Irodov 3.313: Heat from a Parabolic Flux – Amit & Deepali Sinha Foundation

Problem Irodov 3.313

A stationary wire loop has resistance $R$ . During a time interval $\tau$ the magnetic flux through it changes as $\Phi = at(\tau - t)$ , where $t$ runs from $0$ to $\tau$ and $a$ is a constant. Neglecting the loop’s inductance, find the heat generated in the loop over this interval.

Given

loop resistance $R$
$\Phi = at(\tau - t)$ , $0 \le t \le \tau$
inductance neglected

Find

the heat

Q

generated during

\tau

Solution

Step 1The induced emf

The given flux through the loop, a parabola that rises from zero and returns to zero at t = τ, peaking at t = τ/2 (drawn for a > 0). The emf is minus its slope. — The given flux through the loop, a parabola that rises from zero and returns to zero at $t = \tau$ , peaking at $t = \tau/2$ (drawn for $a > 0$ ). The emf is minus its slope.

By Faraday’s law the emf round the loop is minus the rate of change of the flux through it. Differentiating the given law:

\mathcal{E}_i = -\frac{d\Phi}{dt} = -\frac{d}{dt}\bigl(a\tau t - a t^2\bigr)

\mathcal{E}_i = -a(\tau - 2t)

(1)

The emf changes sign at $t = \tau/2$ , where the flux stops growing and starts to fall: the current reverses there.

Step 2Joule heating, instant by instant

With the inductance neglected, the current at each instant is the emf over the resistance, and the heating power is $\mathcal{E}_i^2/R$ whatever the sign of the current. Using (1):

P = \frac{\mathcal{E}_i^2}{R} = \frac{a^2(\tau - 2t)^2}{R} \tag{2}

Step 3The heat over the interval

Integrating the power (2) from $0$ to $\tau$ :

Q = \frac{a^2}{R}\int_0^{\tau}(\tau - 2t)^2\,dt

Substituting $u = \tau - 2t$ , so $dt = -du/2$ and $u$ runs from $\tau$ down to $-\tau$ :

\int_0^{\tau}(\tau - 2t)^2\,dt = \frac{1}{2}\int_{-\tau}^{\tau}u^2\,du = \frac{1}{2}\cdot\frac{2\tau^3}{3} = \frac{\tau^3}{3}

Q = \frac{a^2\tau^3}{3R}

(3)

Check

Units: $a\tau^2$ is a flux (the peak flux is $a\tau^2/4$ ), in V s, so $a^2\tau^3/R$ is $(\text{V\,s})^2/(\Omega\,\text{s})$ , which is V $^2$ s/ $\Omega$ , joules. Mean square: the emf runs linearly from $-a\tau$ to $a\tau$ , and the mean of the square of a linear ramp across a range symmetric about zero is a third of the end value squared, $a^2\tau^2/3$ ; times $\tau/R$ gives the same $Q$ .

Remark

The net change of flux is zero, so the net charge driven round the loop, $\Delta\Phi/R$ , is zero too: the charge sent one way in the first half comes back in the second. The heat, though, is not zero, because it goes with the square of the current. For a fixed peak flux $\Phi_{\mathrm{m}} = a\tau^2/4$ , $Q = 16\Phi_{\mathrm{m}}^2/(3R\tau)$ grows as the same excursion is made faster.