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Magnetic Field of a Rotating Charged Sphere

Introduction

In this article I will derive the magnetic field of a rotating sphere with uniform charge density \rho . I will do this by considering the Maxwell equations inside and outside the sphere, and making an ansatz of the form {C}{r}^{\alpha}{\sin{{\left(\theta\right)}}} or {C}{r}^{\alpha}{\cos{{\left(\theta\right)}}} . This ansatz will provide us with both the particular and the homogeneous solutions. The whole derivation will be carried out inside a spherical coördinate system where \theta is the angle with the z-axis, \phi is the angle with the x-axis and {r} is the distance from the origin. An alternative way of finding the magentic field is to consider the sphere as made up of rotating charged shells and to use integration.

Ampère's law

Now suppose the sphere is spinning around its symmetry axis with a constant angular velocity omega . The current density {j} is then given by: {\mathbf{{j}}}=\rho{\mathbf{{v}}}=\rho{\left({\mathbf{\omega}}\times{\mathbf{{r}}}\right)}\qquad\Rightarrow\qquad{j}=\rho\omega{r}{\sin{{\left(\theta\right)}}} {\mathbf{{j}}} only has a \hat{{\mathbf{\phi}}} component since every charged volume element in the sphere traces out a circle around the rotation axis. So if we let {j} be the strength of the current we have {\mathbf{{j}}}={j}\hat{\phi} . Ampère's law written out in spherical coördinates gives us: \frac{{1}}{{{r}{\sin{{\left(\theta\right)}}}}}{\left[\frac{\partial}{{\partial\theta}}{\left({\sin{{\left(\theta\right)}}}{B}_{\phi}\right)}-\frac{{\partial{B}_{\theta}}}{{\partial\phi}}\right]}={0} \frac{{1}}{{r}}{\left[\frac{{1}}{{\sin{{\left(\theta\right)}}}}\frac{{\partial{B}_{{r}}}}{{\partial\phi}}-\frac{\partial}{{\partial{r}}}{\left({r}{B}_{\phi}\right)}\right]}={0} \frac{{1}}{{r}}{\left[\frac{\partial}{{\partial{r}}}{\left({r}{B}_{\theta}\right)}-\frac{{\partial{B}_{{r}}}}{{\partial\theta}}\right]}=\mu_{{0}}{j} \frac{{1}}{{r}^{{2}}}\frac{\partial}{{\partial{r}}}{\left({r}^{{2}}{B}_{{r}}\right)}+\frac{{1}}{{{r}{\sin{{\left(\theta\right)}}}}}\frac{\partial}{{\partial\theta}}{\left({\sin{{\left(\theta\right)}}}{B}_{\theta}\right)}+\frac{{1}}{{{r}{\sin{{\left(\theta\right)}}}}}\frac{{\partial{B}_{\phi}}}{{\partial\phi}}={0} The derivative of every coördinate of B with respect to \phi has to be zero due to the complete symmetry of the problem in the \phi coördinate. So \frac{{\partial{B}_{{i}}}}{{\partial\phi}}={0} for all three coördinates. Another symmetry argument gives furthermore that {B}_{\phi}={0} . Plugging in these equations in Ampère's law we see that only the last two equations do not reduce to {0}={0} . Writing these out gives \frac{{B}_{\theta}}{{r}}+\frac{{\partial{B}_{\theta}}}{{\partial{r}}}-\frac{{1}}{{r}}\frac{{\partial{B}_{{r}}}}{{\partial\theta}}=\mu_{{0}}{j} \frac{{2}}{{r}}{B}_{{r}}+\frac{{\partial{B}_{{r}}}}{{\partial{r}}}+\frac{{\cos{{\left(\theta\right)}}}}{{{r}{\sin{{\left(\theta\right)}}}}}{B}_{\theta}+\frac{{1}}{{r}}\frac{{\partial{B}_{\theta}}}{{\partial\theta}}={0}

Particular solution

Let us focus on the first equation. This is the point at which we make use of our educated guess. Using {B}_{\theta}={C}_{{1}}{r}^{\alpha}{\sin{{\left(\theta\right)}}} and {B}_{{r}}={C}_{{2}}{r}^{\alpha}{\cos{{\left(\theta\right)}}} we end up with the equation {C}_{{1}}{r}^{{\alpha-{1}}}{\sin{{\left(\theta\right)}}}+\alpha{C}_{{1}}{r}^{{\alpha-{1}}}{\sin{{\left(\theta\right)}}}+{C}_{{2}}{r}^{{\alpha-{1}}}{\sin{{\left(\theta\right)}}}=\mu_{{0}}{j} But we know that {j}=\rho\omega{r}{\sin{{\left(\theta\right)}}} . Filling this in and dividing by {r}^{{\alpha-{1}}}{\sin{{\left(\theta\right)}}} yields: {C}_{{1}}+\alpha{C}_{{1}}+{C}_{{2}}=\rho\omega{r}^{{-\alpha+{2}}} The left side is independent of {r} , so the right side should also be independent of {r} and we conclude \alpha={2} . Now the second equation is: {2}{C}_{{2}}{r}^{{\alpha-{1}}}{\cos{{\left(\theta\right)}}}+\alpha{C}_{{2}}{r}^{{\alpha-{1}}}{\cos{{\left(\theta\right)}}}+{2}{C}_{{1}}{r}^{{\alpha-{1}}}{\cos{{\left(\theta\right)}}}={0} Simply divide by {r}^{{\alpha-{1}}}{\cos{{\left(\theta\right)}}} . Knowing the value of \alpha , we are now left with two linear equations in {C}_{{1}}{\quad\text{and}\quad}{C}_{{2}} . {3}{C}_{{1}}+{C}_{{2}}=\rho\omega\qquad\wedge\qquad{2}{C}_{{1}}+{4}{C}_{{2}}={0} And we conclude {C}_{{1}}=\frac{{2}}{{5}}\rho\omega and {C}_{{2}}=-\frac{{\rho\omega}}{{5}} . So a particular solution is given by {B}_{\theta}=\frac{{2}}{{5}}\mu_{{0}}\rho\omega{r}^{{2}}{\sin{{\left(\theta\right)}}}\qquad\wedge\qquad{B}_{{r}}=-\frac{{1}}{{5}}\mu_{{0}}\rho\omega{r}^{{2}}{\cos{{\left(\theta\right)}}}

Homogeneous solution

We now set out to find the homogeneous solution: a solution to the differential equations with the right hand side replaced by zeros. We need these homogeneous solutions to satisfy the boundary conditions considered below. Plugging in {B}_{\theta}={C}_{{1}}{r}^{\alpha}{\sin{{\left(\theta\right)}}} and {B}_{{r}}={C}_{{2}}{r}^{\alpha}{\cos{{\left(\theta\right)}}} into the differential equations derived from Ampère's law and dividing away common factors we have: {\left({1}+\alpha\right)}{C}_{{1}}+{C}_{{2}}={0}\qquad\wedge\qquad{2}{C}_{{1}}+{\left({2}+\alpha\right)}{C}_{{2}}={0} If these equations are linearly independent we would only find a trivial solution. We calculate the determinant and set this equal to zero to allow for non-trivial solutions. \begin{vmatrix} 1 + \alpha & 1 \\ 2 & 2 + \alpha \\ \end{vmatrix} = \alpha^2 + 3 \alpha = 0 \quad \Rightarrow \quad \alpha = 0 \quad \lor \quad \alpha = -3 By chosing these values of \alpha the two equations become linearly dependent. In other words, we are left with only one equation. This is not a problem however as the homogeneous solutions are determined up to a constant factor, and we need to know only one of {C}_{{1}},{C}_{{2}} to determine the relationship between {B}_{\theta} and {B}_{{r}} . For example, using \alpha=-{3} we find from the first equation {C}_{{2}}={2}{C}_{{1}} . Renaming {C}_{{1}}\Rightarrow{C} we find that for any value of {C} an homogeneous solution is {B}_{\theta}={C}{r}^{{-{3}}}{\sin{{\left(\theta\right)}}}\qquad\wedge\qquad{B}_{{r}}={2}{C}{r}^{{-{3}}}{\cos{{\left(\theta\right)}}} A same procedure in the case of \alpha={0} gives us {B}_{\theta}={C}{\sin{{\left(\theta\right)}}}\qquad\wedge\qquad{B}_{{r}}=-{C}{\cos{{\left(\theta\right)}}}

Boundary conditions

The boundary conditions in this case are: The first boundary condition implies that the first homogeneous solution ( \alpha=-{3} ) cannot exists inside the sphere. Likewise the second boundary condition implies the second homogeneous solution ( \alpha={0} ) cannot exist outside the sphere. Also, outside the sphere, we have no current density so the particular solution does not apply here. Writing down the intermediate result by using some yet unknown coefficients {A} , {C} , we have inside the sphere: {B}_{\theta}=\frac{{2}}{{5}}\mu_{{o}}\omega\rho{r}^{{2}}{\sin{{\left(\theta\right)}}}+{C}{\sin{{\left(\theta\right)}}} {B}_{{r}}=-\frac{{1}}{{5}}\mu_{{o}}\omega\rho{r}^{{2}}{\cos{{\left(\theta\right)}}}+{C}{\sin{{\left(\theta\right)}}} And outside: {B}_{\theta}={A}{r}^{{-{3}}}{\sin{{\left(\theta\right)}}} {B}_{{r}}={2}{A}{r}^{{-{3}}}{\cos{{\left(\theta\right)}}} Letting {R} be the radius of the sphere, we take {r}={R} and find for continuity: \frac{{{2}{R}^{{5}}\mu_{{0}}\rho\omega+{5}{C}{R}^{{3}}}}{{5}}={A}\qquad\qquad\theta\ \text{component} \frac{{{R}^{{5}}\mu_{{0}}\rho\omega+{5}{C}{R}^{{3}}}}{{5}}=-{2}{A}\qquad\qquad{r}\ \text{component} Solving these equations yields: {A}=\frac{{{R}^{{5}}\mu_{{0}}\rho\omega}}{{15}}\qquad\wedge\qquad{C}=-\frac{{{R}^{{2}}\mu_{{0}}\rho\omega}}{{3}}

Final answer

Now we can conclude, that inside the sphere: {B}_{\theta}=\frac{{μ_{{0}}ρω{\left({6}{r}^{{2}}-{5}{R}^{{2}}\right)}{\sin{{\left(θ\right)}}}}}{{15}} {B}_{{r}}=\frac{{μ_{{0}}ρω{\left({5}{R}^{{2}}-{3}{r}^{{2}}\right)}{\cos{{\left(θ\right)}}}}}{{15}} And outside the sphere: {B}_{\theta}=\frac{{μ_{{0}}ρω{R}^{{5}}{\sin{{\left(θ\right)}}}}}{{{15}{r}^{{3}}}} {B}_{{r}}=\frac{{{2}μ_{{0}}ρω{R}^{{5}}{\cos{{\left(θ\right)}}}}}{{{15}{r}^{{3}}}}