# Finding the relationship between magnetic momentum and angular momentum

• Argelium
In summary, the conversation discusses finding the relationship between magnetic momentum and angular momentum for a spinning cylinder with a distributed charge on its surface. The conversation also addresses the use of the letter "I" to represent two different things, and the need to obtain an expression for the current in terms of variables such as surface charge density and angular velocity. Additionally, the conversation mentions the need to calculate magnetic momentums for both the lateral surface area and the top and bottom of the cylinder, and then summing them to obtain the final result.
Argelium

## Homework Statement

A cylinder with radius ##R## and height ##h## which has a distributed charge on its surface with density ##\sigma## spins over its axis with angular velocity ##\omega##.

If the cylinder has a mass density ##\rho##, find the relationship between magnetic momentum and angular momentum.

## Homework Equations

$$L=I\omega$$
$$m_{mag}=IA$$

## The Attempt at a Solution

So my professor gave a hint that there should be no radius or height terms. However I attempted this:

First of all we must calculate two magnetic momentums, the one due to lateral surface area and another due to the cylinder heads.

For the first one consider a ##dI## on the surface area.

$$dI=\frac{dq \omega}{2\pi} = \frac{\omega\sigma}{2\pi}Rd\theta dz$$

Then the magnetic momentum of the sides is

$$m=\omega\sigma R^2h \int d\theta dz$$

$$m=\omega RhQ$$

Then for each heads we can find the magnetic momentum to be

$$\frac{1}{4}\omega R^2 Q$$

So summing them all (both vectors have the same direction: parallel to angular velocity) we obtain

$$m_{tot}= \omega R(hQ+\frac{RQ}{2}$$

Doing the calculations yields that

$$L=\frac{R}{2h+R}\frac{M}{Q} m$$

Where ##M## is the mass.

Am I missing something or is my work correct?

From the very start, you used the letter ## I ## to represent two different things. In the first way you used ## I ## it represents moment of inertia. ## I=\sum m_i r_i^2 ##. In the second case, it is electrical current, i.e. ## I=dq/dt ##. I think that caused some incorrect results. Suggestion is to get an expression for the current ## I ## in terms of ## \sigma##, ## \omega ##, ## R ##, and ## h ##. Then write an expression for the magnetic moment ## \vec{m}=IA \hat{z} ##. ## \\ ## You should also be able to write an expression for the angular momentum ## \vec{L} ##, by computing the moment of inertia for the cylinder, and then ## \vec{L}=I \omega \hat{z} ##, where ## I ## is the moment of inertia.

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From the very start, you used the letter ## I ## to represent two different things. In the first way you used ## I ## it represents moment of inertia. ## I=\sum m_i r_i^2 ##. In the second case, it is electrical current, i.e. ## I=dq/dt ##. I think that caused some incorrect results. Suggestion is to get an expression for the current ## I ## in terms of ## \sigma##, ## \omega ##, ## R ##, and ## h ##. Then write an expression for the magnetic moment ## \vec{m}=IA \hat{z} ##. ## \\ ## You should also be able to write an expression for the angular momentum ## \vec{L} ##, by computing the moment of inertia for the cylinder, and then ## \vec{L}=I \omega \hat{z} ##, where ## I ## is the moment of inertia.

Ok well, I'm on mobile so that's why I wasn't very detailed (and messed up the notation, the current is ##i##, whoops)

As for the expression for the current, i don't see why it is necessary since the current is already predetermined by the charge (well ,the surface charge density and the surface area of the cylinder) and the angular velocity.

So if we take a ##di## we get it in terms of surface area (see the post for the details of my procedure) and then we integrate over the whole surface area. Now, I'm not 100% sure but to do that we must take two steps: for the lateral surface area and the top disks of the cylinder.

So once we do that two calculations of magnetic momentum (the laterals and the top and bottom) we sum them (they are parallel) and the result follows.

Summing up, my main doubt is whether calculating both the laterals and the top and bottom is what must be done (I think so since we're talking about the total surface area)

Argelium said:
Ok well, I'm on mobile so that's why I wasn't very detailed (and messed up the notation, the current is ##i##, whoops)

As for the expression for the current, i don't see why it is necessary since the current is already predetermined by the charge (well ,the surface charge density and the surface area of the cylinder) and the angular velocity.

So if we take a ##di## we get it in terms of surface area (see the post for the details of my procedure) and then we integrate over the whole surface area. Now, I'm not 100% sure but to do that we must take two steps: for the lateral surface area and the top disks of the cylinder.

So once we do that two calculations of magnetic momentum (the laterals and the top and bottom) we sum them (they are parallel) and the result follows.

Summing up, my main doubt is whether calculating both the laterals and the top and bottom is what must be done (I think so since we're talking about the total surface area)
One minor, but important, oversight on my part=I overlooked the heads. For those, you need to do a ## d \vec{m}= A(r) \, \sigma \frac{2\pi r \, dr}{T } \hat{z} ##, where ## \omega=\frac{2 \pi}{T} ##, and ## A(r)=\pi r^2 ##, and integrate over ## r ##, and of course remember that there are two head surfaces. ## \\ ## Additional item: I would refrain from using the letter ## Q ##. Simply use ## \sigma ## and ## h ## and ## R ##. ## \\ ## With surface charge ## \sigma ## included on the head surfaces, I do expect the ratio you get for ## \frac{\vec{m}}{\vec{L}} ## very well might depend upon ## h ## and ## R ##. Without any surface charge on the heads, IMO, then it makes more sense to use total charge ## Q ## and total mass ##M ## and see what you get.

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A follow-in: The result for angular momentum is elementary, so I will post it here: I get ## \vec{L}=\frac{1}{2} MR^2 \omega \hat{z} ##.
Using ## Q_{total \, heads} ## and ## Q_{outer \, surface } ## as parameters, (I'm going to post a result here, because that is the only way to discuss the answer I got), I get ## \vec{m}= [Q_{total \, heads} \frac{R^2 \omega}{4}+Q_{outer \, surface} \frac{R^2 \omega}{2} ]\hat{z} ##. Using ## Q_{total}=Q_{total \, heads}+Q_{outer \, surface} ##, the ## R^2 \omega ## factors out of everything, including ## \vec{L} ##, but we can not use ## Q_{total} ## in the final expression. The problem, if the hint of the professor is going to work, must either have the charge on the heads or on the outer surface, but not both at the same time. Anyway, that's what I got. Perhaps you can either confirm, or find that I made an algebraic error.

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A follow-in: The result for angular momentum is elementary, so I will post it here: I get ## \vec{L}=\frac{1}{2} MR^2 \omega \hat{z} ##.
Using ## Q_{total \, heads} ## and ## Q_{outer \, surface } ## as parameters, (I'm going to post a result here, because that is the only way to discuss the answer I got), I get ## \vec{m}= [Q_{total \, heads} \frac{R^2 \omega}{4}+Q_{outer \, surface} \frac{R^2 \omega}{2} ]\hat{z} ##. Using ## Q_{total}=Q_{total \, heads}+Q_{outer \, surface} ##, the ## R^2 \omega ## factors out of everything, including ## \vec{L} ##, but we can not use ## Q_{total} ## in the final expression. The problem, if the hint of the professor is going to work, must either have the charge on the heads or on the outer surface, but not both at the same time. Anyway, that's what I got. Perhaps you can either confirm, or find that I made an algebraic error.

So basically what I had (and you yourself, although I have a factor of ##h## which does not appear in yours), but with the caveat that the charge in the heads is not the same charge as in the lateral (which should have been obvious). This yields a not so pretty solution. Anyways, as you said the only way the solution does not depend on the geometry of the cylinder is that there is only charge on the tops or only charge on the sides. But I think my doubts are gone! Thanks!

## 1. What is the relationship between magnetic momentum and angular momentum?

The relationship between magnetic momentum and angular momentum is that they are both related to the rotation of an object. Angular momentum is a measure of the rotational motion of an object, while magnetic momentum is a measure of the strength of a magnetic field produced by the rotation of an object.

## 2. How are magnetic momentum and angular momentum measured?

Magnetic momentum is typically measured in units of ampere-meters squared (A·m²), while angular momentum is measured in units of kilogram-meters squared per second (kg·m²/s). These units represent the strength and rotational motion of an object, respectively.

## 3. What factors affect the relationship between magnetic momentum and angular momentum?

The relationship between magnetic momentum and angular momentum can be affected by various factors such as the mass and speed of the rotating object, the strength of the magnetic field, and the distance between the object and the magnetic field.

## 4. Can the relationship between magnetic momentum and angular momentum be changed?

Yes, the relationship between magnetic momentum and angular momentum can be changed by altering any of the factors mentioned above. For example, increasing the speed of rotation or the strength of the magnetic field will result in a stronger relationship between the two.

## 5. What are the real-world applications of understanding the relationship between magnetic momentum and angular momentum?

The understanding of the relationship between magnetic momentum and angular momentum has various real-world applications in fields such as engineering, physics, and technology. It is crucial in the design and operation of motors, generators, and other electrical devices that rely on the interaction between magnetic fields and rotational motion.

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