# Finding the inductance of a rotating cylinder shell

• Eitan Levy
In summary: If you construct the standard rectangular Amperian loop that encloses all of this current and has one...The inductance is the same as the flux multiplied by the length of the loop.
Eitan Levy
Homework Statement
An infinite cylinder shell with a radius equal to R is rotating with a radial velocity ω around an axis located in its center. The density of the charge on the shell is σ. Find the inductance of the cylinder per unit-length.
Relevant Equations
I=dq/dt
0.5LI^2=U
First, the correct answer is μ0*π*R^2.

I tried to look at the cylinder like it was a solenoid, this technique was used in my class.

Then I tried to find the current of the solenoid, to do that I looked at a piece of a solenoid with a legnth of dz, then:

I=dq/dt=(2πRσ*dz)/(2π/ω)=ω*R*σ*dz.

The density of the loops in that case would be n=1/dz. Therefore the magnetic field inside is B=μ0*ω*R*σ.

The problematic part was to find the inductance. Both ways I tried failed.

First, I tried to use L=(magnetic flux)/(current).

The magnetic flux, if we look at a piece with a length equals to L, would be μ0*ω*σ*π*R^3*(L/dz). But dividing this by the current gives a very off answer and I can't get rid of dz.

Another approach was to use U=0.5LI^2 but it suffers from the same problem.

Obviously I am doing something very wrong, but can't understand what exactly. Can anyone help please?

What is wrong is the structure of this equation
Eitan Levy said:
I=dq/dt=(2πRσ*dz)/(2π/ω)=ω*R*σ*dz.
It violates the rule that if you have d[something] on one side of an equation, you should have d[something else] on the other side. Better write the "current per unit length" as dI/dz = ω*R*σ and then divide the flux per unit length by this.

etotheipi and Delta2
kuruman said:
What is wrong is the structure of this equation

It violates the rule that if you have d[something] on one side of an equation, you should have d[something else] on the other side. Better write the "current per unit length" as dI/dz = ω*R*σ and then divide the flux per unit length by this.

My professor used the exact same equation and that's how he wrote it. Not trying to disregard you of course but I prefer to stay with this writing. Still even with your alternative I can't understand how to solve the problem.

OK, I see where your professor is coming from. You need to write an expression for the magnetic field inside the solenoid, then find the flux and from that the inductance. You know that the field inside a very long solenoid is ##B=\mu_0 n I## where ##n## is the number of turns per unit length and ##I## the current in one turn. How can you adapt this expression to this particular problem? Hint: ##nI## is the current per unit length. Alternatively, you can use Ampere's law to derive the magnetic field inside the cylinder for this case.

kuruman said:
OK, I see where your professor is coming from. You need to write an expression for the magnetic field inside the solenoid, then find the flux and from that the inductance. You know that the field inside a very long solenoid is ##B=\mu_0 n I## where ##n## is the number of turns per unit length and ##I## the current in one turn. How can you adapt this expression to this particular problem? Hint: ##nI## is the current per unit length. Alternatively, you can use Ampere's law to derive the magnetic field inside the cylinder for this case.
What is wrong with the expression I got? I am pretty sure it is correct.

Eitan Levy said:
What is wrong with the expression I got? I am pretty sure it is correct.
If you are pretty sure it is correct why did you say that it "gives a very off answer and I can't get rid of dz"?

kuruman said:
If you are pretty sure it is correct why did you say that it "gives a very off answer and I can't get rid of dz"?
Haha, valid question. I think that the expression for the magnetic field is correct because I have used this technique many times. Do you think that the expression for the magnetic field is not correct? Is nI not equal to ω*R*σ?

Edit: It is also the value that appears in the formula sheet for this exact problem for B, which makes me think it's not the issue. The issue is probably me trying to apply it wrongly to the inductance formulas.

Edit 2: What I am getting as an answer, by trying to work with both energy and with the formula L=(magnetic flux)/(current), is L= (μ0*π*l*R^2)/(dz^2), which oddly enough yields the requested answer if I plug dz=l.

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##dz## and ##l## are the same and represent the length of a cylinder segment. Using the energy seems to be the most straightforward way to proceed. The charge on an area element ##dA## on the surface of the cylinder is ##dq=\sigma dA=\sigma ~R~ d\phi~ l##. Here ##l## is the length of the element along the axis. The current is $$\frac{dq}{dt}=\sigma ~R~ \frac{d\phi}{dt}~ l=\sigma~ R~ \omega~ l.$$For what you want to do all that is not needed, but I included it for completeness. As you know, the inductance is a geometric property and does not depend on the current.

If you construct the standard rectangular Amperian loop that encloses all of this current and has one side inside and the other outside the cylinder, Ampere's law says $$B~l=\mu_0 \frac{dq}{dt}~\Rightarrow~\frac{B}{dq/dt}=\frac{\mu_0}{l}.$$If you do follow the energy method, then $$\frac{1}{2}\frac{B^2}{\mu_0}\pi R^2 l=\frac{1}{2}L\left(\frac{dq}{dt}\right)^2.$$Use the last equation to solve for the inductance ##L## and incorporate the result from the previous equation.

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kuruman said:
##dz## and ##l## are the same and represent the length of a cylinder segment. Using the energy seems to be the most straightforward way to proceed. The charge on an area element ##dA## on the surface of the cylinder is ##dq=\sigma dA=\sigma ~R~ d\phi~ l##. Here ##l## is the length of the element along the axis. The current is $$\frac{dq}{dt}=\sigma ~R~ \frac{d\phi}{dt}~ l=\sigma~ R~ \omega~ l.$$If you construct the standard rectangular Amperian loop that encloses all of this current and has one side inside and the other outside the cylinder, Ampere's law says $$B~l=\mu_0 \frac{dq}{dt}~\Rightarrow~\frac{B}{dq/dt}=\frac{\mu_0}{l}.$$If you do follow the energy method, then $$\frac{1}{2}\frac{B^2}{\mu_0}\pi R^2 l=\frac{1}{2}L\left(\frac{dq}{dt}\right)^2.$$Use the last equation to solve for the inductance ##L## and incorporate the result from the previous equation.

I can't understand why l and dz are the same. dz is supposed to be a length so small so that there will be only a single loop in it. l may supposedly be anything. Have I misinterpreted the meaning of dz?

I understand how to get to the solution if l=dz, but can't understand why would that be the case.

Also, thank you so much for your help.

Eitan Levy said:
I can't understand why l and dz are the same. dz is supposed to be a length so small so that there will be only a single loop in it. l may supposedly be anything. Have I misinterpreted the meaning of dz?
You don't even need ##dz## if you go the energy way. Read and understand the second part in post #8 that starts with "If you construct the standard Amperian loop ##\dots##" The energy method allows you to bypass messing with the current. The inductance is a geometric property that does not depend on the current.

kuruman said:
You don't even need ##dz## if you go the energy way. Read and understand the second part in post #8 that starts with "If you construct the standard Amperian loop ##\dots##"
I understand I can escape from dz, but still would like to understand why l=dz is a valid substitution. My professor loves the dz technique, this is why I am trying to understand it better.

Eitan Levy said:
I understand I can escape from dz, but still would like to understand why l=dz is a valid substitution. My professor loves the dz technique, this is why I am trying to understand it better.
OK, in your own words explain to me what ##dz## and ##l## stand for in regards to this particular problem and your method for solving it. Remember that you have an infinite cylinder.

kuruman said:
OK, in your own words explain to me what ##dz## and ##l## stand for in regards to this particular problem and your method for solving it. Remember that you have an infinite cylinder.

Because this is not a real solenoid, but an infinite cylinder which is similar to one, there aren't really any "loops". Is this the reason I could sub any length for dz and still remain with a correct answer? Because dz is actually an arbitrary size?

In my mind dz could only be a very small size, is that not the case?

Eitan Levy said:
In my mind dz could only be a very small size, is that not the case?
Small relative to what along the same dimension? The radius doesn't count.

I think you've discovered that there is an ambiguity in how the "inductance per unit length" is defined for the rotating cylinder.

First, consider a long, ideal solenoid rather than the rotating cylinder. The energy formula is ##U = \frac 1 2 L I^2_{wire}##, where ##I_{wire}## is the current in the wire of the solenoid. Divide both sides by the length ##l## of the solenoid,

##U/l = \frac 1 2 (L/l) I^2_{wire}##.

Or

##\mathscr{U} = \frac 1 2 \mathscr{L} I^2_{wire}##

where ##\mathscr{U}, \mathscr{L}## are the energy per unit length and the inductance per unit length, respectively. Note that the current is still ##I_{wire}##, which is not equal to the current per unit length of the solenoid.

If you try to carry this over to the rotating cylinder, there is a problem with interpreting ##I_{wire}##. If you arbitrarily take a small length ##dz## of the cylinder to represent the "size of the wire", then ##I_{wire}## is proportional to ##dz##. Thus, solving the energy equation for ##\mathscr{L}## yields a result that is inversely proportional to ##dz^2##. This is not very satisfying, since it makes the result dependent on the arbitrary choice of ##dz##.

As you noticed, their answer corresponds to taking ##dz## to be a unit length of the cylinder. This is equivalent to defining ##\mathscr{L}## for the cylinder such that

##\mathscr{U} = \frac 1 2 \mathscr{L} \mathscr{I}^2##

where ##\mathscr{I}## is the current in a unit length of the cylinder. [Edit: See comment * below regarding the units of ##\mathscr I##.] There is a certain naturalness in defining ##\mathscr{L}## this way. (I think @kuruman is essentially defining it this way in post #8.) But, this definition no longer corresponds directly to the ideal solenoid: ##\mathscr{U} = \frac 1 2 \mathscr{L} I^2_{wire} = \frac 1 2\mathscr{L} \mathscr{I}^2/n^2##, where ##n## is the number of turns in a unit length of the solenoid.

The problem statement was not clear in the meaning of "inductance per unit length" for the rotating cylinder.

-------
*[comment] ##\mathscr I## should be thought of as the amount of current in a unit length of the cylinder rather than the current per unit length. The difference is in the units. The current in a unit length would be in amps, while the current per unit length would be in amps/meter.

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Eitan Levy

## 1. What is the purpose of finding the inductance of a rotating cylinder shell?

The inductance of a rotating cylinder shell is important in understanding the behavior of electromagnetic fields in rotating systems. It is also used in the design and analysis of rotating electrical machines, such as motors and generators.

## 2. How is the inductance of a rotating cylinder shell calculated?

The inductance of a rotating cylinder shell can be calculated using the formula L = μ0N^2A/d, where μ0 is the permeability of free space, N is the number of turns, A is the cross-sectional area of the cylinder, and d is the length of the cylinder.

## 3. What factors affect the inductance of a rotating cylinder shell?

The inductance of a rotating cylinder shell is affected by the permeability of the material, the number of turns, the cross-sectional area, and the length of the cylinder. The speed of rotation and the frequency of the applied voltage also play a role.

## 4. How does the inductance of a rotating cylinder shell change with rotation speed?

The inductance of a rotating cylinder shell increases with rotation speed due to the increased magnetic field strength and flux density. However, at very high speeds, the inductance may start to decrease due to the skin effect.

## 5. Can the inductance of a rotating cylinder shell be measured experimentally?

Yes, the inductance of a rotating cylinder shell can be measured experimentally using techniques such as LCR meters or bridge circuits. However, the accuracy of the measurement may be affected by factors such as stray capacitance and eddy currents.

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