Irodov 3.314: A Thick Ring as Nested Loops – Amit & Deepali Sinha Foundation

Problem Irodov 3.314

A ring of square cross-section, made of a conducting material of resistivity $\rho$ , sits coaxially in the middle of a long solenoid. The ring has thickness $h$ along the axis, inner radius $a$ and outer radius $b$ . The magnetic induction inside the solenoid changes with time as $B = \beta t$ , with $\beta$ a constant. Neglecting the ring’s inductance, find the current induced in it.

Cross-section of the solenoid with the conducting ring, inner radius a, outer radius b, thickness h along the axis. B = β t is drawn for β > 0; its sense is not given, drawn out of the page. — Cross-section of the solenoid with the conducting ring, inner radius $a$ , outer radius $b$ , thickness $h$ along the axis. $\vect B = \beta t$ is drawn for $\beta > 0$ ; its sense is not given, drawn out of the page.

Given

resistivity $\rho$
thickness $h$
radii $a$ , $b$
$B = \beta t$
ring coaxial, in the middle of a long solenoid
inductance neglected

Find

the induced current

I

Solution

Step 1The eddy field inside the solenoid

By symmetry about the axis the eddy field runs along coaxial circles with the same magnitude all round each, so Faraday’s law on a circle of radius $r$ , which encloses the flux $\pi r^2\beta t$ , gives

E\cdot 2\pi r = \frac{d}{dt}\bigl(\pi r^2\beta t\bigr) = \pi r^2\beta

E = \frac{\beta r}{2} \tag{1}

Step 2Current density and an element of the ring

By Ohm’s law in local form, using (1), the current density at radius $r$ is

j = \frac{E}{\rho} = \frac{\beta r}{2\rho} \tag{2}

It grows outward: the outer layers of the ring enclose more flux per unit length of their path.

A thin coaxial element of the ring, radius r, width dr (out to the dashed circle), carrying dI; clockwise for B out of the page and growing. — A thin coaxial element of the ring, radius $r$ , width $dr$ (out to the dashed circle), carrying $dI$ ; clockwise for $\vect B$ out of the page and growing.

The element between $r$ and $r + dr$ has cross-section $h\,dr$ , so it carries

dI = j\,h\,dr = \frac{\beta h}{2\rho}\,r\,dr

Step 3The total current

Adding the elements from the inner radius to the outer one:

I = \frac{\beta h}{2\rho}\int_a^b r\,dr = \frac{\beta h}{2\rho}\cdot\frac{b^2 - a^2}{2}

I = \frac{\beta h\,(b^2 - a^2)}{4\rho}

(3)

For $\beta > 0$ the current circulates clockwise as seen with $\vect B$ pointing at the viewer (Lenz): its own field opposes the growth of $\vect B$ .

Check

Units: $\beta$ is T/s, which is V/m $^2$ ; times $h(b^2 - a^2)$ in m $^3$ and over $\rho$ in $\Omega\,$ m, it is V/ $\Omega$ , amperes. A thin ring, $b = a + \delta$ with $\delta \ll a$ , gives $I \approx \beta h\,2a\delta/(4\rho)$ , which is the emf $\pi a^2\beta$ over the resistance $\rho\,2\pi a/(h\delta)$ of a wire of section $h\delta$ , as it must. For the square section the problem names, $h = b - a$ and $I = \beta(b - a)^2(b + a)/(4\rho)$ .

Remark

The thick ring is a set of nested loops that do not exchange current, each driven by its own emf and limited by its own resistance. Here a single loop at the mean radius $(a + b)/2$ , with the whole section $h(b - a)$ , happens to give the exact current, because the current density (2) is linear in $r$ . The shortcut fails for the heat, which goes as $j^2$ : the outer layers, carrying the denser current, heat more per unit volume than the inner ones. The same picture gives the eddy currents in a solid core, and is why cores are laminated.