Irodov 3.223: Radial Wires Add Nothing at the Center

Problem Irodov 3.223

A closed wire loop carries a current $I$ clockwise around a point $O$ . Find the magnetic induction at $O$ for two shapes of the loop.

The loop is an arc of radius $b$ centered on $O$ spanning the angle $\varphi$ , joined by two radial straight pieces to an arc of radius $a < b$ , also centered on $O$ , that covers the remaining angle $2\pi - \varphi$ . The radii $a$ , $b$ and the angle $\varphi$ are known.
The loop is an arc of radius $a$ centered on $O$ covering three quarters of a full turn, closed by a square of side $b$ that has $O$ at one corner: the two sides of the square that meet at $O$ run from the ends of the arc outward, and the two far sides complete the loop. The radius $a$ and the side $b$ are known.

(a) The arc of radius b over the angle φ and the arc of radius a over the rest, joined by two radial pieces; the current runs clockwise. — (a) The arc of radius $b$ over the angle $\varphi$ and the arc of radius $a$ over the rest, joined by two radial pieces; the current runs clockwise.

(b) The three-quarter arc of radius a and the square of side b with O at its corner; the current runs clockwise. — (b) The three-quarter arc of radius $a$ and the square of side $b$ with $O$ at its corner; the current runs clockwise.

Given

$I$
(a) $a$ , $b$ , $\varphi$
(b) $a$ , $b$

Find

B

O

in each case

Solution

Step 1What each kind of piece contributes

The Biot-Savart law gives the field of a current element $I\,d\vect l$ at a point a displacement $\vect r$ away.

d\vect B = \frac{\mu_0 I}{4\pi}\,\frac{d\vect l\times\vect r}{r^3} \tag{1}

On a straight piece that lies along a line through $O$ , every $d\vect l$ is parallel to its $\vect r$ , so the cross product vanishes and the piece adds nothing. On an arc of radius $R$ centered on $O$ , every $d\vect l$ is perpendicular to $\vect r$ and every element pushes $\vect B$ the same way, so the magnitudes simply add. With $dl = R\,d\theta$ over an arc of angle $\theta$ ,

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

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

(2)

(a) Every piece of the clockwise loop gives B into the page at O (the cross), or nothing. — (a) Every piece of the clockwise loop gives $\vect B$ into the page at $O$ (the cross), or nothing.

(b) The same for the arc and the square: B at O is into the page. — (b) The same for the arc and the square: $\vect B$ at $O$ is into the page.

Step 2Loop (a): two arcs, two radial pieces

Both straight pieces in (a) are radial, so only the two arcs count. The outer arc has $\theta = \varphi$ at $R = b$ ; the inner one has $\theta = 2\pi - \varphi$ at $R = a$ . Adding the two values of (2):

B = \frac{\mu_0 I}{4\pi}\,\frac{\varphi}{b} + \frac{\mu_0 I}{4\pi}\,\frac{2\pi - \varphi}{a}

B = \frac{\mu_0 I}{4\pi}\left(\frac{2\pi - \varphi}{a} + \frac{\varphi}{b}\right)

(3)

Step 3Loop (b): the arc

In (b) the two sides of the square that meet at $O$ are radial and add nothing. The arc of radius $a$ covers three quarters of a turn, $\theta = 3\pi/2$ , so by (2)

B_{\text{arc}} = \frac{\mu_0 I}{4\pi}\,\frac{3\pi/2}{a} = \frac{\mu_0 I}{4\pi}\,\frac{3\pi}{2a} \tag{4}

Step 4Loop (b): one far side of the square

The top side seen from O: the perpendicular b from O lands at the side's near end, so the side subtends angles from 0 to 45° about it; an element dx and its r. — The top side seen from $O$ : the perpendicular $b$ from $O$ lands at the side’s near end, so the side subtends angles from $0$ to $45^\circ$ about it; an element $dx$ and its $\vect r$ .

Measure $x$ along the top side from its near end, the foot of the perpendicular from $O$ . The element $dx$ sits at distance $r = (x^2 + b^2)^{1/2}$ from $O$ , and the sine of the angle between $d\vect l$ and $\vect r$ is $b/r$ , so (1) gives

dB = \frac{\mu_0 I}{4\pi}\,\frac{b\,dx}{(x^2 + b^2)^{3/2}}

Integrating over the side, from $x = 0$ to $x = b$ :

\begin{aligned} B_{\text{side}} &= \frac{\mu_0 I}{4\pi}\int_0^{b}\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]_0^{b}\\ &= \frac{\mu_0 I}{4\pi}\,\frac{1}{b}\cdot\frac{b}{\sqrt{2}\,b} \end{aligned}

B_{\text{side}} = \frac{\mu_0 I}{4\pi}\,\frac{\sqrt 2}{2b}

(5)

The right side of the square is the mirror image of the top side in the square’s diagonal through $O$ , so it sits at the same distance and subtends the same angles and gives the same value, and its current (downward, with $O$ to its left) also points $\vect B$ into the page.

Step 5Loop (b): the total

Adding the arc (4) and two far sides (5):

B = \frac{\mu_0 I}{4\pi}\,\frac{3\pi}{2a} + 2\cdot\frac{\mu_0 I}{4\pi}\,\frac{\sqrt 2}{2b}

B = \frac{\mu_0 I}{4\pi}\left(\frac{3\pi}{2a} + \frac{\sqrt 2}{b}\right)

(6)

Check

In (a), $\varphi \to 2\pi$ leaves a full circle of radius $b$ , and (3) gives $(\mu_0 I/4\pi)(2\pi/b) = \mu_0 I/(2b)$ , the field at the center of a circular loop; $\varphi \to 0$ gives the same with $a$ . Setting $\varphi = \pi/2$ in (3) gives the arc term $(2\pi - \pi/2)/a = 3\pi/(2a)$ , which is exactly the three-quarter arc of radius $a$ in (b), so the two parts agree on their common piece.

Remark

The factor in the arc term is worth a second look: the three-quarter arc contributes $3\pi/(2a)$ , while $3\pi/(4a)$ would belong to an arc of $3\pi/4$ , only three eighths of a circle; the check above, with part (a) at $\varphi = \pi/2$ , settles it. More generally, (2) says an arc centered on $O$ counts only through the angle it subtends over its radius, and a radial wire counts not at all, so any loop built from such pieces is read off by inspection.