Irodov 3.310: Charges Fix an Unequal Ring – Amit & Deepali Sinha Foundation

Problem Irodov 3.310

A thin insulated wire ring is fitted tightly over the winding of a long solenoid of cross-sectional radius $a$ . One half of the ring has $\eta$ times the resistance of the other half. The magnetic induction inside the solenoid changes with time as $B = bt$ , with $b$ a constant. Find the magnitude of the electric field strength in the ring.

Cross-section of the solenoid, radius a, with the ring on its winding: two halves of resistance R and η R joined at the dots. B = b t is drawn for b > 0; its sense is not given, drawn out of the page. — Cross-section of the solenoid, radius $a$ , with the ring on its winding: two halves of resistance $R$ and $\eta R$ joined at the dots. $\vect B = b\,t$ is drawn for $b > 0$ ; its sense is not given, drawn out of the page.

Given

solenoid radius $a$
ring tight on the winding
resistances of the halves in the ratio $\eta$
$B = bt$
each half uniform along its length (assumed)

Find

the electric field strength

E

in the ring

Solution

Step 1The eddy field and the emf

By symmetry the eddy field of the solenoid is tangential and the same all round a coaxial circle, so Faraday’s law on the ring’s circle of radius $a$ , through which the whole flux $\pi a^2 B$ passes, gives

E_{\mathrm{ed}}\cdot 2\pi a = \frac{d}{dt}\bigl(\pi a^2 bt\bigr) = \pi a^2 b

\begin{gathered} E_{\mathrm{ed}} = \frac{ab}{2}\\[4pt] \mathcal{E}_i = \pi a^2 b \end{gathered} \tag{1}

This eddy field is the same at every point of the ring, whatever the ring is made of.

Step 2The current

Let $R$ be the resistance of one half; the other half has $\eta R$ . The emf drives one current round the series loop:

I = \frac{\mathcal{E}_i}{R + \eta R} = \frac{\pi a^2 b}{(1 + \eta)R} \tag{2}

Step 3The total field in each half

In each half, uniform along its length $\pi a$ (assumed), Ohm’s law ties the field along the wire to the drop $IR$ or $\eta IR$ across it. Using (2):

\begin{gathered} E_1 = \frac{IR}{\pi a} = \frac{\pi a^2 b}{(1 + \eta)\pi a}\\[4pt] E_2 = \frac{\eta IR}{\pi a} = \frac{\eta\,\pi a^2 b}{(1 + \eta)\pi a} \end{gathered}

\begin{gathered} E_1 = \frac{ab}{1 + \eta}\\[4pt] E_2 = \frac{\eta\, ab}{1 + \eta} \end{gathered}

(3)

Both point along the current. They differ, while the eddy field (1) is the same in both halves: the difference must come from another field, that of charges.

Drawn for η > 1: the current I (clockwise, Lenz), the sign of the surface charge it leaves near each junction, and the charges' field E_q, against I in the half R and along it in the half η R. — Drawn for $\eta > 1$ : the current $I$ (clockwise, Lenz), the sign of the surface charge it leaves near each junction, and the charges’ field $\vect E_q$ , against $I$ in the half $R$ and along it in the half $\eta R$ .

Step 4The field of the charges

The current leaves charge on the surface of the wire, positive toward one junction and negative toward the other, until its field, added to the eddy field, makes the field in each half what (3) requires. Subtracting (1) from the total, along the current:

\begin{gathered} E_{q1} = \frac{ab}{1 + \eta} - \frac{ab}{2} = -\frac{ab}{2}\,\frac{\eta - 1}{\eta + 1}\\[4pt] E_{q2} = \frac{\eta\,ab}{1 + \eta} - \frac{ab}{2} = \frac{ab}{2}\,\frac{\eta - 1}{\eta + 1} \end{gathered}

The two have the same magnitude and opposite signs along the current; in space they point the same way, from the positive junction to the negative one.

E_q = \frac{ab}{2}\,\frac{\eta - 1}{\eta + 1}

(4)

This is the electrostatic field in the ring, the same in magnitude in both halves. For $\eta > 1$ the junction where the current enters the half $\eta R$ is the positive one. For $\eta < 1$ the two junctions swap signs and the expression turns negative; the magnitude is then $ab\,|\eta - 1|/\bigl(2(\eta + 1)\bigr)$ .

Check

The charges’ field is a potential field, so its circulation round the closed ring must vanish: $E_{q1}\pi a + E_{q2}\pi a = 0$ , as it does. For $\eta = 1$ the ring is uniform, $E_q = 0$ , and the total field is the eddy field $ab/2$ everywhere, as (3) also gives. For $\eta \to \infty$ , $E_q \to ab/2$ : the field in the good half nearly cancels, $E_1 \to 0$ , while the poor half carries the whole emf.

Remark

A voltmeter connected across the two junctions does not read a unique voltage: the potential difference of the charges, $E_q\pi a$ , is a single number, but the eddy field adds its own contribution along the meter’s leads, which depends on how they run round the solenoid. In a changing magnetic field “the voltage between two points” needs a path.