Irodov 3.234: A Hole Holds a Uniform Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.234

A long straight uniform wire of circular cross-section contains a long cylindrical cavity of circular cross-section. The cavity’s axis is parallel to the wire’s axis, and the vector from the wire’s axis to the cavity’s axis, perpendicular to both, is $\vect l$ . A direct current of uniform density $\vect j$ flows along the wire. Find the magnetic induction inside the cavity, and in particular for $\vect l = 0$ .

The wire in cross-section with the cavity (white) whose axis O' is displaced by l from the wire's axis O; the current density j is out of the page. — The wire in cross-section with the cavity (white) whose axis $O'$ is displaced by $\vect l$ from the wire’s axis $O$ ; the current density $\vect j$ is out of the page.

Given

round wire, uniform $\vect j$
round cavity, axis parallel, displaced by $\vect l$

Find

\vect B

inside the cavity; the case

\vect l = 0

Solution

Step 1Fill the hole with two opposite currents

The current distribution is unchanged if the cavity is filled with the same current density $\vect j$ and, on top of it, with $-\vect j$ . The wire then becomes a full solid cylinder carrying $\vect j$ , about the axis $O$ , plus a solid cylinder the size of the cavity carrying $-\vect j$ , about the axis $O'$ . The field is the sum of the fields of the two.

Step 2The field inside each full cylinder

By symmetry the field inside a cylinder carrying a uniform current circles its axis, with a magnitude that depends only on the distance $r$ from it. Ampère’s law on a circle of radius $r$ about the axis:

\begin{aligned} B\cdot 2\pi r &= \mu_0 j\pi r^2\\ B &= \frac{\mu_0 j r}{2} \end{aligned}

The direction is $\vect j\times\vect r$ by the right-hand rule, and that vector has length $jr$ ; here $\vect r$ is the perpendicular vector from that cylinder’s axis to the field point:

\vect B = \frac{\mu_0}{2}\,\vect j\times\vect r \tag{1}

Step 3The sum at a point of the cavity

A point P in the cavity is at r_1 from O and r_2 from O', with r_1 = l + r_2. The field in the cavity is the same everywhere, along j· l. — A point $P$ in the cavity is at $\vect r_1$ from $O$ and $\vect r_2$ from $O'$ , with $\vect r_1 = \vect l + \vect r_2$ . The field in the cavity is the same everywhere, along $\vect j\times\vect l$ .

A point $P$ of the cavity lies inside both full cylinders, at $\vect r_1$ from $O$ and $\vect r_2$ from $O'$ . Applying (1) to each, with $-\vect j$ for the second:

\vect B = \frac{\mu_0}{2}\,\vect j\times\vect r_1 + \frac{\mu_0}{2}\,(-\vect j)\times\vect r_2 = \frac{\mu_0}{2}\,\vect j\times(\vect r_1 - \vect r_2)

The difference of the two position vectors is the fixed displacement between the axes, the same for every point $P$ :

\vect r_1 - \vect r_2 = \vect l

\vect B = \frac{\mu_0}{2}\,\vect j\times\vect l

(2)

The field inside the cavity is uniform: magnitude $\mu_0 j l/2$ , perpendicular to both the wire and $\vect l$ , in the right-hand sense of $\vect j$ turned toward $\vect l$ . For a coaxial cavity the displacement vanishes, and so does the field:

\vect B = 0 \ \text{in the cavity, for}\ \vect l = 0

(3)

Check

The case $\vect l = 0$ agrees with Ampère’s law directly: a circle about the axis inside a coaxial cavity encloses no current, so $B\cdot 2\pi r = 0$ , the familiar result that no field exists inside a cylindrical tube carrying an axial current. Units: $\mu_0 j l$ is $(\text{T}\cdot\text{m/A})(\text{A/m}^2)(\text{m}) = \text{T}$ .

Remark

The cavity is the magnetic twin of the off-center spherical cavity in a uniformly charged ball, where $\vect E = \rho\vect l/(3\varepsilon_0)$ is uniform for the same reason: each interior field is linear in the position, so the difference of two of them with shifted centers is a constant. The uniform field does not depend on the size of either cylinder, only on how far apart their axes are.