Irodov 3.298: Half a Loop Makes an Alternator – Amit & Deepali Sinha Foundation

Problem Irodov 3.298

A wire bent into a semicircle of radius $a$ has its two ends on a straight axis $OO'$ , which runs along the semicircle’s diameter, and it is closed through a fixed circuit lying on the other side of the axis. The semicircle rotates about $OO'$ with angular velocity $\omega$ in a uniform magnetic field of induction $B$ perpendicular to the axis, while the rest of the circuit stays at rest. The total resistance of the circuit is $R$ . Neglecting the magnetic field of the induced current, find the mean thermal power generated in the circuit over one rotation period.

The semicircular wire of radius a turning at ω about the axis OO' through its ends, and the fixed circuit with resistance R that closes it. The field B, into the page, is perpendicular to the axis. — The semicircular wire of radius $a$ turning at $\omega$ about the axis $OO'$ through its ends, and the fixed circuit with resistance $R$ that closes it. The field $\vect B$ , into the page, is perpendicular to the axis.

Given

$a$
$\omega$
$B$ perpendicular to $OO'$
$R$ (whole circuit)
induced field neglected

Find

\langle P\rangle

over a period

Solution

Step 1The flux through the circuit

The fixed part of the circuit encloses a constant flux $\Phi_0$ ; only the half disc bounded by the rotating semicircle and the axis changes its orientation. Count the time from the position drawn, where the semicircle lies in the plane of the circuit and its normal is along $\vect B$ . After a time $t$ it has turned by $\omega t$ , and its flux is its area $\pi a^2/2$ times the component of $\vect B$ along its normal:

\Phi = \Phi_0 + \frac{\pi a^2}{2}\,B\cos\omega t \tag{1}

Seen along the axis OO': the semicircle edge on, turned by ω t out of the plane of the fixed circuit (dashed), so that its normal n makes the angle ω t with B. — Seen along the axis $OO'$ : the semicircle edge on, turned by $\omega t$ out of the plane of the fixed circuit (dashed), so that its normal $\vect n$ makes the angle $\omega t$ with $\vect B$ .

Step 2The emf and the current

Faraday’s law applied to (1); the constant $\Phi_0$ drops out:

\mathcal{E}_i = -\frac{d\Phi}{dt} = \frac{\pi a^2 B\omega}{2}\,\sin\omega t \tag{2}

With the field of the induced current neglected, the circuit’s inductance plays no part, and the current is $\mathcal{E}_i/R$ at every instant.

Step 3The mean Joule power

The instantaneous power dissipated in the circuit is $\mathcal{E}_i^2/R$ . Over a period the mean of $\sin^2\omega t$ is $1/2$ , so with (2)

\langle P\rangle = \frac{1}{R}\left(\frac{\pi a^2B\omega}{2}\right)^2\left\langle\sin^2\omega t\right\rangle = \frac{1}{R}\left(\frac{\pi a^2B\omega}{2}\right)^2\cdot\frac{1}{2}

\langle P\rangle = \frac{\left(\pi\omega a^2B\right)^2}{8R}

(3)

Check

Dimensions: $\omega a^2B$ is $\text{s}^{-1}\cdot\text{m}^2\cdot\text{T} = \text{Wb/s} = \text{V}$ , and $\text{V}^2/\Omega = \text{W}$ . The emf amplitude in (2) is the general one for a flat loop turning about an axis in its plane perpendicular to $\vect B$ , $BS\omega$ , with $S = \pi a^2/2$ the area of the half disc. The same mean power is the mean mechanical power needed to keep the semicircle turning against the Ampere forces on its current, as energy conservation requires.

Remark

The mean power goes as the square of the swept area: a full circular loop of radius $a$ turning the same way would dissipate four times as much, $(\pi\omega a^2B)^2/2R$ . Neglecting the induced field is the condition $\omega L \ll R$ for the circuit’s inductance $L$ ; at higher frequencies the inductance limits the current, shifts it in phase behind the emf, and the mean power falls below this value.