Rails: Drag, Terminal Speed and Energy

The rail circuit

Two long parallel rails, a distance \(l\) apart, lie across a uniform field \(\vec B\). At one end they are joined through a resistance \(R\). A metal rod rests across them and is pulled along at a steady speed \(v\). The rod is a cell of emf \(Blv\), and the rails and the resistance close the circuit. So a current \(I = Blv/R\) flows, ignoring the small resistance of the rod and rails.

The rod now carries a current across the field, so the field pushes on it with \(BIl\). In Fig 6.1 the current runs up the rod, and up crossed with into the page points to the left, against the motion. It is Lenz's law in mechanical form: a drag that opposes the motion causing it. To keep the rod at a steady speed, the pull must equal that drag.

The equality is no accident. The pull equals the drag \(BIl\), so its power is \(Fv = BIlv\). Regroup it as \(I(Blv) = I\varepsilon\), the power the rod's emf delivers to the circuit. With \(\varepsilon = IR\), that is \(I^2R\). Every joule of work done by the pull becomes a joule of heat in the resistance. The field only passes the energy on.

A rod let go

Give the rod of mass \(m\) a speed \(v_0\) and let it go on smooth rails. Now nothing pulls it. The only force along the rails is the drag \(B^2l^2v/R\), so Newton's second law gives \(m\,dv/dt = -B^2l^2v/R\). The faster the rod, the harder it is braked. As it slows, the drag weakens.

A rate of fall in proportion to the value itself gives an exponential decay. The speed is \(v = v_0\,e^{-t/\tau}\), with the time constant \(\tau = mR/B^2l^2\). In each time \(\tau\) the speed falls to \(1/e\), about \(0.37\), of its value. The rod never quite stops, but after a few \(\tau\) it has almost stopped.

With the first chip of Fig 6.2, \(\tau\) \(= 0.2 \times 2/(0.25 \times 0.16)\) \(= 10\,\text{s}\). After \(10\,\text{s}\) the rod has slowed from \(5\) to \(1.84\,\text{m/s}\) and covered \(31.6\,\text{m}\). After \(40\,\text{s}\) it has covered \(49\,\text{m}\) and is nearly at rest. The total distance is exactly \(v_0\tau = 50\,\text{m}\). The next derivation shows why.

A steady pull and a terminal speed

Now pull the rod from rest with a steady force \(F\). At first the rod is slow, the drag is small, and the rod speeds up. As the speed grows, so does the drag \(B^2l^2v/R\). The acceleration falls, and the speed rises ever more slowly towards the value at which the drag equals the pull. There the net force is zero and the speed stays constant: the terminal velocity.

The second chip of Fig 6.2 pulls the rod with \(0.4\,\text{N}\). The terminal speed is \(0.4 \times 2/(0.25 \times 0.16)\) \(= 20\,\text{m/s}\), and the current then is \(0.5 \times 0.4 \times 20/2 = 2\,\text{A}\). The drag is \(0.5 \times 2 \times 0.4 = 0.4\,\text{N}\), equal to the pull. The approach is the same exponential, \(v = 20(1 - e^{-t/10})\): \(12.6\,\text{m/s}\) after \(10\,\text{s}\).

Stand the rails upright and let a rod fall between them, and gravity is the steady pull, \(F = mg\). A rod of \(40\,\text{g}\) on these rails feels \(0.4\,\text{N}\) and falls at a terminal \(20\,\text{m/s}\). At terminal speed the power of gravity, \(mgv_t\), all becomes heat. It is the copper-tube magnet of lesson 3 again, a steady fall against magnetic drag.

A rod driven by a cell

Replace the pull by a cell of emf \(E_0\) in series with the resistance. At rest, the cell drives \(I = E_0/R\) through the rod. The field pushes the rod with \(BIl\), and the rod starts to move. It is now a motor: electrical energy in, motion out.

As soon as the rod moves, it has its own emf \(Blv\). By Lenz's law that emf opposes the current that caused the motion. So the net emf in the circuit is \(E_0 - Blv\), and the current \((E_0 - Blv)/R\) falls as the rod speeds up. The rod stops accelerating when \(Blv = E_0\). Then there is no current and no force, and the rod coasts on at the steady speed \(E_0/Bl\).

The third chip of Fig 6.2 puts a \(2\,\text{V}\) cell with the \(2\,\Omega\). At the start \(I = 1\,\text{A}\), and the force is \(0.5 \times 1 \times 0.4 = 0.2\,\text{N}\), so the rod accelerates at \(1\,\text{m/s}^2\). It levels off at \(2/(0.5 \times 0.4) = 10\,\text{m/s}\), where \(Blv = 2\,\text{V}\). The emf that a moving motor makes against its supply is its back emf.

A capacitor or an inductor across the rails

Close the rails with a capacitor \(C\) instead of a resistance, and move the rod at a steady \(v\). The rod's emf charges the capacitor, as in the chapter Capacitance and Dielectrics. A current flows only while the charge builds up. When the capacitor's voltage reaches \(Blv\), the current stops. The final charge is \(Q = CBlv\), and after that there is no current, so no drag and no force is needed.

A current then flows only while the speed changes, because \(Q = CBlv\) gives \(I = CBl\,dv/dt\). On upright rails with a capacitor, a falling rod of mass \(m\) feels a drag \(BIl = B^2l^2C\,dv/dt\). So \(mg - B^2l^2C\,a = ma\), and \(a = mg/(m + B^2l^2C)\), a constant. The rod falls with a steady acceleration a little below \(g\), with no terminal speed.

Now close the rails with a pure inductor, a coil of no resistance, met again in the next chapter. Give the rod a push. Its emf drives a growing current through the coil, and the drag on the rod slows it. The energy of the motion goes into the magnetic field of the coil. When the rod stops, the coil's current is largest. The coil keeps that current flowing, which pushes the rod back the other way. With no resistance nothing is lost, so the energy swaps back and forth and the rod oscillates.

On sloping rails only the part of the velocity across the field counts. A rod \(0.5\,\text{m}\) long slides down rails inclined at \(37°\) at \(2\,\text{m/s}\), in a vertical field of \(0.5\,\text{T}\). The vertical part of its velocity lies along the field and gives nothing. The horizontal part, \(v\cos 37° = 1.6\,\text{m/s}\), is across the field. So \(\varepsilon = 0.5 \times 0.5 \times 1.6 = 0.4\,\text{V}\).