Irodov 3.312: The Squeeze Peaks at Half Current – Amit & Deepali Sinha Foundation

Problem Irodov 3.312

A thin wire ring of radius $a$ and resistance $r$ sits inside a long solenoid, coaxial with it. The solenoid has length $l$ and cross-sectional radius $b$ . At some moment the solenoid is connected to a source of constant voltage $V$ ; the total resistance of that circuit is $R$ . Neglecting the ring’s inductance, find the largest value of the radial force per unit length acting on the ring.

Cross-section of the long solenoid, radius b, with the coaxial wire ring of radius a inside it, before the source is connected. — Cross-section of the long solenoid, radius $b$ , with the coaxial wire ring of radius $a$ inside it, before the source is connected.

Given

ring radius $a$ , resistance $r$
solenoid length $l$ , radius $b$
source voltage $V$
circuit resistance $R$
ring inductance neglected
solenoid current zero at the connection (assumed)

Find

the largest radial force per unit length

F_{1\,\mathrm{max}}

on the ring

Solution

Step 1The solenoid’s current and inductance

Let the solenoid have $N$ turns, so $n = N/l$ per unit length. Being long, it has the uniform inside field $B = \mu_0 n I$ (Ampere’s law on a rectangle with one side inside along the axis), and its inductance is the flux $N B\pi b^2$ per unit current:

L = \mu_0 n^2\pi b^2 l \tag{1}

Taking $t = 0$ at the connection, with no current then (assumed), the loop equation $V = RI + L\,dI/dt$ gives the familiar rise

I = \frac{V}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}

and its rate, differentiating:

L\frac{dI}{dt} = V e^{-t/\tau} \tag{2}

Step 2The ring’s current

The ring encloses the flux $\pi a^2 B$ of the uniform inside field. With its inductance neglected, its current is the emf over its resistance:

i = \frac{1}{r}\,\frac{d}{dt}\bigl(\pi a^2\mu_0 n I\bigr) = \frac{\pi a^2\mu_0 n}{r}\,\frac{dI}{dt} \tag{3}

By Lenz’s rule it circulates opposite to the solenoid’s current.

Step 3The force on unit length of the ring

Drawn for B out of the page and growing (its sense is not given): the ring's current i runs clockwise, against the winding's, and the force on each element of the ring points to the axis. — Drawn for $\vect B$ out of the page and growing (its sense is not given): the ring’s current $i$ runs clockwise, against the winding’s, and the force on each element of the ring points to the axis.

Every element of the ring carries $i$ at right angles to the axial field $B$ , so the force per unit length is $iB$ , directed along the radius. Using (3):

F_1 = iB = \frac{\pi a^2\mu_0 n}{r}\,\frac{dI}{dt}\cdot\mu_0 n I = \frac{\pi a^2\mu_0}{r}\,\bigl(\mu_0 n^2\bigr)\, I\,\frac{dI}{dt}

The unknown $n$ enters only as $\mu_0 n^2$ , which (1) replaces by $L/(\pi b^2 l)$ :

F_1 = \frac{\mu_0 a^2}{r b^2 l}\; I\cdot L\frac{dI}{dt} \tag{4}

Step 4When the force is largest

Substituting the current and (2) into (4), with $y = e^{-t/\tau}$ falling from 1 to 0:

F_1 = \frac{\mu_0 a^2}{r b^2 l}\cdot\frac{V}{R}(1 - y)\cdot V y = \frac{\mu_0 a^2 V^2}{r R b^2 l}\; y(1 - y)

The product $y(1 - y)$ is largest, $1/4$ , at $y = 1/2$ , that is at $t = \tau\ln 2$ , when the solenoid’s current has reached half its final value:

F_{1\,\mathrm{max}} = \frac{\mu_0 a^2 V^2}{4 r R b^2 l}

(5)

The force is directed toward the axis: it compresses the ring.

Check

Units: $\mu_0 a^2V^2/(rRb^2l)$ is H/m times V $^2$ over $\Omega^2$ and m, that is H $\,$ A $^2$ /m $^2$ ; since $LI^2$ is an energy, H $\,$ A $^2$ is J, and J/m $^2$ is N/m, a force per unit length. Limits: at $t = 0$ the field is still zero, and as $t \to \infty$ the field stops changing and the ring’s current dies, so in both $F_1 = 0$ , and the maximum must lie between, as it does.

Remark

The inductance, and with it the number of turns, has dropped out: a solenoid with more turns makes a stronger field but builds it more slowly, and the two effects cancel exactly in the product $I\,dI/dt$ . The time of the peak, $\tau\ln 2$ , does depend on $L$ ; only the height of the peak does not.