# Magnetic field in the case of a thin magnetized cylinder (1 Viewer)

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#### Granger

1. The problem statement, all variables and given/known data
Consider a cylinder of thickness a=1 mm and radius R = 1 cm that is uniformly magnetized across z axis being its magnetization M= 10^5 A./m. Calculate the bound currents on the cylinder and, doing convenient approximations, the B field on the axis of the cylinder for z=0.

2. Relevant equations

3. The attempt at a solution
So I had no problems with the first question. The bound current is only at the surface since the magnetization is constant. It's also only in the lateral of the cylinder because M is parallel to the exterior normal on the top and on the bottom. We conclude that $$\vec{J}=10^5 \vec{e}_{\theta} A/m$$

Now the second part is what is giving me trouble. The magnetization is across the z axis, so the vectors B and H will also be in that direction. That means the problem doesn't have symmetry right? So that don't let me apply any of the versions of Ampere's law listed above.
Should I apply Biot-Savart law? But I'm not in vacuum. Doesn't that make it invalid? Wouldn't that be like applying Coulomb's law in a dielectric? They talk about making approximations, what should I do? Also the fact that they don't give us any information about the permeability or the susceptibility is making me confused too? Can someone please help me to clarify this problem?
Thanks!

#### Charles Link

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The surface current is the current per unit length. You have a thickness of 1mm, so you can calculate the entire current which is a ring of current of radius 1cm. The $B$ field at the center of the ring can readily be computed from Biot-Savart. There is no additional calculation required. This is the $B$ field in the material from these surface currents. $\\$ Edit: And note: The approximation here is because the ring is 1mm thick, all of the current isn't precisely at z=0.

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#### Granger

The surface current is the current per unit length. You have a thickness of 1mm, so you can calculate the entire current which is a ring of current of radius 1cm. The $B$ field at the center of the ring can readily be computed from Biot-Savart. There is no additional calculation required. This is the $B$ field in the material from these surface currents.
But all of my current is that superficial bound current, right? There is no free current. So my full current will be $#10^5 \times 2\pi \times 1 \times 10^{-2} \times 1 \times 10^{-3} = 6.3 A#$

As for the B field. I'm having a bit of trouble applying Biot Savart.

So I have

Having dl the vector across the direction of the current I (so it has the direction of$\vec{e}_{\theta}$?) we have
being equal to $d\theta (-\vec{e}_{z})$. Then I'll have

$$d_B=\frac{\mu_0 I d \theta}{4\pi r^2}$$

We substitute r with R since we are on the center of the cylinder And integrating across a very thin disk:
$$B=\frac{\mu_0 I 2\pi}{4\pi r^2}$$

And then over the cylinder we just multiply by its thickness of 1mm I obtain a field of $50.3 \muT$. However the right answer should have been $6.3 mT$ across positive z axis. What am I misunderstanding?

#### Granger

My other guess was that this degenerated in the case of a ring of charge (current only superficial on the sides and neglect thickness). But that would give me a field of $$B=\frac{\mu_0 I}{2R}$$ and that would lead me up to a field 39.584 mT...

#### Charles Link

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What did you get for $I$? Your second case here is correct. In your above post, you left off the $R$ in the computation of $dl=R \, d \theta$. What did you use for $\mu_o$? This second formula should get you the correct answer immediately.

#### Granger

Oh so he ring approximation is the way to go right?
For $I$ I got 6.3 A (as I described above multiplied the current density by $2 \pi R l$ where R is the radius and l the thickness.
For $\mu_0$ I used $4\pi \times 10^{-7}$

#### Charles Link

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The units might be giving some difficulty. In the formula for magnetic surface current density per unit length $\vec{K}_m=\vec{M} \times \hat{n}$ when using $B=\mu_o (H+M)$. $\vec{K}_m=10^5 \, A/m$, and with $d=.001 \, m$, $I=100 \, A$. $\\$ Edit: Now I think I see what you did... You need to look more closely at what the magnetic surface current per unit length means. (the length is measured across (and not along) the direction of current. But to get current $I$, you do not multiply by the circumference.

#### Granger

The units might be giving some difficulty. In the formula for magnetic surface current density per unit length $\vec{K}_m=\vec{M} \times \hat{n}$ when using $B=\mu_o (H+M)$. $\vec{K}_m=10^5 \, A/m$, and with $d=.001 \, m$, $I=100 \, A$. $\\$ Edit: Now I see what you did... You need t look moe closely at what teh magnetic surface current per unit length means. (the length is measured across (and not along) the direction of current.
Ok that gives me the right answer. But now I understand what I'm not figuring out. Why are you only using the thickness of the cylinder to calculate I. Shouldn't we also use the perimeter? I know in terms of units we should only use one (to cancel out the m^-1)... But why using the thickness and not the perimeter?

#### Charles Link

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That's the definition of current per unit length. Meanwhile, in employing Biot-Savart you use the circumference because you are integrating $I \, dl$. In computing the current $I$ in a wire, you do not need to know the length of the wire.

#### Granger

That's the definition of current per unit length. Meanwhile, in employing Biot-Savart you use the circumference because you are integrating $I \, dl$. In computing the current $I$ in a wire, you do not need to know the length of the wire.
But isn't $2 \pi L$ also a length? Its the "path" the the current takes, right?
Oh so because I'm integrating $I \, dl$ in Biot Savart if I used that perimeter when computing I it would be like multiplying by the same thing twice...

#### Charles Link

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Yes, the formula you have for a ring is derived from Biot-Savart. Both ways you get the same answer. If you compute Biot-Savart correctly above using $dl=R \, d \theta$, you get the same result as your formula.

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