Irodov 3.303: Pulled Toward a Terminal Speed – Amit & Deepali Sinha Foundation

Problem Irodov 3.303

A U-shaped conductor lies in a horizontal plane: two long parallel bars joined at one end by a cross piece. A straight connector $AB$ of length $l$ , mass $m$ and resistance $R$ lies across the bars and can slide along them without friction. A uniform vertical magnetic field of induction $B$ , pointing down (into the page when the arrangement is seen from above), fills the region. From $t = 0$ a constant horizontal force $F$ , directed along the bars away from the cross piece, acts on the connector and moves it without turning. Find the connector’s velocity as a function of time $t$ . The inductance of the circuit and the resistance of the U-shaped conductor are negligible.

Top view: the U-shaped conductor closed at the left, the connector AB of resistance R across it, the constant force F pulling it to the right, and the vertical field B, drawn into the page. — Top view: the U-shaped conductor closed at the left, the connector $AB$ of resistance $R$ across it, the constant force $\vect F$ pulling it to the right, and the vertical field $\vect B$ , drawn into the page.

Given

$l$
$m$
$R$
$B$ vertical
$F = \text{const}$ from $t = 0$
at rest at $t = 0$ (assumed)
no friction

Find

v(t)

Solution

Step 1The current at speed

v

The connector is at rest when the force starts acting (assumed, since no initial velocity is given). At speed $v$ it carries the motional emf of a rod of length $l$ moving across the field, and the circuit’s only resistance is the connector’s own $R$ :

\begin{aligned} \mathcal{E}_i &= Blv \\ I &= \frac{\mathcal{E}_i}{R} = \frac{Blv}{R} \end{aligned} \tag{1}

By Lenz’s rule the current opposes the growth of the inward flux through the circuit, so it circulates counter-clockwise, upward from $B$ to $A$ in the connector.

The connector at speed v: the induced current I, counter-clockwise, and the Ampere force F_A against v, beside the applied force F. — The connector at speed $v$ : the induced current $I$ , counter-clockwise, and the Ampere force $\vect F_{\mathrm{A}}$ against $\vect v$ , beside the applied force $\vect F$ .

Step 2The equation of motion

The current in the connector feels the Ampere force $I\vect l\times\vect B$ , of magnitude $IlB$ , directed against the motion. Newton’s second law along the bars, with (1):

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

Dividing by $m$ and naming the constant rate $\alpha$ :

\begin{align*} \alpha &= \frac{B^2l^2}{mR} \\ \frac{dv}{dt} &= \frac{F}{m} - \alpha v \tag{2}\end{align*}

Step 3Solving with

v(0) = 0

Separating the variables in (2) and integrating from rest at $t = 0$ to speed $v$ at time $t$ :

\begin{aligned} \int_0^v\frac{dv'}{F/m - \alpha v'} &= \int_0^t dt' \\ -\frac{1}{\alpha}\ln\frac{F/m - \alpha v}{F/m} &= t \\ 1 - \frac{\alpha m v}{F} &= e^{-\alpha t} \end{aligned}

Solving for $v$ :

v = \frac{F}{\alpha m}\left(1 - e^{-\alpha t}\right), \quad \alpha = \frac{B^2l^2}{mR}

(3)

Step 4The graph

The speed rises from zero with the initial slope $F/m$ and approaches the terminal value $F/(\alpha m) = FR/(B^2l^2)$ , reaching $1 - 1/e$ of it at $t = 1/\alpha$ .

The connector's speed against time: a rise from rest toward the terminal speed F/α m, with time constant 1/α. — The connector’s speed against time: a rise from rest toward the terminal speed $F/\alpha m$ , with time constant $1/\alpha$ .

Check

Early times, $\alpha t \ll 1$ : $1 - e^{-\alpha t} \approx \alpha t$ , so $v \approx Ft/m$ , free acceleration before any current has built up. Late times: $v \to FR/(B^2l^2)$ , where $F$ equals the drag $\alpha m v = B^2l^2v/R$ of (2), as a steady state requires. Units: $B^2l^2/(mR)$ is $\text{T}^2\text{m}^2/(\text{kg}\,\Omega)$ ; with $\text{T} = \text{kg}/(\text{A}\,\text{s}^2)$ and $\Omega = \text{kg}\,\text{m}^2/(\text{A}^2\text{s}^3)$ it is $\text{s}^{-1}$ , a rate.

Remark

In the steady state the whole power $Fv$ of the applied force goes into Joule heat in the connector, $I^2R = B^2l^2v^2/R$ : the arrangement is a linear generator loaded by its own resistance, and $F/v = B^2l^2/R$ is the magnetic drag coefficient it presents, like a viscosity. A smaller resistance gives a stronger brake and a lower terminal speed.