# Magnetic field inside and outside of a magnetized cylinder

## Homework Statement

An infinitely long cylinder of radius R carries a "frozen-in" magnetization, parallel to the axis, ## \vec M = ks \hat z ##. There are no free currents. Find the magnetic field inside and outside the cylinder by two different methods:

(a) Locate all the bound currents and calculate the field they produce.

(b) Use Ampere's law (in the form of 6.20) to find ## \vec H ##, and then get ## \vec B ## from 6.18.

## Homework Equations

(6.20) $$\oint \vec H \cdot d \vec l = I_{f, enc}$$

(6.18) $$\vec H = \frac{1}{\mu_{0}} \vec B - \vec M$$

(*) $$\vec J_{B} = \nabla \times \vec M$$

(**) $$\vec K_{B} = \vec M \times \hat n$$

## The Attempt at a Solution

Let s be the distance from the z axis. Then using the form of the curl for cylindrical coordinates,
$$\vec J = [ \frac {\partial } {\partial s} (ks) ] \hat \phi = -k \hat \phi,$$
and
$$\vec K = ks \hat \phi.$$

Since bound currents are circulating on the surface, the object behaves somewhat like a solenoid, so the magnetic field due to the surface current is

$$\vec B = \mu_{0} nI, I = KL \rightarrow \vec B = \mu_{0} n(kR)(2\pi R)$$

I am not very sure about how to handle the volume current, but I think it should act like concentric solenoids, so I integrated from 0 to R, with I = JA

$$- k \mu_{0} n \int_0^R \pi s^2 \, ds = -\frac{1}{3} k \mu_{0} n R^3$$

And like a solenoid the magnetic field outside is zero.

But when I do part (b), 6.20 and 6.18 imply that ## \vec B = \mu_{0} \vec M , ## which is clearly not the same as what I got for part (a). Where did I go wrong?

TSny
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Let s be the distance from the z axis. Then using the form of the curl for cylindrical coordinates,
$$\vec J = [ \frac {\partial } {\partial s} (ks) ] \hat \phi = -k \hat \phi,$$
and
$$\vec K = ks \hat \phi.$$
This looks correct except for the sign. Check the formula for curl in cylindrical coordinates.

Since bound currents are circulating on the surface, the object behaves somewhat like a solenoid, so the magnetic field due to the surface current is

$$\vec B = \mu_{0} nI, I = KL \rightarrow \vec B = \mu_{0} n(kR)(2\pi R)$$
In the solenoid formula ## \vec B = \mu_{0} nI, ## what does ##n## represent and what does ##I## represent? Thus, what does ##nI## represent? Your result for B should not include ##n##.

I am not very sure about how to handle the volume current, but I think it should act like concentric solenoids
Yes.

so I integrated from 0 to R
You want to find B at an arbitrary value of s. So, you need to think about your range of integration.

with I = JA

$$- k \mu_{0} n \int_0^R \pi s^2 \, ds = -\frac{1}{3} k \mu_{0} n R^3$$
Again, ##n## should not appear in the result. You will need to think about what to use for ##nI## for one of your concentric solenoids.

And like a solenoid the magnetic field outside is zero.
Yes, the field outside any concentric solenoid is zero. This is helpful in setting up the limits of integration for summing the contributions from the concentric solenoids.

In the solenoid formula ⃗B=μ0nI,B→=μ0nI, \vec B = \mu_{0} nI, what does nnn represent and what does III represent? Thus, what does nInInI represent? Your result for B should not include nnn.
n is the turns per unit length, N/L. If I = KL then I*n = KL*(N/L) = KN. But I'm not sure how to interpret this since the cylinder is infinite.

TSny
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Gold Member
n is the turns per unit length, N/L. If I = KL then I*n = KL*(N/L) = KN. But I'm not sure how to interpret this since the cylinder is infinite.
Yes, in ## \vec B = \mu_{0} nI##, ##n## is the number of turns per unit length.

How would you describe the meaning of ##I## in the formula? How would you describe the meaning of ##nI##?

I is the current in one of the turns, so nI is the current per unit length? Then is that just equal to K?

TSny
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I is the current in one of the turns, so nI is the current per unit length? Then is that just equal to K?
Yes.

So, since J is dI/dA, does nI = J/A?

TSny
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Gold Member
So, since J is dI/dA,
Can you describe the shape and orientation of your area dA?

does nI = J/A?
On the left you have current per unit length. On the right, you have current per unit area divided by an area. Can't be.

(I've got to turn in for the night. Will check back tomorrow.)

Or since I'm integrating, would I just say J = nI in my integral (because it is "acting" like a surface current in the integrand)?

Good night.

TSny
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Gold Member
##J## is a current per unit area. ##nI## is a current per unit length.

For a concentric cylindrical shell of thickness ##ds##, think about how to express the current per unit length of the cylinder in terms of ##J## and ##ds##.

If the thickness is ds, then the cross sectional area is ds * horizontal length, so J = current per unit length * ds = nIds? Which makes my integral evaluate to ##\mu_{0}ks,## so the sum of the fields due to J and K is ##2\mu_{0}ks.## So my result from 6.18 still doesn't quite match.

TSny
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If the thickness is ds, then the cross sectional area is ds * horizontal length, so J = current per unit length * ds = nIds? Which makes my integral evaluate to ##\mu_{0}ks,## so the sum of the fields due to J and K is ##2\mu_{0}ks.## So my result from 6.18 still doesn't quite match.
I suspect that you didn't use the correct limits of integration for the contribution due to J. And also maybe you didn't use the correct value of s for the contribution due to K. Can you show your reasoning in more detail?

This is my new integral with In = Jds

$$\mu_{0} \int_0^s J ds$$

$$= \mu_{0} \int_0^s k ds$$

$$= \mu_{0} ks$$

and the contribution due to K with nI = K is

$$B = \mu_{0} nI =\mu_{0} K = \mu_{0} ks$$

TSny
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Gold Member
This is my new integral with In = Jds

$$\mu_{0} \int_0^s J ds$$

The limits of integration are not correct. Suppose you are finding B at a point that is a distance s from the central axis. Which concentric shells contribute to B at that point: the shells with radius less than s, or the shells with radius greater than s?

and the contribution due to K with nI = K is

$$B = \mu_{0} nI =\mu_{0} K = \mu_{0} ks$$
What value of s should you use here? In other words, when using the formula ##\vec K = \vec M \times \hat n##, where should ##\vec M## be evaluated?

Since the field outside of a solenoid is zero, only the shells with radius greater than s contribute. So the limits of integration are actually from s to R. And since K is on the surface, s=R and $$B_{K}=\mu_{0)kR$$. This almost gets me to the correct answer except I end up with $$\mu_{0)kR + (\mu_{0)kR - \mu_{0)ks)$$. If the integration went from R to s instead then the signs would be flipped and I would get $$\mu_{0)ks = B),$$ like I got for part (b). But I'm not sure how to justify integrating from R to s as opposed to s to R.

TSny
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Gold Member
Did you take into account that the directions of ##\vec J## and ##\vec K## are opposite?

$$\vec J = [0 - \frac {\partial } {\partial s} (ks) ] \hat \phi = -k \hat \phi$$

so the terms cancel after all. Thanks!

TSny
Homework Helper
Gold Member
OK. Good work!

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