Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Exercise 4.1.10

About the further exercises of Section 4.1

Twenty further exercises in the spirit of the problems, the worked examples and the home exercises, graded from easy to harder within each group. Exercises 4.1.1–4.1.10 belong to the theory of Lecture 8 (the Lorentz force, Biot–Savart's law, flux, force and the vector potential) and Exercises 4.1.11–4.1.20 to the exercise session of Lecture 9 (Ampère's law, coils and rotating charges, measuring and shaping fields). They use only

F = q (E + v \times B), F = I \int d l \times B, B = \frac{μ _{0}}{4 π} \int \frac{I d l ^{'} \times R}{R ^{3}}, B = \nabla \times A, \nabla \cdot B = 0, \oint_{C} H \cdot d l = I_{enc}, Φ = \int_{S} B \cdot d S,

the two results of the section, $B = μ_{0} I / (2 π ρ)$ for a long straight wire and $B_{z} = μ_{0} I b^{2} / [2 (b^{2} + h^{2})^{3/2}]$ on the axis of a loop, and superposition.

Exercise 4.1.10

Field on the axis of a square loop

cf. Home exercise 4.1.1

A square loop of side $a$ in the $x y$ -plane, centred at the origin, carries $I$ . (a) Find $B$ at the point $P (0, 0, h)$ on its axis, using the segment formula of Exercise 4.1.4. (b) Check the limits $h = 0$ (Home exercise 4.1.1) and $h ≫ a$ (dipole). (c) Evaluate for $a = 20$ cm, $I = 5$ A, $h = 10$ cm, and compare with the circular loop of radius $b = a /2$ at the same height.

Adapted from: B. M. Notaroš, Electromagnetics, Problem 4.1.

Try it yourself first.

Strategy

Each side is a finite segment seen from P at the distance $D = h^{2} + a^{2} /4$ from its midpoint, so Exercise 4.1.4 gives its field. The contributions of the four sides are tilted symmetrically about the axis, so only their axial components survive.

Take the side $y^{'} = a /2$ . The perpendicular from P to the line containing it meets it at the midpoint $(0, a /2, 0)$ , at the distance $D = h^{2} + a^{2} /4$ . Seen from P the two ends are at the angles $\pm θ$ with $sin θ = \frac{a /2}{D ^{2} + a ^{2} /4} = \frac{a /2}{h ^{2} + a ^{2} /2}$ , so by Exercise 4.1.4 the side produces

B_{side} = \frac{μ _{0} I}{4 π D} 2 sin θ = \frac{μ _{0} I a}{4 π D h ^{2} + a ^{2} /2},

directed perpendicular to the plane containing P and the side, i.e. tilted from the axis by the angle $ψ$ with $cos ψ = (a /2) / D$ . The components perpendicular to the axis cancel between opposite sides, and the axial components of the four sides add:

B_{z} = 4 B_{side} cos ψ = 4 \frac{μ _{0} I a}{4 π D h ^{2} + a ^{2} /2} \frac{a /2}{D} = \frac{μ _{0} I a ^{2}}{2 π D ^{2} h ^{2} + a ^{2} /2} .

Answer

B (0, 0, h) = \frac{μ _{0} I a ^{2}}{2 π ( h ^{2} + a ^{2} /4 ) h ^{2} + a ^{2} /2} a_{z}

$h = 0$ : $\frac{μ _{0} I a ^{2}}{2 π ( a ^{2} /4 ) ( a / 2 )} = \frac{2 2 μ _{0} I}{π a}$ , the result of Home exercise 4.1.1. $h ≫ a$ : $B_{z} \to \frac{μ _{0} I a ^{2}}{2 π h ^{3}} = \frac{μ _{0} m}{2 π h ^{3}}$ with $m = I a^{2}$ , the dipole field with the moment current $\times$ area, the same far field as the loop of Problem 4.1.2 with $π b^{2} \to a^{2}$ .

$D^{2} = 0.01 + 0.01 = 0.02$ m $^{2}$ and $h^{2} + a^{2} /2 = 0.03 = 0.1732$ m:

B_{z} = \frac{4 π \times 1 0 ^{- 7} \times 5 \times 0.04}{2 π \times 0.02 \times 0.1732} = 1.15 \times 1 0^{- 5} T = 11.5 μ T,

against $28.3$ μT at the centre ( $41 %$ ) and $\frac{μ _{0} I b ^{2}}{2 ( b ^{2} + h ^{2} ) ^{3/2}} = \frac{4 π \times 1 0 ^{- 7} \times 5 \times 0.01}{2 \times 0.0 2 ^{1.5}} = 11.1$ μT for the circle of radius $10$ cm.

Answer

B_{z} (h = 10 cm) = 11.5 μ T (circle of radius a /2 : 11.1 μ T)

Check

Units: $μ_{0} I a^{2} / D^{3}$ is T. At the centre the inscribed circle beats the square by $10 %$ (its current is nearer). Half a side away the square is ahead by $4 %$ (it carries more current, $4 a$ against $π a$ , and the extra length at the corners starts to count). Far away only $I \times$ area matters and the square, with $a^{2} > π a^{2} /4$ , wins by $27 %$ . The dipole formula $μ_{0} I a^{2} /2 π h^{3}$ would give $40$ μT at $h = 10$ cm, three and a half times too much, since $h = a /2$ is not "far".

Exercise 4.1.10: Field on the axis of a square loop