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

Problem 4.2.1

Interactive: Forces Between Parallel Currents

Problem 4.2.1

Force between two parallel currents

Determine the force per unit length between two infinitely long parallel conducting wires carrying currents $I_{1}$ and $I_{2}$ in the same direction. The wires are separated by a distance $d$ .

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 6-21.

Try it yourself first.

Strategy

Wire 1 makes the field of an infinite line current (Section 4.1), and wire 2 feels it through $I d l \times B$ . Choose the coordinates so that the cross product is a one-liner.

1. Geometry

Put wire 1 on the $z$ -axis and wire 2 parallel to it at $y = d$ , both carrying their current in the $+ z$ direction (Figure 4.2.16a).

2. Field of wire 1 at wire 2

Ampère's law on a circle of radius $ρ$ round wire 1 gives $2 π ρ H_{ϕ} = I_{1}$ , so that at the position of wire 2

B_{1} = \frac{μ _{0} I _{1}}{2 π d} a_{ϕ} = - \frac{μ _{0} I _{1}}{2 π d} a_{x} at y = d,

since on the positive $y$ -axis the azimuthal unit vector is $a_{ϕ} = - a_{x}$ (the field of a $+ z$ current circulates anticlockwise seen from $+ z$ , so at a point on the $+ y$ side it points along $- x$ ).

3. Force on wire 2

For a thin wire $J_{2} d V \to I_{2} d l a_{z}$ , and Eq. (4.2.14) gives per unit length

\frac{d F _{12}}{d z} = I_{2} a_{z} \times B_{1} = - \frac{μ _{0} I _{1} I _{2}}{2 π d} (a_{z} \times a_{x}) = - \frac{μ _{0} I _{1} I _{2}}{2 π d} a_{y} .

Answer

\frac{F}{l} = \frac{μ _{0} I _{1} I _{2}}{2 π d} [N/m], attractive for parallel currents

(a) The field of wire 1 at wire 2 and the forces.

(b) F/l versus the separation for 100\,\mathrm A in each wire. Problem 4.2.1. (a) The circular field lines of wire 1 pass wire 2 pointing into the page, and I_2\vec a_z\times\vec B_1 points towards wire 1. (b) The 1/d law: 100\,\mathrm A at 1\,\mathrm{cm} gives 0.2\,\mathrm{N/m}, at 10\,\mathrm{cm} only 0.02\,\mathrm{N/m}. — **Figure 4.2.16.** (b) $F / l$ versus the separation for $100 A$ in each wire. Problem 4.2.1. (a) The circular field lines of wire 1 pass wire 2 pointing into the page, and $I_{2} a_{z} \times B_{1}$ points towards wire 1. (b) The $1/ d$ law: $100 A$ at $1 cm$ gives $0.2 N/m$ , at $10 cm$ only $0.02 N/m$ .

Drag one wire and watch $F / l$ follow $1/ d$ . Try 100 A at 1 cm.

InteractiveLoading simulation…

Check

The minus sign points the force on wire 2 towards wire 1 (along $- a_{y}$ ): parallel currents attract. Repeating the argument for wire 1 in the field of wire 2, which at wire 1 is $+ μ_{0} I_{2} / (2 π d) a_{x}$ , gives $F_{21} = + μ_{0} I_{1} I_{2} / (2 π d) a_{y}$ per unit length, so Newton's third law is satisfied, $F_{21} = - F_{12}$ . If one current is reversed the product $I_{1} I_{2}$ changes sign and the wires repel. Units: $μ_{0} I^{2} / d$ is (H/m)(A $^{2}$ )/m $=$ (J/A $^{2}$ )(A $^{2}$ )/m $^{2}$ $=$ N/m. The $1/ d$ law has no length scale of its own, so it never saturates: $I_{1} = I_{2} = 100 A$ at $d = 1 cm$ gives $2 \times 1 0^{- 7} \times 1 0^{4} /0.01 = 0.2 N/m$ , at $10 cm$ only $0.02 N/m$ , but at $10 kA$ and $10 cm$ it is $200 N/m$ , which is why bus-bars are braced (Figure 4.2.16b). The formula was the definition of the ampere until 2019: two wires $1 m$ apart carrying $1 A$ attract with exactly $2 \times 1 0^{- 7} N$ per metre.

Problem 4.2.1: Force between two parallel currents