Irodov 3.233: Inside a Wire, B Is Half j Cross r – Amit & Deepali Sinha Foundation

Problem Irodov 3.233

A long, straight, uniform wire of circular cross-section of radius $R$ carries a direct current of uniform density $\vect j$ along its length. A point’s position is given by the vector $\vect r$ drawn to it from the wire’s axis, perpendicular to the axis. The permeability is $1$ everywhere. Find the magnetic induction vector at that point.

The wire in cross-section, radius R, with the current density j out of the page; r runs from the axis O, perpendicular to it, to the field point. — The wire in cross-section, radius $R$ , with the current density $\vect j$ out of the page; $\vect r$ runs from the axis $O$ , perpendicular to it, to the field point.

Given

round wire, radius $R$
uniform $\vect j$ along the wire
$\vect r$ from the axis
$\mu = 1$

Find

\vect B

as a function of

\vect r

Solution

Step 1The magnitude from Ampère’s law

By the wire’s symmetry about its axis, and the mirror argument for an axial vector, $\vect B$ is tangent to circles about the axis with a magnitude that depends only on $r$ . Take such a circle of radius $r$ as the Amperian contour. Inside the wire it encloses the current $j\pi r^2$ , outside the whole $j\pi R^2$ :

B\cdot 2\pi r = \begin{cases} \mu_0 j\pi r^2, & r \le R\\ \mu_0 j\pi R^2, & r \ge R \end{cases}

B = \begin{cases} \dfrac{\mu_0 j r}{2}, & r \le R\\[6pt] \dfrac{\mu_0 j R^2}{2r}, & r \ge R \end{cases} \tag{1}

Step 2The direction as a cross product

Amperian circles inside and outside the wire (dashed; the outer one's sense marked). B is tangent to each, counterclockwise for j out of the page: along j· r. — Amperian circles inside and outside the wire (dashed; the outer one’s sense marked). $\vect B$ is tangent to each, counterclockwise for $\vect j$ out of the page: along $\vect j\times\vect r$ .

By the right-hand rule about $\vect j$ , $\vect B$ circles counterclockwise seen from the tip of $\vect j$ . At the end of $\vect r$ that tangent direction is the direction of $\vect j\times\vect r$ , and since $\vect j$ and $\vect r$ are perpendicular,

|\vect j\times\vect r| = j r

Each magnitude in (1) is therefore the length of a multiple of $\vect j\times\vect r$ :

\vect B = \begin{cases} \dfrac{\mu_0}{2}\,\vect j\times\vect r, & r \le R\\[8pt] \dfrac{\mu_0}{2}\,\dfrac{R^2}{r^2}\,\vect j\times\vect r, & r \ge R \end{cases}

(2)

The magnitude is $\mu_0 j r/2$ inside, growing linearly from zero on the axis, and $\mu_0 j R^2/(2r)$ outside; the direction is perpendicular to both the wire and $\vect r$ , in the right-hand sense of the current.

Check

At $r = R$ both forms give $\mu_0 j R/2$ , so $\vect B$ is continuous at the surface, as it must be with no surface current. Outside, with $I = j\pi R^2$ , the magnitude is $\mu_0 j R^2/(2r) = \mu_0 I/(2\pi r)$ , the field of a thin wire carrying the whole current.

Remark

Inside, $\vect B$ is linear in the position, like the velocity field $\vect{\omega}\times\vect r$ of a rigid rotation, with $\mu_0\vect j/2$ playing the role of the angular velocity; the curl of such a field is twice that vector, which gives back $\nabla\times\vect B = \mu_0\vect j$ . The vector form is what makes superposition easy: two such fields with different centers subtract to a uniform one, which is why an off-axis cylindrical cavity in a wire holds a uniform field.