Potential difference across the ends of a solenoid

[songoku]

Homework Statement

A disc of area $A$ rotates at angular frequency $\omega$ at the center of long solenoid of length L and having N turns, carrying current I. The plane of the disc is normal to magnetic flux. The rotation rate is adjusted so that the emf generated between the center of the disc and its rim is 10% of the potential difference across the ends of the solenoid. Which expression gives the potential difference across the ends of the solenoid.
a) $10 \mu_0 NIA \omega$
b) $1.6 \mu_0 NIA \omega / L$
c) $0.1 \mu_0 NIA \omega$
d) $0.016 \mu_0 NIA \omega /L$

Relevant Equations

B at center of solenoid = $\mu_0 NI / L$

$E = - \frac{d \phi}{dt }$

Emf between the center of the disc and its rim
= $- \frac{d \phi}{dt }$
= $-B.A \frac{d \cos (\omega t)}{dt}$
=$\mu_0 \frac{N}{L} I A \omega \sin (\omega t)$

Potential difference across the ends of the solenoid = $10\mu_0 \frac{N}{L} I A \omega \sin (\omega t)$

Is this correct? How to continue until I get the answer on the options?

Thanks

 

[haruspex]

I don't understand how you get the trig terms. There is nothing in the question that varies with time or angle.
Hint: consider a small radial element of the disc of length dr. What is the p.d. along it?

 

[Delta2]

Read about Faraday Homopolar generator to find out how to calculate EMF of the disc.

 

[songoku]

I don't understand how you get the trig terms. There is nothing in the question that varies with time or angle.

I use: $\phi = BA \cos \theta$ and I imagine $\theta$ will change as the disc rotates. Or maybe from the question " The plane of the disc is normal to magnetic flux" means that $\theta$ is constant? If $\theta$ is constant, my imagination about the direction rotation is wrong.

I imagine it like this: the plane of the disc is vertical and it rotates with vertical axis of rotation through the center of the disc. Or maybe the rotation is just like the rotation of bicycle wheel (with the plane always perpendicular to magnetic field produced by solenoid)?

What change causes the change in magnetic flux (to produced induced emf)? I thought the only possible change is $\theta$ since the value of B and A are constant. If the rotation of the disc is just like bicycle wheel, wouldn't the magnetic flux on the disc be constant, hence no emf?

Hint: consider a small radial element of the disc of length dr. What is the p.d. along it?

I need to use integration to solve this problem? I am not sure what is the p.d. along a small radial element of the disc since I have no idea what causes the emf to be produced.

Thanks

 

[songoku]

Read about Faraday Homopolar generator to find out how to calculate EMF of the disc.

I will google it

 

[haruspex]

the plane always perpendicular to magnetic field produced by solenoid)?

Yes.

the magnetic flux on the disc be constant, hence no emf?

See the answer at https://physics.stackexchange.com/questions/466143/emf-induced-in-a-rotating-disc, where it points out the difference between electric Lorentz force and magnetic Lorentz force.

I am not sure what is the p.d. along a small radial element of the disc since I have no idea what causes the emf to be produced.

Just think of it as a wire moving across the field. If it is at radius r from the axis of rotation and pointing away from the axis, how fast is it moving across the field?

 

[Delta2]

If the rotation of the disc is just like bicycle wheel, wouldn't the magnetic flux on the disc be constant, hence no emf

Yes the rotation is like a bicycle wheel with the magnetic field perpendicular to the plane of the wheel. Seemingly it violates Faraday's law of induction.

To apply Faraday's law correctly, consider a radius of the "bicycle wheel" and calculate how much surface area S it spans in time t as this radius rotates with constant angular velocity $\omega$. Then calculate $\frac{dS}{dt}$ and the emf will be $$\mathcal{E}=\frac{d\Phi}{dt}=\mathbf{B}\frac{dS}{dt}$$.

The other approach is that of @haruspex where you go by Lorentz force law and integration of infinitesimal EMFs $Bvdr=B\omega rdr$.

 

[haruspex]

I need to use integration to solve this problem?

As @Delta2 notes, thinking in terms of area swept out by a complete radius does the integration for you.

 

[Delta2]

As @Delta2 notes, thinking in terms of area swept out by a complete radius does the integration for you.

My approach involves two integrations actually.

• One with respect to the spatial variables to find the area of a cyclic section of angle $\phi$ and radius $R$, which of course is already well known and the result is $\frac{\phi}{2}R^2$, and
• One integration with respect to the time variable to find the angle $\phi$. But the latter integration is going to be "neutralized" when we take the time derivative of the flux, so no actual integrations needed at all.

 

[songoku]

Just think of it as a wire moving across the field. If it is at radius r from the axis of rotation and pointing away from the axis, how fast is it moving across the field?

speed = $\omega$ . r ?

My approach involves two integrations actually.

• One with respect to the spatial variables to find the area of a cyclic section of angle $\phi$ and radius $R$, which of course is already well known and the result is $\frac{\phi}{2}R^2$, and
• One integration with respect to the time variable to find the angle $\phi$. But the latter integration is going to be "neutralized" when we take the time derivative of the flux, so no actual integrations needed at all.

I think I understand integration of first part but I don't understand the second part. I thought there is only one integration with respect to $\phi$ to find the area swept by "radius of bicycle wheel" and after that we take the time derivative to obtain emf induced.

Thanks

 

[Delta2]

The integration of the second part is more easy than you think, it is just that $$\phi=\int \omega dt$$ because by the definition of $\omega$ it is $$\omega=\frac{d\phi}{dt}$$ , what do you think?
In this example $\omega$ is constant in time so things are even more simple.

 

[songoku]

The integration of the second part is more easy than you think, it is just that $$\phi=\int \omega dt$$ because by the definition of $\omega$ it is $$\omega=\frac{d\phi}{dt}$$ , what do you think?
In this example $\omega$ is constant in time so things are even more simple.

What I had in mind is like this:

$$\phi = \int B.dA = B \int dA = \frac 1 2 Br^2 \phi$$

Then:

$$E=-\frac{d\phi}{dt}=-\frac{d (\frac 1 2 Br^2 \phi)}{dt}=-\frac 1 2 Br^2 \omega$$

How to apply the second integration in my working? Thanks

 

[Delta2]

You are right, no need to do second integration e hehe.

Btw to correct a small typo in your work, you use the same letter $\phi$ for the flux and the angle. You should use capital $\Phi$ for the flux.

 

[songoku]

Thank you very much haruspex and Delta2. Happy new year

 

[Delta2]

Happy New Year to you too!