GSO001 Problem 11

Let $v$ be the speed of the rod, $I$ the current, and $q$ the charge stored in the capacitor at time $t$ .

When the rod moves at speed $v$ , the induced EMF by Faraday's law is $V = vBL$ . By Kirchhoff's voltage law:

vBL - RI - \frac{q}{C} = 0 \quad \cdots (1)

The current $I = \frac{dq}{dt}$ . The Ampere force on the rod is directed opposite to motion with magnitude $IBL$ .

The equation of motion of the rod:

m \frac{dv}{dt} = - IBL \quad \cdots (2)

From equation (2), integrating over time:

m dv = - B L I dt = - B L dq

Integrating from $t=0$ (where $v = v_0, q = 0$ ) to $t \to \infty$ (where $v = v_f$ , $I = 0$ , and the final charge is $q_f$ ):

m(v_f - v_0) = - B L q_f \quad \cdots (3)

In the final state, $I = 0$ and $v_f$ is constant, so from equation (1):

v_f B L - R \cdot 0 - \frac{q_f}{C} = 0 \quad \Rightarrow \quad q_f = C B L v_f \quad \cdots (4)

Substituting (4) into (3):

m(v_f - v_0) = - B L (C B L v_f) = - C B^2 L^2 v_f

(m + C B^2 L^2) v_f = m v_0

v_f = \frac{m}{m + C B^2 L^2} v_0

Substituting numerical values. First, compute $C B^2 L^2$ :

C B^2 L^2 = (2.0 \times 10^{-3}) \times (2.0)^2 \times (0.50)^2 = 2.0 \times 10^{-3} \times 4.0 \times 0.25 = 2.0 \times 10^{-3} \text{ kg}

With $m = 1.2 \times 10^{-2} \text{ kg} = 12 \times 10^{-3} \text{ kg}$ :

v_f = \frac{12 \times 10^{-3}}{12 \times 10^{-3} + 2.0 \times 10^{-3}} \times 10 = \frac{12 \times 10^{-3}}{14 \times 10^{-3}} \times 10 = \frac{12}{14} \times 10 = \frac{60}{7} \text{ m/s}

This fraction cannot be reduced further, so $p = 60, q = 7$ . The required value is $p + q = 60 + 7 = 67$ .

(Note: The value of resistance $R$ does not affect the final speed; it only determines the rate of energy dissipation and the time to reach the final state.)

GSO001 Problem 11

Conducting Rod on Parallel Rails Connected to a Capacitor

Problem Statement

Constraints

Input Format

Solution