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Argh! I didn't get that the tilting is also time dependent. Then it's of course even more complicated. Of course where the quasistatic approximation is valid depends on the largest "frequency" in the problem, i.e., the corresponding wavelength must be larger than the distance from the system considered.
 

Delta2

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Argh! I didn't get that the tilting is also time dependent. Then it's of course even more complicated. Of course where the quasistatic approximation is valid depends on the largest "frequency" in the problem, i.e., the corresponding wavelength must be larger than the distance from the system considered.
Well now that I reread the statement, it says that
Homework Statement: Approximately calculate the voltage U(t) inside a conductive ring.
Homework Equations: none were given

The next task is to take this exact disk and rotate it slowly by the angle theta(t) around the x-axis.
so the tilting is done slowly, I guess it means slowly enough so that the quasistatic approximation is valid.
 
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Assuming the disk and the pick up loop lie on the same plane with parallel axis of symmetry when ##\theta=0##, I suggest you use the law of Biot and Savart to compute the B field at the center of the pick up loop:$$dB(r)=\frac {\mu_0}{4 \pi}\frac {dI(r') \times {(\hat{r}-\hat{r'})}}{|r-r'|^2}$$ where r is the coordinate of the B field and r' is the coordinate of the current element. You can derive or it is shown here http://pleclair.ua.edu/ph126/Homework/HW5_SOLN.pdf that$$dI(r')=\sigma \omega r'dr'$$where ##\sigma## is the charge density. Therefor I get (omitting some algebra)$$B(L)=\frac {\mu_0 \sigma \omega}{4 \pi} \int_0^R \frac {r'dr'}{(L-2r')^2}=\frac {\mu_0 \sigma \omega}{16 \pi}(\frac{2R}{L-2R}+log(\frac{L}{L-2R}) \hat { z}$$
The disk and the loop are co_planar so the field lines threading the loop are perpendicular to the surface area of the loop and we assume that there is an approximately uniform distribution across the area. If we advance theta by ##\frac {\pi}{2}##, the field lines are parallel to the surface area of the loop meaning that the flux through the loop is zero. We write the flux (with A being the surface area) $$ \Phi= A \frac {\mu_0 \sigma \omega}{16 \pi}(\frac{2R}{L-2R}+log(\frac{L}{L-2R})cos(\theta(t))$$and thus$$e(t)=\frac{d \Phi}{dt}=A \frac {\mu_0 \sigma \omega}{16 \pi}(\frac{2R}{L-2R}+log(\frac{L}{L-2R})sin(\dot{\theta})$$
 

TSny

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I thought of calculating the B-field in the z-direction first without an angle. $$\vec B (\vec r) = \frac{\mu_0}{4 \pi} \frac{3 \vec r (\vec m \cdot \vec r ) - \vec m |r|^2}{|r|^5}=\frac{-\mu_0 Q \omega R^2}{16 \pi L^3} \vec e_z$$
The problem begins here as I don't know if I can just add a cos(theta(t)) to the equation to get the right magnetic field that acts on the conductive ring.
One way to justify that you can "just add a ##\cos \vartheta##" to take into account the tilt of ##\vec m## is to work in a spherical coordinate system.

Wikipedia gives the field of the dipole in spherical coordinates as
1570928809131.png

Here ##\theta = 0## corresponds to being on the axis of the dipole. So, the coordinate system is oriented as shown below when ##\vec m## is oriented as shown. O is the origin of the coordinate system. The azimuthal angle ##\phi## of the coordinate system is not relevant to this problem and is not shown.

1570929394704.png


Note that ##\theta## here is not the angle ##\vartheta## given in the figure of the problem statement. But there is a simple relation between the two angles. From this, it is easy to get an expression for the component of ##\vec B## that is perpendicular to the ring as a function of the tilt angle ##\vartheta## of ##\vec m##. This will justify your "just add a ##\cos \vartheta##".

It should then be easy to get an expression for the magnetic flux ##\Phi## through the ring as a function of ##\vartheta## under the assumption that the ring is small enough that the B field can be taken to be uniform over the area of the ring.
 
Last edited:
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I feel compelled to correct an error in my previous post. The answer should actually read$$e(t)=\frac{A \dot {\theta} \omega \rho \mu_0 }{16 \pi}(\frac {2R}{L-2R}+ log(\frac {L}{L-2R}))sin(\dot {\theta}t)$$
 

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