Induction brake

We first calculate the induction voltage of the coil $\begin{aligned} U &= -\frac{d}{dt} N \int \vec B(x) \cdot d\vec A \\ &= -\frac{d}{dt} N l \int_{x(t) - l}^{x(t)} B(x') \cdot d x' \\ &= -\frac{d}{dt} N l \begin{cases} \frac{1}{2} \beta x(t)^2 & x(t) < l \\ \beta (x(t) - l/2) l & x(t) \geq l \end{cases}\\ &= \begin{cases} -N l \beta x(t) v(t) & x(t) < l \\ - N l^2 \beta v(t) & x(t) \geq l \end{cases} \end{aligned}$ Due to the electrical resistance we get an electric current $I = U / R$ and thus an electric power $P = UI$ . The electric power comes from the loss of kinetic energy of the cart. Therefore, a decelerating force $\vec F = - F \vec e_x$ acts on the cart with $\begin{aligned} P &= \frac{U^2}{R} = F v \\ \Rightarrow \quad F &= \frac{U^2}{R v} = \begin{cases} \frac{(N l \beta)^2}{R} x^2 v & x < l \\ \frac{(N l^2 \beta)^2}{R} v & x \geq l \end{cases} \end{aligned}$ The equation of motion must be solved separately for the cases $x <l$ and $x \geq l$ . In the first case, these can easily be integrated over time $\begin{aligned} & & m \frac{d v}{dt} &= -\frac{(N l \beta)^2}{R} x^2 v \\ & & &= -\frac{(N l \beta)^2}{3R} \frac{d}{dt} \left( x^3 \right) \\ \Rightarrow & & m (v - v_0) &= -\frac{(N l \beta)^2}{3 R} x^3 \\ \Rightarrow & & v_1 = v(t_1) &\stackrel{x=l}{=} v_0 -\frac{(N \beta)^2 }{3 m R} l^5 \\ & & &= \left(3 - \frac{(300 \cdot 1)^2}{3 \cdot 0.1 \cdot 3} \cdot (0.1)^5 \right) \,\frac{\text{m}}{\text{s}} \\ & & &= 2 \,\frac{\text{m}}{\text{s}} \end{aligned}$ where $t_1$ denotes the time when the car reaches the position $x = l$ . Because of $v_1> 0$ the car has not stopped yet, so we have to solve the equations of motion also for the case $x> 0$ . For this we use the separation of the variables: $\begin{aligned} & & m \frac{dv}{dt} &= -\frac{(N l^2 \beta)^2}{R} v \\ \Rightarrow & & \frac{dv}{v} &= - \frac{dt}{\tau}, \quad \text{where } \tau := \frac{m R}{(N l^2 \beta)^2} = \frac{1}{30} \,\text{s}\\ \Rightarrow & & \int_{v_1}^{v(t)} \frac{dv}{v} &= \log \frac{v(t)}{v_1} = - \frac{t - t_1}{\tau} \\ \Rightarrow & & v(t) &= v_1 \exp \left( - \frac{t - t_1}{\tau} \right), \quad \text{where } t \geq t_1 \end{aligned}$ The entire distance covered then results through integration over time $\begin{aligned} s &= x(t_1) + \int_{t_1}^{\infty} v(t) dt \\ &= l + \left[ - v_1 \tau \exp \left( - \frac{t - t_1}{\tau} \right) \right]_{t_1}^{\infty} \\ &= l + v_1 \tau \\ &= \left(0.1 + 2 \cdot \frac{1}{30} \right)\, \text{m}\\ &\approx 16.7 \,\text{cm} \end{aligned}$