Exercise 4.1.6: Two parallel wires with opposite currents

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 4.1.6

Two parallel wires with opposite currents

cf. Worked example 4.1.1

Two long parallel wires at $x = \pm d /2$ , $y = 0$ carry $I = 50$ A in opposite directions ( $+ z$ at $x = + d /2$ , $- z$ at $x = - d /2$ ), with $d = 10$ cm. Find $B$ (a) at the midpoint O, (b) at the point $P (0, h, 0)$ on the perpendicular bisector, $h = 5$ cm, (c) at a point on the $x$ -axis outside the pair, $x ≫ d$ , to leading order.

Try it yourself first.

Strategy

Superposition of two long-wire fields, Eq. (4.1.21), each circling its own wire by the right-hand rule. Between antiparallel wires the two circulations run the same way and add. Far away the $1/ ρ$ terms cancel and only a $1/ ρ^{2}$ remainder survives.

Each wire gives $B = μ_{0} I / (2 π r)$ along $a_{ϕ}$ about its own axis (Problem 4.1.1), and the fields add vectorially (Figure 4.1.31).

At O both wires are at $r = d /2$ . For the $+ z$ wire at $x = + d /2$ , the point O lies in the $- x$ direction, and $a_{ϕ} = a_{z} \times (- a_{x}) = - a_{y}$ . For the $- z$ wire at $x = - d /2$ , O lies in the $+ x$ direction and $a_{ϕ}$ reversed by the current sign is $- (a_{z} \times a_{x}) = - a_{y}$ again. The two fields are parallel:

Answer

B (O) = - 2 \frac{μ _{0} I}{2 π ( d /2 )} a_{y} = - \frac{2 μ _{0} I}{π d} a_{y} = - 0.40 a_{y} mT

Both wires are at $r = (d /2)^{2} + h^{2} = 7.07$ cm from P, each giving $B_{1} = μ_{0} I / (2 π r) = 0.141$ mT. The field of the right wire at P is perpendicular to the line from $(d /2, 0)$ to P. By symmetry the $x$ -components of the two fields cancel and the $y$ -components add, each being $B_{1} cos β$ with $cos β = (d /2) / r$ :

B_{y} = - 2 B_{1} \frac{d /2}{r} = - \frac{μ _{0} I d}{2 π r ^{2}} = - \frac{4 π \times 1 0 ^{- 7} \times 50 \times 0.10}{2 π \times 0.0707 1 ^{2}} = - 2.0 \times 1 0^{- 4} T .

Answer

B (P) = - \frac{μ _{0} I d}{2 π [ ( d /2 ) ^{2} + h ^{2} ]} a_{y} = - 0.20 a_{y} mT

(Check: $h \to 0$ gives $- 2 μ_{0} I / (π d) a_{y}$ , the result of (a). The perpendicular bisector is itself a field line, so the field points along it, in the $- a_{y}$ direction, at every height.)

On the $x$ -axis at $x > d /2$ the two fields are antiparallel ( $+ a_{y}$ from the near wire, $- a_{y}$ from the far one):

B_{y} = \frac{μ _{0} I}{2 π} (\frac{1}{x - d /2} - \frac{1}{x + d /2}) = \frac{μ _{0} I}{2 π} \frac{d}{x ^{2} - d ^{2} /4} \approx \frac{μ _{0} I d}{2 π x ^{2}} (x ≫ d) .

Answer

B \approx \frac{μ _{0} I d}{2 π x ^{2}} for x ≫ d : a 1/ x^{2} decay, not 1/ x

Exercise 4.1.6. Computed field lines of two long antiparallel currents (the wires run perpendicular to the page, d/2=0.5 in the units of the plot). Between the wires the two circulations run the same way and the field is strong, on the perpendicular bisector it points along the bisector, and far away the lines of the two wires nearly cancel: the field decays as 1/x^2. — **Figure 4.1.31.** Exercise 4.1.6. Computed field lines of two long antiparallel currents (the wires run perpendicular to the page, $d /2 = 0.5$ in the units of the plot). Between the wires the two circulations run the same way and the field is strong, on the perpendicular bisector it points along the bisector, and far away the lines of the two wires nearly cancel: the field decays as $1/ x^{2}$ .

Check

Units: $μ_{0} I / d$ is T. A go-and-return pair is the magnetic analogue of a dipole line: the $1/ ρ$ terms cancel and only the $1/ ρ^{2}$ term, proportional to the spacing $d$ , survives. This is the reason twisted or closely spaced supply leads radiate little field, and why the three-phase line of Worked example 4.1.3 gives much less than the single-conductor estimate.