Find magnetic field at center of rotating sphere

[Jaccobtw]

Homework Statement

A sphere of radius r=5cm and volume charge density ρ=14μC/m^3
spins at ω=200rad/s. What is the magnetic field at the center of the sphere in Tesla? Hints: this requires a great deal of integration, and the resulting field is very small.

Relevant Equations

Magnetic field at a point on the center axis of a loop of current:
$$B = \frac{\mu_0 I}{2\pi(y +R^2)^{3/2}}\hat{j}$$

if a sphere rotates, it's like multiple currents going around in a circle. I can find the magnetic field of each of those currents at the center point of the circle and add them together. We can integrate with respect to y and R. y ranges from 0 to 5 cm away from the center of the loop and the radius also ranges from 0 to 5 cm. However, I'm not sure how to incorporate volume charge density into the equation. Do you guys have any ideas?

 

[haruspex]

Homework Statement:: A sphere of radius r=5cm and volume charge density ρ=14μC/m^3
spins at ω=200rad/s. What is the magnetic field at the center of the sphere in Tesla? Hints: this requires a great deal of integration, and the resulting field is very small.
Relevant Equations:: Magnetic field at a point on the center axis of a loop of current:
$$B = \frac{\mu_0 I}{2\pi(y +R^2)^{3/2}}\hat{j}$$

if a sphere rotates, it's like multiple currents going around in a circle. I can find the magnetic field of each of those currents at the center point of the circle and add them together. We can integrate with respect to y and R. y ranges from 0 to 5 cm away from the center of the loop and the radius also ranges from 0 to 5 cm. However, I'm not sure how to incorporate volume charge density into the equation. Do you guys have any ideas?

There isn't going to be a current around the loop unless it actually contains some charge, and that will only be true if it has volume. So define some cross section element for the loop in suitable coordinates, maybe spherical or cylindrical.

 

[Jaccobtw]

There isn't going to be a current around the loop unless it actually contains some charge, and that will only be true if it has volume. So define some cross section element for the loop in suitable coordinates, maybe spherical or cylindrical.

I chose spherical coordinates
$$dq = \rho dV$$
$$dV = rdrd\theta d\phi$$

or would dV be equal to $$dV = r^{2}sin\theta drd\theta d\phi$$ ?

 

[kuruman]

I chose spherical coordinates
$$dq = \rho dV$$
$$dV = rdrd\theta d\phi$$

or would dV be equal to $$dV = r^{2}sin\theta drd\theta d\phi$$ ?

Which of the $dV$ expressions has dimensions of length cubed?

 

[Jaccobtw]

Which of the $dV$ expressions has dimensions of length cubed?

Dont they both have dimensions of length cubed?

 

[Jaccobtw]

Which of the $dV$ expressions has dimensions of length cubed?

I guess the second one. I'm just not sure where the expression $r^2 sin(\theta)$ comes from

 

[Delta2]

I think the setup of the integral comes easier to mind if you consider the integral form of biot savart law: It will be for the magnetic field at the center, $$\vec{B}=\frac{\mu_0}{4\pi}\int\frac{\vec{r}}{|\vec{r}|^3}\times\vec{J(\vec{r})}d^3\vec{r}$$

If we try to calculate this integral in spherical coordinates, it will be $d^3\vec{r}=r^2\sin\theta drd\phi d\theta$ where $r=|\vec{r}|$ and also $\vec{J(\vec{r})}=\rho\vec{v}=\rho(\vec{\omega}\times\vec{r})$.
You will have to calculate a double cross product $\vec{r}\times (\vec{\omega}\times\vec{r})$ in spherical coordinates, and be careful in the expression of $\vec{\omega}$ as vector in spherical coordinates.

 

[Jaccobtw]

I think the setup of the integral comes easier to mind if you consider the integral form of biot savart law: It will be for the magnetic field at the center, $$\vec{B}=\int\frac{\vec{r}}{|\vec{r}|^3}\times\vec{J(\vec{r})}d^3\vec{r}$$

If we try to calculate this integral in spherical coordinates, it will be $d^3\vec{r}=r^2\sin\theta drd\phi d\theta$ where $r=|\vec{r}|$ and also $\vec{J(\vec{r})}=\rho\vec{v}=\rho(\vec{\omega}\times\vec{r})$.
You will have to calculate a double cross product $\vec{r}\times (\vec{\omega}\times\vec{r})$ in spherical coordinates, and be careful in the expression of $\vec{\omega}$ as vector in spherical coordinates.

I think I might get it. $dr$ is one dimension, $rd\theta$ is another dimension, and $rsin \theta d\phi$ is the third dimension. I had to draw a 3d picture. What do you think?

 

[kuruman]

I think I might get it. $dr$ is one dimension, $rd\theta$ is another dimension, and $rsin \theta d\phi$ is the third dimension. I had to draw a 3d picture. What do you think?

You are correct. It helps to have that volume element memorized especially when you do Quantum Mechanics.

 

[Delta2]

I think I might get it. $dr$ is one dimension, $rd\theta$ is another dimension, and $rsin \theta d\phi$ is the third dimension. I had to draw a 3d picture. What do you think?

Yes, you refer to the volume element $d^3\vec{r}$ (or $dV$) in spherical coordinates.
To calculate the aforementioned double cross product , take advantage that $\vec{\omega}$ is a constant vector and use the identity $$\vec{a}\times(\vec{b}\times\vec{a})=(\vec{a}\cdot\vec{a})\vec{b}-(\vec{a}\cdot\vec{b})\vec{a}$$.

So the integral will split into two terms one term will be like $$\int f(r,\theta)\vec{\omega}dr d\theta d\phi$$ and the other$$\int g(r,\theta)\vec{r} dr\ d\theta d\phi$$.

If I did it correctly the first term gives $\frac{\mu_0}{2}\rho\omega R^2\hat z$ and the second term gives $\frac{\mu_0}{6}\rho\omega R^2\hat z$ (you also might need extra help with the second term its not simple as the first term) and subtracting them gives the end result $$\vec{B}=\frac{\mu_0}{3}\rho\omega R^2\hat z$$