Irodov 3.228: Arc and Wire Fields Add at Right Angles

Problem Irodov 3.228

Take right-handed axes with $z$ vertical and $x$ , $y$ horizontal, with the origin at the point $O$ . A wire carrying a current $I = 8.0$ A contains a circular arc of radius $R = 100$ mm lying in the vertical $yz$ plane and centred on $O$ ; the current enters the arc at the point $(0, -R, 0)$ and runs up over its top. The straight parts of the wire are very long. Find the magnetic induction at $O$ for three shapes of the wire.

The current arrives along a straight wire parallel to the $x$ axis that ends at $(0, -R, 0)$ , flowing in the $-x$ direction; the arc is a half circle ending at $(0, R, 0)$ ; from there a straight wire parallel to the $x$ axis carries the current away in the $+x$ direction.
The current arrives as in (a) and the arc is the same half circle; from $(0, R, 0)$ a straight wire runs vertically downward, in the $-z$ direction.
The current arrives as in (a), at the point $(0, -R, 0)$ of a complete circle of wire of radius $R$ centred on $O$ in the $yz$ plane. From that point the circle runs two ways to its lowest point $(0, 0, -R)$ : three quarters of the circle over the top and down the far side, and the remaining lower quarter directly down (this quarter is hidden behind the horizontal plane in the figure). From the lowest point a straight wire parallel to the $x$ axis carries the current away in the $+x$ direction.

(a) Half circle of radius R in the vertical plane, between two long wires parallel to x; the current comes in along -x and leaves along +x. — (a) Half circle of radius $R$ in the vertical plane, between two long wires parallel to $x$ ; the current comes in along $-x$ and leaves along $+x$ .

(b) The same half circle; the current leaves down a long vertical wire.

(c) A complete circle: from (0,-R,0) the current runs over the top to the lowest point and also down the lower quarter (dashed, hidden behind the horizontal plane); it leaves along +x from the lowest point. — (c) A complete circle: from $(0,-R,0)$ the current runs over the top to the lowest point and also down the lower quarter (dashed, hidden behind the horizontal plane); it leaves along $+x$ from the lowest point.

Given

$I = 8.0$ A
$R = 100$ mm
straight parts very long
(c) uniform wire (assumed)

Find

B

O

in each of (a), (b), (c)

Solution

Step 1The field of each kind of piece

Every piece is either an arc centred on $O$ or a long straight wire that ends at the foot of the perpendicular dropped to it from $O$ , at the distance $R$ . The Biot-Savart law gives the field of an element $I\,d\vect l$ at a displacement $\vect r$ from it:

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

On an arc centred on $O$ , every $d\vect l$ is perpendicular to its $\vect r$ and every element gives $d\vect B$ along the same normal to the arc’s plane, so an arc of angle $\theta$ gives

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)

On a straight wire, measure $s$ along it from its end, the foot of the perpendicular; the element $ds$ is at distance $(s^2 + R^2)^{1/2}$ , and the sine of its angle with $\vect r$ is $R/(s^2+R^2)^{1/2}$ . All elements give $d\vect B$ along the same normal to the plane holding the wire and $O$ :

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

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

(3)

Step 2The direction of each contribution

Write $\vect i$ , $\vect j$ , $\vect k$ for the unit vectors along $x$ , $y$ , $z$ ( $x$ toward the viewer of the Problem figures, $y$ to the right, $z$ up; $\vect i\times\vect j = \vect k$ ), and $B_w = \mu_0 I/(4\pi R)$ for the value of (3). The sense of each piece is the sense of $d\vect l\times\vect r$ for one of its elements, with $\vect r$ pointing from the element to $O$ . For the arc take the element at its top $(0, 0, R)$ ; for a straight wire, only the part of $\vect r$ perpendicular to the wire matters, since the part along $d\vect l$ drops out of the cross product. Piece by piece, $d\vect l\times\vect r$ has the direction

\begin{aligned} &\text{arc, at the top:} && \vect j\times(-\vect k) = -\vect i\\ &\text{incoming wire, } y = -R\text{:} && (-\vect i)\times\vect j = -\vect k\\ &\text{outgoing wire (a), } y = R\text{:} && \vect i\times(-\vect j) = -\vect k\\ &\text{outgoing wire (b), vertical at } y = R\text{:} && (-\vect k)\times(-\vect j) = -\vect i\\ &\text{outgoing wire (c), } z = -R\text{:} && \vect i\times\vect k = -\vect j \end{aligned}

An arc traversed from $(0,-R,0)$ over the top gives its field along $-\vect i$ ; an arc traversed the other way round the circle gives it along $+\vect i$ . The figures of Steps 3 to 5 draw these contributions head to tail at $O$ , all three seen from behind the $yz$ plane, so that no vector lies along the image of an axis. Numerically, the unit of every term is

B_w = \frac{\mu_0 I}{4\pi R} = \frac{(10^{-7}\text{ T}\cdot\text{m/A})(8.0\text{ A})}{0.100\text{ m}} = 8.0\ \mu\text{T} \tag{4}

Step 3Shape (a): a half circle and two parallel wires

(a) The contributions at O head to tail, with the axes seen from behind the yz plane: the half circle's B_0 along -x, the two wires' B_1 and B_2 both along -z; their sum B. — (a) The contributions at $O$ head to tail, with the axes seen from behind the $yz$ plane: the half circle’s $\vect B_0$ along $-x$ , the two wires’ $\vect B_1$ and $\vect B_2$ both along $-z$ ; their sum $\vect B$ .

The half circle has $\theta = \pi$ in (2), and each wire gives $B_w$ along $-\vect k$ . The two parts of the sum are perpendicular, so the magnitude is the root of the sum of their squares; with (4),

\begin{aligned} \vect B &= B_w\left(-\pi\,\vect i - 2\,\vect k\right)\\ B &= (8.0\ \mu\text{T})\,(\pi^2 + 4)^{1/2} \end{aligned}

B = \frac{\mu_0 I}{4\pi R}\,(\pi^2 + 4)^{1/2} = 30\ \mu\text{T}

(5)

The field lies in the $xz$ plane, at $\arctan(2/\pi) \approx 32^\circ$ from the $-x$ direction toward $-z$ .

Step 4Shape (b): a half circle, a parallel wire and a vertical wire

(b) Seen from behind the yz plane, as in (a): the half circle's B_0 and the vertical wire's B_2 both along -x; the incoming wire's B_1 along -z; their sum B. — (b) Seen from behind the $yz$ plane, as in (a): the half circle’s $\vect B_0$ and the vertical wire’s $\vect B_2$ both along $-x$ ; the incoming wire’s $\vect B_1$ along $-z$ ; their sum $\vect B$ .

Now the outgoing vertical wire adds $B_w$ along $-\vect i$ , the same way as the arc, and only the incoming wire gives a part along $-\vect k$ . The magnitude follows as before, and the square expands:

\begin{aligned} \vect B &= B_w\left(-(\pi + 1)\,\vect i - \vect k\right)\\ B &= (8.0\ \mu\text{T})\,\left((\pi + 1)^2 + 1\right)^{1/2} = (8.0\ \mu\text{T})\,(\pi^2 + 2\pi + 2)^{1/2} \end{aligned}

B = \frac{\mu_0 I}{4\pi R}\,(\pi^2 + 2\pi + 2)^{1/2} = 34\ \mu\text{T}

(6)

The field lies in the $xz$ plane, at $\arctan\bigl(1/(\pi+1)\bigr) \approx 14^\circ$ from the $-x$ direction toward $-z$ .

Step 5Shape (c): the current splits round the circle

From $(0,-R,0)$ to the lowest point the current has two paths in parallel: the three-quarter arc of length $3\pi R/2$ over the top, and the quarter arc of length $\pi R/2$ directly down. For a uniform wire (assumed) each resistance is proportional to its length, and the same voltage drives both, so the currents are inversely proportional to the lengths:

\begin{aligned} I_{3/4} &= I\,\frac{\pi R/2}{\pi R/2 + 3\pi R/2} = \frac{I}{4}\\ I_{1/4} &= I\,\frac{3\pi R/2}{\pi R/2 + 3\pi R/2} = \frac{3I}{4} \end{aligned}

The three-quarter arc is traversed over the top, so by (2) with $\theta = 3\pi/2$ its field is along $-\vect i$ ; the quarter arc runs the other way round the circle, with $\theta = \pi/2$ , and its field is along $+\vect i$ :

\vect B_{\text{arcs}} = \frac{\mu_0}{4\pi R}\left(-\frac{I}{4}\cdot\frac{3\pi}{2} + \frac{3I}{4}\cdot\frac{\pi}{2}\right)\vect i = 0

(c) Seen from behind the yz plane, as in (a). The two arcs' fields cancel and are not drawn; the incoming wire's B_1 along -z and the outgoing wire's B_2 along -y, and their sum B. — (c) Seen from behind the $yz$ plane, as in (a). The two arcs’ fields cancel and are not drawn; the incoming wire’s $\vect B_1$ along $-z$ and the outgoing wire’s $\vect B_2$ along $-y$ , and their sum $\vect B$ .

Only the two straight wires remain, each carrying the whole current $I$ and each giving $B_w$ , along $-\vect k$ and $-\vect j$ :

\begin{aligned} \vect B &= B_w\left(-\vect j - \vect k\right)\\ B &= (8.0\ \mu\text{T})\cdot 2^{1/2} \end{aligned}

B = \frac{\mu_0 I}{4\pi R}\,2^{1/2} = 11\ \mu\text{T}

(7)

The field lies in the $yz$ plane, the plane of the circle, midway between the $-y$ and $-z$ directions.

Check

Removing the straight wires leaves the arcs alone. In (a) that is $\pi B_w = \mu_0 I/(4R)$ , half the field $\mu_0 I/(2R)$ at the center of a full loop, as a half circle must give. In (c) the cancellation does not depend on where the current enters and leaves the ring: the field of an arc is proportional to its current times its length, and in two parallel branches of a uniform wire current times resistance, hence current times length, is the same for both. Units: $\mu_0 I/R$ is $(\text{T}\cdot\text{m/A})(\text{A})/\text{m} = \text{T}$ .

Remark

Whether a straight wire adds to the arc as a plain number or as a Pythagorean leg depends here on whether it lies in the arc’s plane or runs perpendicular to it. The vertical wire of (b) lies in that plane, so its field at $O$ is normal to the plane, like the arc’s, and the two add as $\pi + 1$ . A wire perpendicular to the plane, like the horizontal wires here, gives a field lying in the plane, at right angles to the arc’s. In (c) the trap is to treat the drawn circle as a single three-quarter arc: once the current divides between the two branches, the ring contributes nothing at its center, and the whole field comes from the leads.