Exercise 5.3.4: Linear motor on rails

At rest there is no motional emf: $i_0=V_0/R=120\mathrm{A}$, $F_0=B{i}_{0}l=1\times 120\times 0.5=60\mathrm{N}$, $a_0=F_0/m=300m/{\mathrm{s}}^2$. The force $il\times B$ pushes the bar along the rails, away from the source (check with the right-hand rule for the sense of $i$ in the figure).

Once the bar moves with velocity $v$ it generates the back-emf $Blv$, which opposes $V_0$ (Lenz), so the loop equation and Newton's law read

The acceleration vanishes when $Blv=V_0$:

The equation is linear, $dv/dt=(v_t-v)/\tau$ with

With $i=0$ the source delivers no power and nothing is dissipated: the bar coasts. With friction or a load force $F_L$ the bar settles at $v=v_t-F_{L}R/(B^2{l}^2)$, the source supplies $V_{0}i$, of which $Blvi=F_{L}v$ is mechanical power and $i^{2}R$ is heat, the power balance of every DC motor.

Units: $mR/(B^2{l}^2)$ is $\mathrm{kg}\cdot \Omega /({\mathrm{T}}^2{\mathrm{m}}^2)$, and since $\mathrm{T}\cdot {\mathrm{m}}^2=\mathrm{V}\cdot \mathrm{s}$ and ${\mathrm{V}}^2{\mathrm{s}}^2/\Omega =\mathrm{J}\cdot \mathrm{s}=\mathrm{kg}\cdot {\mathrm{m}}^2/s$, the result is seconds. The same $Blv$ appears as emf in the generator of Home exercise 5.3.2 and as back-emf here, and $\tau =mR/(B^2{l}^2)$ is also the stopping time of the released generator bar.