Irodov 3.292: Half Disc Turning Through the Edge – Amit & Deepali Sinha Foundation

Problem Irodov 3.292

A uniform magnetic field of induction $B$ , directed into the page, fills the half-plane to the right of a straight boundary line; there is no field to the left of it. A wire loop shaped as a half disc of radius $a$ (a semicircular arc closed by its diameter) lies in the plane of the page on the boundary of the field, with its center $O$ on the boundary line. At $t = 0$ the loop is set turning clockwise about the axis through $O$ perpendicular to the page (a line of $\vect B$ on the boundary), with a constant angular acceleration $\beta$ . Taking the clockwise sense round the loop as the positive sense of the emf, find the emf induced in the loop as a function of time $t$ , and sketch its graph.

The half-disc loop of radius a centered on the axis O, which lies on the boundary of the field B (into the page, right of the dashed line only). The loop turns clockwise; φ is its angle from the boundary. The arrow on the arc marks the positive sense of the emf. — The half-disc loop of radius $a$ centered on the axis $O$ , which lies on the boundary of the field $\vect B$ (into the page, right of the dashed line only). The loop turns clockwise; $\varphi$ is its angle from the boundary. The arrow on the arc marks the positive sense of the emf.

Given

$a$
$B$ into the page, right of the boundary
$O$ on the boundary
from rest at $t = 0$ , loop outside the field (assumed)
angular acceleration $\beta$
positive sense clockwise

Find

\mathcal{E}_i(t)

and its graph

Solution

Step 1How far the loop has turned

The loop starts from rest at $t = 0$ and turns with constant angular acceleration $\beta$ , so its angle from the starting position and its angular velocity are

\varphi = \frac{\beta t^2}{2}, \qquad \omega = \frac{d\varphi}{dt} = \beta t \tag{1}

The starting position is taken with the diameter along the boundary and the half disc outside the field (assumed), so that the angle $\varphi$ is measured from the boundary, as drawn, and the loop enters the field during the first half turn.

During the first half turn the part of the loop inside the field is the sector S between the boundary above O and the half diameter, of angle φ; it grows as the loop turns. — During the first half turn the part of the loop inside the field is the sector $S$ between the boundary above $O$ and the half diameter, of angle $\varphi$ ; it grows as the loop turns.

Step 2The area inside the field

Because $O$ is the center of the half disc and lies on the boundary, the part of the loop inside the field is always a circular sector with its vertex at $O$ . During the first half turn ( $0 < \varphi < \pi$ ) the loop enters the field and the sector has angle $\varphi$ ; during the second ( $\pi < \varphi < 2\pi$ ) it leaves, and the sector has angle $2\pi - \varphi$ :

S = \frac{a^2\varphi}{2} \quad\text{(entering)}, \qquad S = \frac{a^2(2\pi - \varphi)}{2} \quad\text{(leaving)}

The pattern repeats every turn. In either case the area changes at the same rate, with the sign that alternates every half turn, using (1):

\frac{dS}{dt} = \pm\,\frac{a^2}{2}\,\frac{d\varphi}{dt} = \pm\,\frac{a^2\beta t}{2} \tag{2}

with the upper sign while the loop enters the field (the first, third, … half turn) and the lower sign while it leaves.

Step 3Faraday’s law with the given positive sense

The positive sense is clockwise, and by the right-hand rule a clockwise circulation has its normal into the page, along $\vect B$ . So the flux counted with that normal is $\Phi = BS$ , and the emf in the positive sense is $\mathcal{E}_i = -d\Phi/dt$ . With (2), during the $n$ -th half turn

\mathcal{E}_i = -B\,\frac{dS}{dt} = \mp\,\frac{Ba^2\beta t}{2}

The first half turn ( $n = 1$ ) has the minus sign, the second the plus sign, and so on:

\mathcal{E}_i = (-1)^n\,\frac{Ba^2\beta t}{2}

(3)

Here $n = 1, 2, \ldots$ numbers the half turn in progress at time $t$ . The $n$ -th half turn ends when $\varphi = n\pi$ , which by (1) is at

t_n = \sqrt{\frac{2\pi n}{\beta}}

While the loop enters the field the flux into the page grows, and Lenz’s rule makes the induced current counter-clockwise, against the positive sense: that is the minus sign of the first half turn.

Step 4The graph

The magnitude of the emf grows in proportion to $t$ , and its sign flips at every $t_n$ . The instants $t_n$ crowd together as $\sqrt{n}$ , because the loop spins ever faster.

The emf against time: magnitude growing in proportion to t, negative during the first half turn, then alternating in sign, with a jump at each t_n = (2π n/β)^1/2; the t_n crowd together as the loop speeds up. — The emf against time: magnitude growing in proportion to $t$ , negative during the first half turn, then alternating in sign, with a jump at each $t_n = (2\pi n/\beta)^{1/2}$ ; the $t_n$ crowd together as the loop speeds up.

Check

Dimensions: $Ba^2\beta t$ is $\text{T}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{s} = \text{Wb/s} = \text{V}$ , as (3) requires. Over a whole turn the flux returns to its starting value, so the emf must integrate to zero over the turn; with $|\mathcal{E}_i|$ growing as $t$ and the second half turn shorter than the first, the two halves cancel exactly:

\int_0^{t_2}\mathcal{E}_i\,dt = \frac{Ba^2\beta}{2}\left(-\frac{t_1^2}{2} + \frac{t_2^2 - t_1^2}{2}\right) = \frac{Ba^2\beta}{4}\left(-\frac{2\pi}{\beta} + \frac{2\pi}{\beta}\right) = 0

Remark

The jumps at the $t_n$ are an idealisation of the sharp field boundary: in a real magnet the field fades over a fringe region, and each jump becomes a steep but continuous swing. The same loop turning at a constant $\omega$ would give a square wave of fixed height $Ba^2\omega/2$ : a crude alternator, whose output here grows because the rotation keeps speeding up.