Irodov 3.235: Reading the Current Off the Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.235

A parallel stream of electrons is symmetric about its axis. Inside the stream the magnetic induction at distance $r$ from the axis is $B = br^{\alpha}$ , with $b$ and $\alpha$ positive constants. Find the current density in the stream as a function of $r$ .

The electron stream side on, moving with velocity v along its axis; r is the distance from the axis. — The electron stream side on, moving with velocity $\vect v$ along its axis; $r$ is the distance from the axis.

Given

parallel electron stream, symmetric about its axis
$B = br^{\alpha}$ inside, $b > 0$ , $\alpha > 0$

Find

j(r)

Solution

Step 1Ampère’s law on a circle about the axis

The stream is a set of parallel currents with axial symmetry, so, as for a round wire, its field circles the axis with a magnitude depending only on $r$ . A circle of radius $r$ about the axis encloses the current flowing through the disc it spans, which is built from rings of radius $r'$ and area $2\pi r'\,dr'$ :

B(r)\cdot 2\pi r = \mu_0\int_0^{r} j(r')\,2\pi r'\,dr' \tag{1}

The stream in cross-section about its axis O. Below it, the senses: the electrons' velocity v into the page (cross), the current density j out of it (dot). The Amperian circle of radius r (dashed) and B tangent to it, counterclockwise. — The stream in cross-section about its axis $O$ . Below it, the senses: the electrons’ velocity $\vect v$ into the page (cross), the current density $\vect j$ out of it (dot). The Amperian circle of radius $r$ (dashed) and $\vect B$ tangent to it, counterclockwise.

Step 2Differentiating to isolate the local current density

The right side of (1) is an integral up to $r$ , so its derivative with respect to $r$ is the integrand at $r$ . Dividing by $2\pi$ first and then differentiating:

\frac{d}{dr}\bigl(r B\bigr) = \mu_0\, j(r)\, r

With the given $B = br^{\alpha}$ , the product is $rB = b r^{\alpha+1}$ :

\frac{d}{dr}\bigl(b r^{\alpha+1}\bigr) = b(\alpha+1)\,r^{\alpha} = \mu_0\, j(r)\, r

Dividing by $\mu_0 r$ :

j(r) = \frac{b}{\mu_0}\,(1+\alpha)\,r^{\alpha-1}

(2)

The current density points opposite to the electrons’ velocity, since their charge is negative, and $\vect B$ circles the axis in the right-hand sense of that current.

Check

For $\alpha = 1$ , (2) gives a uniform $j = 2b/\mu_0$ , and indeed inside a cylinder carrying that uniform current Ampère’s law gives back the given field:

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

Units: $b r^{\alpha}$ is a field, so $b r^{\alpha-1}/\mu_0$ is $\text{T}/(\text{m}\cdot\text{T}\cdot\text{m/A}) = \text{A/m}^2$ , a current density.

Remark

The exponent decides where the current is carried. For $\alpha > 1$ the density vanishes on the axis and grows outward, a hollow-cored beam; for $\alpha < 1$ it diverges on the axis as $r^{\alpha-1}$ , yet the current inside radius $r$ , which is $2\pi r B/\mu_0 \propto r^{\alpha+1}$ , still goes to zero there. This is the differential form of Ampère’s law at work: in cylindrical symmetry $(1/r)\,d(rB)/dr = \mu_0 j$ .