GSO000 Problem 3

This is the classic problem of a conducting rod moving in a magnetic field. Since the electrical resistance is negligible, we use the impulse-momentum relation combined with conservation laws.

Let the velocity of the rod at time $t$ be $v(t)$ . The induced EMF $V(t)$ is:

V(t) = v(t)BL

Since the resistance is zero, this EMF equals the potential difference across the capacitor. So the charge stored in the capacitor $Q(t)$ is:

Q(t) = CV(t) = C v(t) BL

The current $I(t)$ is the time rate of change of charge:

I(t) = \frac{dQ(t)}{dt} = CBL \frac{dv(t)}{dt}

The Ampere force $F(t)$ acting on the rod (opposing motion) is:

F(t) = -I(t)BL = -CB^2L^2 \frac{dv(t)}{dt}

From the equation of motion $m \frac{dv(t)}{dt} = F(t)$ :

m \frac{dv(t)}{dt} = -CB^2L^2 \frac{dv(t)}{dt} \implies (m + CB^2L^2)\frac{dv(t)}{dt} = 0

This means $(m + CB^2L^2)v(t)$ is constant over time (a conservation law).

Integrating the equation of motion from $t=0$ to $t \to \infty$ :

m v_f - m v_0 = -BL Q_f

where the final charge is $Q_f = C v_f BL$ . Substituting:

m(v_f - v_0) = -CB^2L^2 v_f

(m + CB^2L^2)v_f = m v_0 \implies v_f = \frac{m}{m + CB^2L^2} v_0

Substituting numerical values: $CB^2L^2 = 0.50 \times (0.20)^2 \times (1.0)^2 = 0.50 \times 0.040 = 0.020\text{ kg}$

v_f = \frac{0.020}{0.020 + 0.020} \times 12 = \frac{1}{2} \times 12 = 6.0\text{ m/s}

Answer: 6

GSO000 Problem 3

Conducting Rod Moving in a Magnetic Field Connected to a Capacitor

Problem Statement

Constraints

Input Format

Solution