Finite Wires and the Point on the Line

A wire that ends

Real wires are not endless. The side of a square coil, a short link in a circuit board and one arm of a bent wire all stop somewhere. For such a piece the sum of the last lesson runs over part of the wire only. The angle \(\phi\) no longer goes from \(-90°\) to \(+90°\).

Let the wire run from B to A, with the current from B towards A. P is at a perpendicular distance \(d\) from the line of the wire. Call \(\theta_1\) the angle at P between the perpendicular and the line to A. Call \(\theta_2\) the angle between the perpendicular and the line to B. Some books write them \(\alpha\) and \(\beta\).

A long wire has \(\theta_1 = \theta_2 = 90°\), so the bracket is \(2\) and the formula gives \(\mu_0 I/2\pi d\) again. A wire of length \(2d\), seen from its midpoint at distance \(d\), has \(\sin\theta_1 = \sin\theta_2 = 1/\sqrt2\). Its field is \(0.71\) of the long-wire value, though the wire is short. Most of the field comes from the part of a wire near P.

Start with P level with the middle of the wire, as in the figure above. Both half-lengths are \(0.3\,\text{m}\) and \(d = 0.4\,\text{m}\), so each line to an end is \(0.5\,\text{m}\) long. Then \(\sin\theta_1 = \sin\theta_2 = 0.3/0.5 = 0.6\). The field is \(10^{-7}\times 10/0.4\times 1.2 = 3\times10^{-6}\,\text{T}\), into the page.

Now raise P level with the upper end A. The angle \(\theta_1\) shrinks to zero and the field falls to \(2.08\,\mu\text{T}\). Raise P further, to \(0.6\,\text{m}\). Now both ends are below the foot of the perpendicular. The elements still all give fields into the page, but the range of \(\phi\) no longer contains zero. So \(\theta_1\) counts as negative, \(-36.9°\), and \(\theta_2 = 66.0°\). The field is only \(0.78\,\mu\text{T}\).

At the end of a long wire

Let a wire start at a point O and run far away in one direction. It is a half-infinite, or semi-infinite wire. Take P on the perpendicular through O, at a distance \(d\). The far end is seen at \(\theta_1 = 90°\), and the near end O is at the foot of the perpendicular, so \(\theta_2 = 0\).

The result is half the field of an endless wire, and the reason is simple. An endless wire is two such halves, joined at O, and each half gives the same field at P. For \(10\,\text{A}\) at \(0.05\,\text{m}\), the endless wire gives \(4\times10^{-5}\,\text{T}\) and the half-wire \(2\times10^{-5}\,\text{T}\).

No field on the line of a current

Take a point P on the line of a straight wire, beyond its end. For every element, the line from the element to P runs along the wire. So \(d\vec l\) is parallel to \(\hat r\), \(\sin\theta = 0\), and each element gives nothing. A straight current gives no field anywhere on its own line.

This fact makes bent wires easy. Most such problems are solved by superposition: cut the wire into straight pieces and arcs, find each field, and add them as vectors. A piece whose line passes through P can simply be dropped.

In the figure above each wire is \(0.1\,\text{m}\) from P, so each gives \(2\times10^{-7}\times 10/0.1 = 2\times10^{-5}\,\text{T}\). With the currents in the same direction the two fields at P are opposite, and the net field is zero. With the lower current reversed both are into the page, and the net field is \(4\times10^{-5}\,\text{T}\).

Fig 2.1 shows the method at work. The horizontal arm is dropped, because P lies on its line. The vertical arm ends at O, and P is on the perpendicular through O at \(d = 0.1\,\text{m}\). So \(B = 10^{-7}\times 20/0.1 = 2\times10^{-5}\,\text{T}\). For the direction, the current runs down and P is on the left of it. Right thumb down, and the fingers go into the page on the left.