Irodov 3.296: Gravity Against the Magnetic Brake – Amit & Deepali Sinha Foundation

Problem Irodov 3.296

Two smooth parallel metal bars a distance $l$ apart lie in a plane inclined at an angle $\alpha$ to the horizontal, running down the slope, and are joined at their upper ends through a resistor $R$ . A metal rod of mass $m$ lies across the bars and slides down them under gravity, keeping contact with both. A uniform magnetic field of induction $B$ is directed perpendicular to the inclined plane. The resistances of the bars, the rod and the contacts are negligible, and so is the self-inductance of the circuit. Find the speed the rod settles to.

The bars inclined at α, joined at the top through R, with the connector of mass m sliding down at v under its weight m g. The field B is normal to the bars' plane. — The bars inclined at $\alpha$ , joined at the top through $R$ , with the connector of mass $m$ sliding down at $\vect v$ under its weight $m\vect g$ . The field $\vect B$ is normal to the bars’ plane.

Given

$m$
$\alpha$
$R$
$l$
$B$ normal to the plane
smooth bars

Find

v

in the steady state

Solution

Step 1The emf and the current

The connector, the two bars and the resistor form a closed circuit whose area grows as the connector slides. The connector moves at $\vect v$ perpendicular both to its own length $\vect l$ and to $\vect B$ , so the motional emf across it is

\mathcal{E} = (\vect v\times\vect B)\cdot\vect l = Blv

With every other resistance negligible, the whole of it drives current through $R$ .

I = \frac{\mathcal{E}}{R} = \frac{Blv}{R} \tag{1}

The induced current I round the circuit through R, and the Ampere force F it brings on the connector, pointing up the slope against the motion. — The induced current $I$ round the circuit through $R$ , and the Ampere force $\vect F$ it brings on the connector, pointing up the slope against the motion.

Step 2The magnetic brake

The current in the connector sits in the field, so the connector feels the Ampere force. Since $\vect B$ is perpendicular to the connector, the force lies in the inclined plane along the bars, and by Lenz’s rule it opposes the motion that produces it, so it points up the slope. Its magnitude follows with (1):

\begin{aligned} \vect F &= I\,\vect l\times\vect B \\ F &= IlB = \frac{B^2l^2v}{R} \end{aligned} \tag{2}

Step 3Newton’s law along the slope

Along the bars only two forces act: the weight’s component $mg\sin\alpha$ down the slope and the magnetic force (2) up it. The normal reaction of the smooth bars and the weight’s normal component balance and drop out. Taking down the slope as positive:

m\frac{dv}{dt} = mg\sin\alpha - \frac{B^2l^2}{R}\,v \tag{3}

The right-hand side falls as $v$ rises, so the acceleration dies away and the speed approaches the value at which it vanishes.

Step 4The steady state

In the steady state $dv/dt = 0$ , and (3) becomes a balance of the two forces:

mg\sin\alpha = \frac{B^2l^2}{R}\,v

Solving for $v$ :

v = \frac{mgR\sin\alpha}{B^2l^2}

(4)

Check

Energy gives the same answer by another route. At constant speed the kinetic energy is fixed, so the power the weight delivers, $mg\sin\alpha\,v$ , must all appear as Joule heat in $R$ . With (1):

mg\sin\alpha\,v = I^2R = \frac{B^2l^2v^2}{R},

whose nonzero root is (4). The limits behave: a horizontal track ( $\alpha = 0$ ) gives no drive and $v = 0$ ; a vanishing field gives an unbounded speed, the free slide on a smooth incline that never settles.

Remark

Equation (3) has the solution $v(t) = v_\infty\left(1 - e^{-t/\tau}\right)$ from rest, with $v_\infty$ the answer and $\tau = mR/(B^2l^2)$ , the same exponential approach as a body falling through a viscous fluid. A larger $R$ means a weaker current, a weaker brake and a higher terminal speed. In the companion arrangement with a capacitor in place of $R$ the current is set by the acceleration rather than the speed, and the connector instead slides with a constant, reduced acceleration.