Irodov 3.232: An Open Line That Acts Closed – Amit & Deepali Sinha Foundation

Problem Irodov 3.232

A circular loop of wire carries a current $I$ . Evaluate the line integral of the magnetic induction, $\int \vect B\cdot d\vect r$ , taken along the loop’s axis from $-\infty$ to $+\infty$ , and explain the result.

The circular loop carrying the current I, and its axis. — The circular loop carrying the current $I$ , and its axis.

Given

round loop, current $I$
the path: the whole axis, from $-\infty$ to $+\infty$

Find

\int\vect B\cdot d\vect r

along the axis, and why

Solution

Step 1The field on the axis

Let the loop have radius $R$ , put its center at the origin, and point the $z$ axis along its axis so that the current circulates counterclockwise seen from $+z$ . Every element $I\,d\vect l$ is perpendicular to the line joining it to the axial point $z$ , at the distance $(R^2+z^2)^{1/2}$ . By symmetry the components of $d\vect B$ across the axis cancel round the loop, and the axial component of each is the fraction $R/(R^2+z^2)^{1/2}$ of its magnitude:

The loop edge on, radius R: an element I d l at the top (out of the page), the vector r from it to the axial point P, and its d B, perpendicular to r, with the axial component dashed. — The loop edge on, radius $R$ : an element $I\,d\vect l$ at the top (out of the page), the vector $\vect r$ from it to the axial point $P$ , and its $d\vect B$ , perpendicular to $\vect r$ , with the axial component dashed.

B_z = \oint\frac{\mu_0 I}{4\pi}\,\frac{dl}{R^2+z^2}\cdot\frac{R}{(R^2+z^2)^{1/2}} = \frac{\mu_0 I}{4\pi}\,\frac{2\pi R\cdot R}{(R^2+z^2)^{3/2}}

B_z = \frac{\mu_0 I R^2}{2\,(R^2+z^2)^{3/2}}

(1)

On the axis $\vect B$ points along $+z$ everywhere, the right-hand sense of the current.

Step 2Integrating along the whole axis

Along the axis $d\vect r = \vect k\,dz$ , so the integrand is $B_z\,dz$ with (1):

\int_{-\infty}^{+\infty}\vect B\cdot d\vect r = \frac{\mu_0 I R^2}{2}\int_{-\infty}^{+\infty}\frac{dz}{(R^2+z^2)^{3/2}}

The substitution $z = R\tan\theta$ maps the infinite range onto $-\pi/2 < \theta < \pi/2$ :

\begin{aligned} dz &= \frac{R\,d\theta}{\cos^2\theta}, \qquad R^2 + z^2 = \frac{R^2}{\cos^2\theta}\\ \int_{-\infty}^{+\infty}\frac{dz}{(R^2+z^2)^{3/2}} &= \frac{1}{R^2}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta = \frac{2}{R^2} \end{aligned}

\int_{-\infty}^{+\infty}\vect B\cdot d\vect r = \mu_0 I

(2)

The radius has dropped out. Taken the other way along the axis, or for the opposite current, the integral is $-\mu_0 I$ .

Step 3Why: the axis closes at infinity

The loop's plane through the axis: the two crossings of the loop (current out of the page above, into it below) and the axis closed by a large semicircle of radius ρ (dashed), its sense marked. — The loop’s plane through the axis: the two crossings of the loop (current out of the page above, into it below) and the axis closed by a large semicircle of radius $\rho$ (dashed), its sense marked.

Close the axis from $z = -\rho$ to $z = \rho$ by a semicircle of radius $\rho$ in a plane through the axis. Far from the loop its field is that of a magnetic dipole, falling off as $1/\rho^3$ , while the semicircle is only $\pi\rho$ long, so its contribution vanishes as $1/\rho^2$ :

\left|\int_{\text{semicircle}}\vect B\cdot d\vect r\right| \le \pi\rho\cdot\frac{\text{const}}{\rho^3} \to 0

The closed contour, axis plus semicircle, encircles the upper crossing of the wire once, in the right-hand sense of its current, and nothing else. The circulation theorem then gives the whole answer at once:

\int_{-\infty}^{+\infty}\vect B\cdot d\vect r = \lim_{\rho\to\infty}\oint\vect B\cdot d\vect r = \mu_0 I

Check

The two routes agree, and the integral in Step 2 really is independent of $R$ : a large loop gives a weaker field ( $B_z(0) = \mu_0 I/2R$ ) spread over a longer stretch of axis (width of order $R$ ), and the product stays $\mu_0 I$ . Units: $\mu_0 I$ is $(\text{T}\cdot\text{m/A})(\text{A}) = \text{T}\cdot\text{m}$ , the unit of $B$ times a length.

Remark

The argument of Step 3 never used the loop’s shape or the line’s position: any straight line that passes once through any closed current $I$ carries $\int\vect B\cdot d\vect r = \pm\mu_0 I$ , and a line that misses the circuit gives zero. Only how the line links the current matters, which is the topological content of the circulation theorem.