Irodov 3.222: Arc and Chord Pull Together – Amit & Deepali Sinha Foundation

Problem Irodov 3.222

A thin closed wire consists of a circular arc of radius $R = 120$ mm centered on a point $O$ and the straight chord joining the arc’s ends; the chord subtends the angle $2\varphi = 90^\circ$ at $O$ , so the arc covers the remaining $360^\circ - 2\varphi$ . A current $I = 5.0$ A circulates around the wire. Find the magnetic induction at $O$ .

The wire: an arc of radius R about O closed by a chord that subtends 2φ at O (dashed radii); the current I circulates clockwise. — The wire: an arc of radius $R$ about $O$ closed by a chord that subtends $2\varphi$ at $O$ (dashed radii); the current $I$ circulates clockwise.

Given

$I = 5.0$ A
$R = 120$ mm
$2\varphi = 90^\circ$
$\mu_0 = 4\pi\times10^{-7}$ H/m

Find

B

O

Solution

Step 1The arc

By the Biot-Savart law an element $I\,d\vect l$ gives $dB = (\mu_0 I/4\pi)\,dl\,\sin\theta/r^2$ , $\theta$ being the angle between $d\vect l$ and $\vect r$ . On the arc every element is at $r = R$ with $\theta = 90^\circ$ , and all of them push $\vect B$ the same way at $O$ , so with $dl = R\,d\alpha$ over the angle $2\pi - 2\varphi$ the arc covers,

B_{\text{arc}} = \frac{\mu_0 I}{4\pi}\int_0^{2\pi - 2\varphi}\frac{R\,d\alpha}{R^2}

B_{\text{arc}} = \frac{\mu_0 I}{4\pi R}\,(2\pi - 2\varphi)

(1)

Step 2The chord

The chord seen from O: the perpendicular from O to it has length b = Rφ, and its ends are seen at φ on either side of the perpendicular (one side labelled). — The chord seen from $O$ : the perpendicular from $O$ to it has length $b = R\cos\varphi$ , and its ends are seen at $\varphi$ on either side of the perpendicular (one side labelled).

The chord is a straight wire at perpendicular distance $b = R\cos\varphi$ from $O$ . Measure $x$ along it from the foot of the perpendicular; an element $dx$ is at $r = (x^2 + b^2)^{1/2}$ with $\sin\theta = b/r$ , and the ends are at $x = \pm b\tan\varphi$ :

\begin{aligned} B_{\text{chord}} &= \frac{\mu_0 I}{4\pi}\int_{-b\tan\varphi}^{b\tan\varphi}\frac{b\,dx}{(x^2 + b^2)^{3/2}} = \frac{\mu_0 I}{4\pi}\left[\frac{x}{b\,(x^2 + b^2)^{1/2}}\right]_{-b\tan\varphi}^{b\tan\varphi}\\ &= \frac{\mu_0 I}{4\pi}\,\frac{2\sin\varphi}{b} = \frac{\mu_0 I}{4\pi}\,\frac{2\sin\varphi}{R\cos\varphi} \end{aligned}

B_{\text{chord}} = \frac{\mu_0 I}{4\pi R}\,2\tan\varphi

(2)

Step 3Arc and chord together

Arc and chord both give B into the page at O for the clockwise current (the cross). — Arc and chord both give $\vect B$ into the page at $O$ for the clockwise current (the cross).

The current circulates clockwise round $O$ along both the arc and the chord, so by the right-hand rule both fields point into the page and their magnitudes (1) and (2) add:

B = \frac{\mu_0 I}{4\pi R}\,(2\pi - 2\varphi + 2\tan\varphi)

Here $\varphi = \pi/4$ and $\tan\varphi = 1$ , so numerically

B = \frac{4\pi\times10^{-7}\times5.0}{2\pi\times0.120}\left(\pi - \frac{\pi}{4} + 1\right)\ \text{T}

B = \frac{\mu_0 I}{2\pi R}\,(\pi - \varphi + \tan\varphi) = 28\ \mu\text{T}

(3)

$\vect B$ points into the page for the clockwise current shown.

Check

At $\varphi = 0$ the chord shrinks to a point and the arc becomes a full circle. Then (3) gives $(\mu_0 I/2\pi R)\,\pi = \mu_0 I/(2R)$ , which is the field at the center of a circular loop. At $\varphi = \pi/2$ the wire is a semicircle closed by its diameter, which passes through $O$ ; the chord term $\tan\varphi$ then diverges, as it must for a point lying on a current-carrying wire.

Remark

The chord does more than its length suggests: at $2\varphi = 90^\circ$ it is only $0.23$ of the wire’s length but gives $\tan\varphi/(\pi - \varphi + \tan\varphi) = 0.30$ of the field, because it passes closer to $O$ ( $R\cos\varphi \approx 0.71R$ ) than the arc does.