EMI BYU 13 - CHATUP707

Solution to Problem 13

Solution: Falling Disc in Magnetic Field

1. Equivalent Capacitor

As the disc falls with velocity $v$ in a horizontal magnetic field $B$, the Lorentz force separates charges across the thickness $d$ of the disc. The disc acts as a capacitor with area $A = \pi r^2$ and separation $d$. The induced potential difference is $V = vBd$. The charge stored is: $$ q = CV = \left( \frac{\varepsilon_0 \pi r^2}{d} \right) (vBd) = \varepsilon_0 \pi r^2 B v $$

2. Charging Current

The charging current $i$ flows as velocity increases: $$ i = \frac{dq}{dt} = \varepsilon_0 \pi r^2 B \frac{dv}{dt} = \varepsilon_0 \pi r^2 B a $$

3. Magnetic Force and Acceleration

This current flows across the thickness of the disc, perpendicular to $B$. It experiences an upward magnetic force $F_m = i d B$: $$ F_m = (\varepsilon_0 \pi r^2 B a) d B = \varepsilon_0 \pi r^2 d B^2 a $$

Applying Newton’s Second Law ($mg – F_m = ma$): $$ mg – \varepsilon_0 \pi r^2 d B^2 a = ma $$ $$ mg = a(m + \varepsilon_0 \pi r^2 d B^2) $$

$$ a = \frac{mg}{m + \varepsilon_0 \pi r^2 d B^2} $$