Irodov 3.226: Arcs, Tangents and Radial Leads – Amit & Deepali Sinha Foundation

Problem Irodov 3.226

A wire carrying a current $I$ consists of a circular arc of radius $R$ centered on a point $O$ and two very long straight parts. Find the magnetic induction at $O$ for three shapes:

a semicircle whose two ends continue as straight wires along the line through $O$ joining them, outward in opposite directions;
three quarters of a circle; the current arrives along a straight wire tangent to the circle at one end of the arc, runs round the arc, and leaves along a straight wire that continues the radius at the other end of the arc outward, away from $O$ ; the two straight wires are parallel and lie on the same side of $O$ ;
a semicircle whose two ends continue as straight wires tangent to it, both running off parallel on the same side, forming a hairpin.

(a) A semicircle of radius R about O, the straight parts on the line through O. — (a) A semicircle of radius $R$ about $O$ , the straight parts on the line through $O$ .

(b) Three quarters of a circle, a tangent wire in, a radial wire out.

(c) A semicircle with two tangent wires: a hairpin.

Given

$I$
$R$
straight parts very long

Find

B

O

in (a), (b), (c)

Solution

Step 1The three kinds of piece

By the Biot-Savart law an element $I\,d\vect l$ gives $dB = (\mu_0 I/4\pi)\,dl\sin\theta/r^2$ at a point a distance $r$ away, where $\theta$ is the angle between $d\vect l$ and $\vect r$ . A straight part on a line through $O$ has $\theta = 0$ all along it and adds nothing. On an arc of radius $R$ about $O$ , $\theta = 90^\circ$ and $r = R$ everywhere, so an arc of angle $\beta$ gives

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

B_{\text{arc}} = \frac{\mu_0 I}{4\pi}\,\frac{\beta}{R}

(1)

Step 2A half-infinite tangent wire

A tangent wire seen from O: the perpendicular from O, of length R, lands on its end; an element dx at distance x along it, and its r. The wire runs on without end. — A tangent wire seen from $O$ : the perpendicular from $O$ , of length $R$ , lands on its end; an element $dx$ at distance $x$ along it, and its $\vect r$ . The wire runs on without end.

A straight wire tangent to the circle at the end of the arc has the perpendicular from $O$ , of length $R$ , landing on its end. Measuring $x$ from that end, an element $dx$ is at $r = (x^2 + R^2)^{1/2}$ with $\sin\theta = R/r$ , and $x$ runs to infinity:

B_{\text{tan}} = \frac{\mu_0 I}{4\pi}\int_0^{\infty}\frac{R\,dx}{(x^2 + R^2)^{3/2}} = \frac{\mu_0 I}{4\pi}\left[\frac{x}{R\,(x^2 + R^2)^{1/2}}\right]_0^{\infty}

B_{\text{tan}} = \frac{\mu_0 I}{4\pi R}

(2)

which is half the field of an infinite wire at distance $R$ , as it must be: the other half of that wire would contribute equally.

Step 3Adding up the three shapes

(a) B at O is into the page. — (a) $\vect B$ at $O$ is into the page.

(c) Into the page: every piece that contributes drives B the same way. — (c) Into the page: every piece that contributes drives $\vect B$ the same way.

In each shape the current goes clockwise round $O$ along every piece that contributes, so every contribution points into the page (right-hand rule) and the magnitudes from (1) and (2) add.

(a) The straight parts are radial; only the semicircle counts, and (1) with $\beta = \pi$ gives

B_a = \frac{\mu_0}{4\pi}\,\frac{\pi I}{R}

(3)

(b) The outgoing wire is radial and adds nothing; the arc has $\beta = 3\pi/2$ , and the incoming wire is a tangent wire:

B_b = \frac{\mu_0 I}{4\pi}\,\frac{3\pi/2}{R} + \frac{\mu_0 I}{4\pi R}

B_b = \frac{\mu_0}{4\pi}\left(1 + \frac{3\pi}{2}\right)\frac{I}{R}

(4)

B_c = \frac{\mu_0 I}{4\pi}\,\frac{\pi}{R} + 2\cdot\frac{\mu_0 I}{4\pi R}

B_c = \frac{\mu_0}{4\pi}\,(2 + \pi)\,\frac{I}{R}

(5)

In all three $\vect B$ is normal to the plane of the wire, into the page for the currents drawn.

Check

The two building blocks reproduce the known fields. A full circle, $\beta = 2\pi$ in (1), gives $\mu_0 I/(2R)$ , the field at the center of a loop; two half-infinite wires joined end to end at the foot of the perpendicular make a complete straight wire, and twice (2) is $\mu_0 I/(2\pi R)$ , the field of a long wire at distance $R$ . In units of $\mu_0 I/(4\pi R)$ the three answers are $3.14$ , $5.71$ and $5.14$ ; (b) is the largest because its arc is the longest.

Remark

Every answer here is read off by counting: an arc of angle $\beta$ about $O$ is worth $\beta$ , a tangent half-infinite wire is worth $1$ and a radial wire $0$ , all in units of $\mu_0 I/(4\pi R)$ . A complete infinite wire at distance $R$ is worth $2$ , the same as an arc of two radians.