Irodov 3.221: Diagonals Set the Frame's Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.221

A rectangular wire frame carries a current $I = 5.0$ A. Each of its diagonals has length $d = 16$ cm, and the two diagonals meet at an angle $\varphi = 30^\circ$ . Find the magnetic induction at the center of the frame.

The rectangular frame with its diagonals of length d crossing at the angle φ at the center O; the current I (sense chosen) is drawn counter-clockwise. — The rectangular frame with its diagonals of length $d$ crossing at the angle $\varphi$ at the center $O$ ; the current $I$ (sense chosen) is drawn counter-clockwise.

Given

$d = 16$ cm
$\varphi = 30^\circ$
$I = 5.0$ A
$\mu_0 = 4\pi\times10^{-7}$ H/m

Find

B

at the center

O

Solution

Step 1The field of one straight side

A straight wire seen from P: the perpendicular b and the angles α at which its two ends are seen on either side of it. — A straight wire seen from $P$ : the perpendicular $b$ and the angles $\alpha$ at which its two ends are seen on either side of it.

A straight wire at perpendicular distance $b$ from a point, whose two ends are seen from it at the angle $\alpha$ on either side of the perpendicular, gives a field found from the Biot-Savart law. Measure $x$ along the wire from the foot of the perpendicular; an element $dx$ is at $r = (x^2 + b^2)^{1/2}$ , and the sine of the angle between $d\vect l$ and $\vect r$ is $b/r$ , so

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

The wire runs from $x = -b\tan\alpha$ to $x = b\tan\alpha$ :

\begin{aligned} B_1 &= \frac{\mu_0 I}{4\pi}\int_{-b\tan\alpha}^{b\tan\alpha}\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\alpha}^{b\tan\alpha}\\ &= \frac{\mu_0 I}{4\pi}\cdot\frac{2\tan\alpha}{b\,(1 + \tan^2\alpha)^{1/2}} \end{aligned}

B_1 = \frac{\mu_0 I}{4\pi}\,\frac{2\sin\alpha}{b}

(1)

Step 2The two kinds of side

The perpendiculars from O: b_1 to a short side, which subtends φ between the diagonals, and b_2 to a long side, which subtends π - φ. — The perpendiculars from $O$ : $b_1$ to a short side, which subtends $\varphi$ between the diagonals, and $b_2$ to a long side, which subtends $\pi - \varphi$ .

The diagonals bisect each other, so every corner is $d/2$ from $O$ . A short side subtends the angle $\varphi$ between the diagonals, a long side the supplementary angle $\pi - \varphi$ . Each perpendicular bisects its side’s angle, so for the short and the long sides

\begin{aligned} \alpha_1 &= \frac{\varphi}{2}, & b_1 &= \frac{d}{2}\cos\frac{\varphi}{2},\\ \alpha_2 &= \beta = \frac{\pi - \varphi}{2}, & b_2 &= \frac{d}{2}\sin\frac{\varphi}{2} \end{aligned}

and $\sin\beta = \cos(\varphi/2)$ . Putting these into (1):

\begin{aligned} B_{\text{short}} &= \frac{\mu_0 I}{4\pi}\,\frac{2\sin(\varphi/2)}{(d/2)\cos(\varphi/2)} = \frac{\mu_0 I}{4\pi}\,\frac{4}{d}\tan\frac{\varphi}{2}\\ B_{\text{long}} &= \frac{\mu_0 I}{4\pi}\,\frac{2\cos(\varphi/2)}{(d/2)\sin(\varphi/2)} = \frac{\mu_0 I}{4\pi}\,\frac{4}{d}\cot\frac{\varphi}{2} \end{aligned}

Step 3All four sides together

The whole frame: every side drives B at O the same way, out of the page for the counter-clockwise current. — The whole frame: every side drives $\vect B$ at $O$ the same way, out of the page for the counter-clockwise current.

The current goes the same way round $O$ along all four sides, so the four fields point the same way and their magnitudes add, two sides of each kind:

B = 2\cdot\frac{\mu_0 I}{4\pi}\,\frac{4}{d}\left(\tan\frac{\varphi}{2} + \cot\frac{\varphi}{2}\right)

The bracket simplifies with $\sin^2 + \cos^2 = 1$ and the double-angle formula:

\tan\frac{\varphi}{2} + \cot\frac{\varphi}{2} = \frac{1}{\sin(\varphi/2)\cos(\varphi/2)} = \frac{2}{\sin\varphi}

so that

B = \frac{\mu_0 I}{4\pi}\,\frac{16}{d\sin\varphi}

Numerically,

B = \frac{4\times4\pi\times10^{-7}\times5.0}{\pi\times0.16\times\sin 30^\circ}\ \text{T}

B = \frac{4\mu_0 I}{\pi d\sin\varphi} = 0.10\ \text{mT}

(2)

$\vect B$ is normal to the plane of the frame, out of the page for the counter-clockwise current drawn (right-hand rule).

Check

At $\varphi = 90^\circ$ the frame is a square of side $a = d/\sqrt 2$ , and (2) gives $4\mu_0 I/(\pi d) = 2\sqrt 2\,\mu_0 I/(\pi a)$ , the known field at the center of a square loop (four sides at $a/2$ , each with $\alpha = 45^\circ$ in (1)). The units are those of $\mu_0 I/d$ , tesla.

Remark

At fixed diagonal the field is smallest for the square, where $\sin\varphi = 1$ , and grows without bound as $\varphi \to 0$ or $\pi$ : the frame flattens into two long wires carrying opposite currents a distance $d\sin(\varphi/2)$ or $d\cos(\varphi/2)$ apart, and the center comes ever closer to both of them.