MEC BYU 2

Solution: Magnetic Field in a Rotating Dielectric Rod

Analysis: A rotating cylinder with surface charge creates a surface current. This configuration is physically equivalent to a solenoid.

Step 1: Calculate Effective Current
The surface charge density is $\sigma$. Consider a small length element $dx$ of the cylinder. The charge on this ring is:

$$ dq = \sigma \cdot dA = \sigma (2\pi r \cdot dx) $$

Since the cylinder rotates with frequency $f = \frac{\omega}{2\pi}$, the equivalent current $dI$ produced by this ring is:

$$ dI = dq \cdot f = (\sigma 2\pi r dx) \cdot \frac{\omega}{2\pi} = \sigma r \omega dx $$

Step 2: Current per Unit Length ($nI$)
In a solenoid, the magnetic field depends on the number of turns per unit length times current ($nI$). Here, we can define the linear current density $K$ (current per unit length):

$$ K = \frac{dI}{dx} = \sigma r \omega $$

Step 3: Magnetic Field Calculation
For a long solenoid ($l \gg r$), the magnetic field inside is uniform and given by $B = \mu nI$. Replacing $nI$ with our calculated $K$:

$$ B_{vacuum} = \mu_0 K = \mu_0 \sigma r \omega $$

However, the rod is made of a dielectric material with relative magnetic permeability $\mu_r$. Therefore, the field inside the material is:

$$ B = \mu_r B_{vacuum} $$