Irodov 3.257: A Wire on a Half Cylinder's Axis – Amit & Deepali Sinha Foundation

Problem Irodov 3.257

A long straight conductor has the cross-section of a thin half-ring of radius $R$ : it is a thin-walled half cylinder, and a direct current $I$ flows along it. A thin wire runs along the axis of the half cylinder, through the point $O$ at the centre of the half-ring, and carries the same current $I$ in the opposite direction. Find the magnetic force between the two conductors per unit length.

The cross-section: the thin half-ring of radius R, a long half cylinder whose current runs normal to the page, and its axis O, where the thin wire runs. — The cross-section: the thin half-ring of radius $R$ , a long half cylinder whose current runs normal to the page, and its axis $O$ , where the thin wire runs.

Given

half-ring radius $R$ , current $I$ along it, spread evenly round the thin wall (assumed)
wire on the axis $O$ , current $I$ opposite

Find

the force between them per unit length

Solution

Step 1The current in an element of the half-ring

The wall is thin and uniform, so the current $I$ is spread evenly round the half-ring (assumed), over the arc $\pi R$ . Measure the angle $\theta$ at $O$ from the half-ring’s line of symmetry. The strip of the wall between $\theta$ and $\theta + d\theta$ is a long straight filament carrying

dI = \frac{I}{\pi R}\,R\,d\theta = \frac{I}{\pi}\,d\theta \tag{1}

Two filaments dI and dI' of the half-ring at θ from its line of symmetry (current into the page) and the forces d F and d F' they exert on the wire at O (current out of the page), each pointing away from its filament. — Two filaments $dI$ and $dI'$ of the half-ring at $\pm\theta$ from its line of symmetry (current into the page) and the forces $d\vect F$ and $d\vect F'$ they exert on the wire at $O$ (current out of the page), each pointing away from its filament.

Step 2The force of one filament on the wire

The filament is a long straight current at the distance $R$ from the wire. By Ampere’s law on a circle of radius $R$ round it, on which $B$ is tangential and the same all round,

dB\cdot 2\pi R = \mu_0\,dI

The wire, carrying $I$ across this field, feels per unit length the force

dF = I\,dB = \frac{\mu_0 I\,dI}{2\pi R} \tag{2}

The two currents are antiparallel, so the force is a repulsion: it points along the line from the filament through $O$ , away from the filament.

Step 3Adding the filaments

Filaments at $+\theta$ and $-\theta$ push the wire with equal forces whose components across the line of symmetry cancel, as the figure shows; the components along it, $dF\cos\theta$ , add. With (2) and (1), integrating over the half-ring from $-\pi/2$ to $\pi/2$ :

F_1 = \int_{-\pi/2}^{\pi/2}\frac{\mu_0 I}{2\pi R}\cdot\frac{I}{\pi}\cos\theta\,d\theta = \frac{\mu_0 I^2}{2\pi^2 R}\Big[\sin\theta\Big]_{-\pi/2}^{\pi/2} = \frac{\mu_0 I^2}{2\pi^2 R}\cdot 2

F_1 = \frac{\mu_0 I^2}{\pi^2 R}

(3)

The force on the wire points along the line of symmetry, away from the half-ring, toward its open side. By Newton’s third law the half-ring is pushed the other way with the same force per unit length.

Check

Two comparisons bracket the result. If the half-ring’s whole current were concentrated in its lowest filament, at the distance $R$ , the force would be $\mu_0 I^2/(2\pi R)$ ; spreading it round the half-ring tilts most filaments away from the symmetry line, so (3) is smaller, by the factor $2/\pi = 0.64$ , the mean of $\cos\theta$ over the half-ring. A complete thin cylinder would push equally from every side and give no force at all, consistent with the zero field inside a uniform tubular current. Units: $\mu_0 I^2/R$ is $(\text{N/A}^2)\cdot\text{A}^2/\text{m} = \text{N/m}$ , a force per unit length.

Remark

The half-ring and the wire together are a return circuit, like a coaxial cable with half its outer conductor removed. In a full coaxial cable the inner conductor feels no force at all; removing half of the outer one leaves a net push of $\mu_0 I^2/(\pi^2 R)$ on each metre, which is why the outer conductor’s symmetry matters mechanically as well as electrically when large currents are switched.