Irodov 3.302: A Brake Proportional to Speed – Amit & Deepali Sinha Foundation

Problem Irodov 3.302

Two long parallel conducting rails a distance $l$ apart are joined at their left ends through a resistor $R$ . A conducting rod $AB$ of mass $m$ lies across them and can slide along them without friction. A uniform magnetic field of induction $B$ is directed perpendicular to the plane of the rails, into the page. At $t = 0$ the rod is set moving to the right with velocity $v_0$ and is then left to itself. The resistances of the rails and the rod, and the self-inductance of the circuit, are negligible. Find:

the distance the rod travels before it stops;
the heat released in the resistor $R$ meanwhile.

The rails joined at the left through R, the rod AB launched to the right at v_0, and the uniform field B into the page. — The rails joined at the left through $R$ , the rod $AB$ launched to the right at $\vect v_0$ , and the uniform field $\vect B$ into the page.

Given

$m$
$l$
$R$
$B$
$v_0$ at $t = 0$
no friction

Find

(a)

s

· (b)

Q

Solution

Step 1The emf and the current

At speed $v$ the rod cuts the field and carries a motional emf; $\vect v$ , $\vect B$ and the rod are mutually perpendicular. The only resistance in the circuit is $R$ , so Ohm’s law gives the current:

\begin{aligned} \mathcal{E}_i &= (\vect v\times\vect B)\cdot\vect l = Blv \\ I &= \frac{\mathcal{E}_i}{R} = \frac{Blv}{R} \end{aligned} \tag{1}

With $\vect v$ to the right and $\vect B$ into the page, $\vect v\times\vect B$ points from $B$ to $A$ , so the current runs up the rod and counter-clockwise round the circuit.

The rod at speed v: the induced current I, from B to A in the rod and counter-clockwise round the circuit, and the Ampere force F on the rod, against v. — The rod at speed $v$ : the induced current $I$ , from $B$ to $A$ in the rod and counter-clockwise round the circuit, and the Ampere force $\vect F$ on the rod, against $\vect v$ .

Step 2The equation of motion

The current in the rod feels the Ampere force $I\vect l\times\vect B$ , of magnitude $IlB$ , directed to the left, against the motion. It is the only horizontal force, so with (1):

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

Step 3(a) The distance to the stop

The distance comes most directly by trading time for distance. Writing $dv/dt = v\,dv/dx$ in (2) and cancelling one factor $v$ :

\begin{aligned} m v\frac{dv}{dx} &= -\frac{B^2l^2}{R}\,v \\ m\,dv &= -\frac{B^2l^2}{R}\,dx \end{aligned}

Integrating from the launch, $x = 0$ and $v = v_0$ , to the stop, $x = s$ and $v = 0$ :

m(0 - v_0) = -\frac{B^2l^2}{R}\,s

s = \frac{m v_0 R}{B^2 l^2}

(3)

The speed falls linearly with the distance covered, $v = v_0(1 - x/s)$ .

Step 4(b) The heat

The rails are smooth and nothing else does work, so all the kinetic energy the rod loses is turned by the current into Joule heat in $R$ , the only resistance. The rod ends at rest, so:

Q = \frac{m v_0^2}{2}

(4)

Check

The heat directly. Integrating (2) in time gives $v = v_0 e^{-t/\tau}$ with $\tau = mR/(B^2l^2)$ . The power in $R$ is $I^2R = B^2l^2v^2/R$ by (1), so

Q = \int_0^\infty \frac{B^2l^2v_0^2}{R}\,e^{-2t/\tau}\,dt = \frac{B^2l^2v_0^2}{R}\cdot\frac{\tau}{2} = \frac{m v_0^2}{2},

which is (4). The distance is also $\int_0^\infty v\,dt = v_0\tau$ , which is (3).

Remark

The stopping distance is proportional to $v_0$ , not to $v_0^2$ as for a constant friction force: each unit of distance removes the same momentum, $B^2l^2\,dx/R$ , whatever the speed, because the drag is proportional to the speed. The magnetic brake removes momentum at a fixed rate per metre; a larger resistance weakens the current and lets the rod coast further.