A rod sliding on rails in a magnetic field makes a motional emf $BLv$ and drives a current. That current sits in the field, so it feels a force. Lenz's law decides which way, and the answer makes the rod behave exactly like an object moving through a fluid with linear drag.
Key idea
A rod of length $L$ moving at $v$ on rails closed by a resistance $R$ feels a magnetic force
$$F = \frac{B^2L^2v}{R}$$
against its motion. Left to coast, $v = v_0e^{-t/\tau}$ with $\tau = \dfrac{mR}{B^2L^2}$, and it travels a total distance $v_0\tau$. Falling on vertical rails, it reaches a terminal speed $v_T = \dfrac{mgR}{B^2L^2}$.
Why it works
Follow the chain: speed makes emf, $\varepsilon = BLv$; emf drives current, $I = BLv/R$; current in a field feels force, $F = ILB = B^2L^2v/R$. The force is proportional to the speed, like drag $F = -bv$ with $b = B^2L^2/R$. Newton's law $m\,dv/dt = -bv$ then gives exponential decay. Energy check: the rod's kinetic energy turns into heat in the resistor, $I^2R$, so the force must oppose the motion.
Worked examples
Worked example: Coasting to a stop
A $0.10\ \text{kg}$ rod $0.40\ \text{m}$ long slides on frictionless rails joined by $2.0\ \Omega$, in a vertical $0.50\ \text{T}$ field. It starts at $3.0\ \text{m/s}$. Find the starting force, the time constant and how far it goes.
- $F_0 = \dfrac{B^2L^2v_0}{R} = \dfrac{(0.50)^2(0.40)^2(3.0)}{2.0} = 0.0600\ \text{N}$, backward.
- $\tau = \dfrac{mR}{B^2L^2} = \dfrac{(0.10)(2.0)}{(0.50)^2(0.40)^2} = 5.00\ \text{s}$.
- Distance: $\int_0^\infty v_0e^{-t/\tau}dt = v_0\tau = 15.0\ \text{m}$.
Answer: $0.0600\ \text{N}$, $5.00\ \text{s}$, $15.0\ \text{m}$
Worked example: Falling at terminal speed
A $50\ \text{g}$ rod $0.50\ \text{m}$ long falls down vertical rails joined by $1.0\ \Omega$, in a horizontal $1.0\ \text{T}$ field. Find its terminal speed and the heating rate in the resistor.
- At terminal speed the magnetic force equals the weight: $\dfrac{B^2L^2v_T}{R} = mg$, so $v_T = \dfrac{(0.050)(9.8)(1.0)}{(1.0)^2(0.50)^2} = 1.96\ \text{m/s}$.
- Heating: $P = mgv_T = (0.050)(9.8)(1.96) = 0.960\ \text{W}$. All the gravitational energy lost each second becomes heat.
Answer: $v_T = 1.96\ \text{m/s}$
Common mistake
"The induced current pushes the rod along." If it did, the rod would speed up, inducing more current and more push, gaining energy from nowhere. Lenz's law (and energy conservation) says the force opposes the motion. Check the direction with the right-hand rule: the force $I\vec{L}\times\vec{B}$ on the induced current always points against $\vec{v}$. This is how eddy-current brakes on some trains and amusement rides slow them smoothly without touching anything.