I'm not understanding how to calculate the magnetic field of a sphere

[LCSphysicist]

Homework Statement

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Relevant Equations

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I need to calculate the magnetic field generated by a static sphere at its center. On the surface of the sphere flows a constant current $K \hat \phi$.

Now, my guess was that the field produced would be equal to the field produced by a lot of rings, that is, i will split the sphere in a lot of rings, so we have:

$$B_{zring} = \mu I b^2/(2(b^2+z^2))^{3/2}$$

The total field will be:
$$\sum_{\theta} \mu K (R sin \theta)^2/(2(R^3))$$ (since b = Rcos and z = Rsin)

That's the real problem, how do i turn it in a integral?
I think there is some problems regarding my interpretation of the question:
Maybe K is the surface current, and not a current (as a corrent in a simple circuit) itself?
Is it possible to solve b this way?​

 

[haruspex]

On the surface of the sphere flows a constant current $$K \phi$$.

What is $\phi$ here? Is this as in spherical polar coordinates? Which direction in those coordinates?

 

[LCSphysicist]

What is $\phi$ here? Is this as in spherical polar coordinates? Which direction in those coordinates?

Ops, yes. $\hat \phi$ would be the unit tengential vector

 

[haruspex]

Ops, yes. \phi would be the unit tengential vectorView attachment 284577

Oh, so it is $K\hat\phi$? And it is a current density, yes?

 

[LCSphysicist]

Oh, so it is $K\hat\phi$?

Just realized now that it makes more sense to be the superificial current density.

If it is the case, $B = \int \mu (Rsin(\theta)^2 K R d\theta/(2*(R)^3) = \int \mu K sin^2 \theta d\theta / 2 = \mu K 2/3$ Is it right??

 

[haruspex]

Just realized now that it makes more sense to be the superificial current density.

If it is the case, $B = \int \mu (Rsin(\theta)^2 K R d\theta/(2*(R)^3) = \int \mu K sin^2 \theta d\theta / 2 = \mu K 2/3$ Is it right??

I don't agree with the last step. What is the integral of $\sin^2$?

 

[Delta2]

Just realized now that it makes more sense to be the superificial current density.

If it is the case, $B = \int \mu (Rsin(\theta)^2 K R d\theta/(2*(R)^3) = \int \mu K sin^2 \theta d\theta / 2 = \mu K 2/3$ Is it right??

I think you also forgot a factor of $\frac{1}{R}$ in the result.

 

[haruspex]

I think you also forgot a factor of $\frac{1}{R}$ in the result.

All the Rs cancelled.
K is a linear current density, so has dimension Q/(TL).

 

[Delta2]

All the Rs cancelled.

Yes sorry now I see the numerator is in fact $(R\sin\theta)^2Rd\theta$. And of course by dimensional analysis you are right also :D