MEC BYU 1

Solution: Magnetic Field of a Rotating Lamina

Analysis: We are given an insulating lamina carrying a charge distributed in the $x-y$ plane. We need to relate the electrostatic potential $V_0$ at the origin to the magnetic field $B$ produced when the lamina rotates.

Let us consider a small charge element $dq$ on the lamina at a distance $r$ from the origin.

Step 1: Electrostatic Potential
The electrostatic potential $V_0$ at the origin due to the stationary charge distribution is given by the superposition principle:

$$ V_0 = \int \frac{1}{4\pi\varepsilon_0} \frac{dq}{r} $$

Rearranging this, we can express the integral of the charge distribution as:

$$ \int \frac{dq}{r} = 4\pi\varepsilon_0 V_0 \quad \text{— (i)} $$

Step 2: Magnetic Field Generation
When the lamina rotates with angular velocity $\omega$ about the z-axis, the charge element $dq$ moves in a circle, creating an equivalent current $dI$. The time period of rotation is $T = \frac{2\pi}{\omega}$.

$$ dI = \frac{dq}{T} = \frac{dq}{(2\pi/\omega)} = \frac{\omega}{2\pi} dq $$

This current loop of radius $r$ produces a magnetic field $dB$ at the center (origin). Using the formula for the magnetic field at the center of a circular loop ($B = \frac{\mu_0 I}{2r}$):

$$ dB = \frac{\mu_0 (dI)}{2r} = \frac{\mu_0}{2r} \left( \frac{\omega dq}{2\pi} \right) = \frac{\mu_0 \omega}{4\pi} \frac{dq}{r} $$

Step 3: Integration
The total magnetic field is the integral of $dB$ over the entire lamina:

$$ B = \int dB = \frac{\mu_0 \omega}{4\pi} \int \frac{dq}{r} $$

Substituting the value of $\int \frac{dq}{r}$ from equation (i):

$$ B = \frac{\mu_0 \omega}{4\pi} (4\pi\varepsilon_0 V_0) $$