Magnetic field inside and outside of a magnetized cylinder

In summary: K_{B} = \vec M \times \hat n $$(In summary, using two different methods, the magnetic field inside the cylinder is found to be zero and the magnetic field outside the cylinder is found to be positive.)The magnetic field inside the cylinder is found to be zero.
  • #1
astrocytosis
51
2

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?
 
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  • #2
astrocytosis said:
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.
 
  • #3
TSny said:
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.
 
  • #4
astrocytosis said:
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##?
 
  • #5
I is the current in one of the turns, so nI is the current per unit length? Then is that just equal to K?
 
  • #6
astrocytosis said:
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.
 
  • #7
So, since J is dI/dA, does nI = J/A?
 
  • #8
astrocytosis said:
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.)
 
  • #9
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.
 
  • #10
##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##.
 
  • #11
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.
 
  • #12
astrocytosis said:
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?
 
  • #13
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

$$ B = \mu_{0} nI =\mu_{0} K = \mu_{0} ks $$
 
  • #14
astrocytosis said:
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?
 
  • #15
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.
 
  • #16
Did you take into account that the directions of ##\vec J## and ##\vec K## are opposite?
 
  • #17
$$\vec J = [0 - \frac {\partial } {\partial s} (ks) ] \hat \phi = -k \hat \phi$$

so the terms cancel after all. Thanks!
 
  • #18
OK. Good work!
 
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Related to Magnetic field inside and outside of a magnetized cylinder

1. What is a magnetic field?

A magnetic field is a region in space where a magnetic force can be detected. It is created by moving electric charges and is represented by magnetic lines of force.

2. How is a magnetic field measured?

A magnetic field can be measured using a device called a magnetometer. This device detects the strength and direction of the magnetic field and can provide a numerical value for its intensity.

3. How does a magnetized cylinder create a magnetic field?

A magnetized cylinder, or bar magnet, has a north and south pole. These poles have opposite magnetic charges and create a magnetic field around the magnet. The strength of the magnetic field is strongest at the poles and decreases as you move further away from the magnet.

4. What is the difference between the magnetic field inside and outside of a magnetized cylinder?

The magnetic field inside a magnetized cylinder is stronger than the field outside. This is because the magnetic field lines are more concentrated within the material of the cylinder. Outside of the cylinder, the field lines are spread out and weaker.

5. How does the direction of a magnetic field change inside and outside of a magnetized cylinder?

Inside a magnetized cylinder, the magnetic field lines run from the north pole to the south pole. Outside of the cylinder, the field lines curve around the magnet from the north to the south pole, creating a closed loop. The direction of the field can also be affected by the orientation of the magnet and the presence of other magnetic objects nearby.

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