MEC BYU 4

Solution: Magnetic Moment and Field of a Rotating Sphere

Step 1: Charge Setup
Since the sphere is a conductor, charge $Q$ resides entirely on the surface. Given potential $V$, the total charge is: $$ Q = 4\pi\varepsilon_0 R V $$ The surface charge density is $\sigma = \frac{Q}{4\pi R^2}$.

We consider an elemental ring on the surface at polar angle $\theta$ with width $R d\theta$.
Radius of the ring: $r = R \sin\theta$.
Area of the ring strip: $dA = (2\pi r)(R d\theta) = 2\pi R^2 \sin\theta d\theta$.
Charge on the ring: $dq = \sigma dA$.
Equivalent Current: $dI = \frac{dq}{T} = dq \left(\frac{\omega}{2\pi}\right)$.

Step 2: Magnetic Field Calculation (Integration)
We calculate the magnetic field $dB$ at the center of the sphere due to this rotating ring using the Biot-Savart law for a current loop on its axis.
The field on the axis of a current loop of radius $r$ at a distance $z$ from the loop center is: $$ B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + z^2)^{3/2}} $$ Here, the distance from the ring wire to the center of the sphere is simply the sphere radius $\sqrt{r^2 + z^2} = R$. Substituting our values: $$ dB = \frac{\mu_0 (dI) (R \sin\theta)^2}{2 R^3} = \frac{\mu_0 dI \sin^2\theta}{2 R} $$ Now, substitute $dI$: $$ dI = \frac{\omega}{2\pi} (\sigma 2\pi R^2 \sin\theta d\theta) = \sigma \omega R^2 \sin\theta d\theta $$ Substitute $dI$ into the $dB$ equation: $$ dB = \frac{\mu_0}{2R} (\sigma \omega R^2 \sin\theta d\theta) \sin^2\theta $$ $$ dB = \frac{\mu_0 \sigma \omega R}{2} \sin^3\theta d\theta $$ Integrate from $\theta = 0$ to $\pi$: $$ B = \frac{\mu_0 \sigma \omega R}{2} \int_{0}^{\pi} \sin^3\theta \, d\theta $$ Using the standard integral $\int_{0}^{\pi} \sin^3\theta \, d\theta = \frac{4}{3}$: $$ B = \frac{\mu_0 \sigma \omega R}{2} \cdot \frac{4}{3} = \frac{2}{3} \mu_0 \sigma \omega R $$ Finally, substitute $\sigma = \frac{Q}{4\pi R^2}$ and $Q = 4\pi\varepsilon_0 R V$: $$ B = \frac{2}{3} \mu_0 \omega R \left( \frac{4\pi\varepsilon_0 R V}{4\pi R^2} \right) = \frac{2}{3} \mu_0 \omega R \left( \frac{\varepsilon_0 V}{R} \right) $$

Step 3: Magnetic Moment Calculation
The magnetic moment $dm$ of the elemental loop is Area $\times$ Current: $$ dm = (\pi r^2) dI = \pi (R \sin\theta)^2 (\sigma \omega R^2 \sin\theta d\theta) $$ $$ dm = \pi \sigma \omega R^4 \sin^3\theta d\theta $$ Integrating from $0$ to $\pi$: $$ m = \pi \sigma \omega R^4 \left( \frac{4}{3} \right) = \frac{4\pi}{3} \sigma \omega R^4 $$ Substitute $\sigma = \frac{Q}{4\pi R^2}$: $$ m = \frac{4\pi}{3} \omega R^4 \left( \frac{Q}{4\pi R^2} \right) = \frac{1}{3} Q R^2 \omega $$ Using $Q = 4\pi\varepsilon_0 R V$: