Irodov 3.404: Where Field Equals Half Its Mean – Amit & Deepali Sinha Foundation

Problem Irodov 3.404

The magnetic induction of an axially symmetric field is a known function $B(r)$ of the distance $r$ from the field’s axis. Use the betatron condition (the induction on the orbit equals half the mean induction inside it) to find the radius of a circular electron orbit. Work the problem out in particular for $B = B_0 - ar^2$ , with $B_0$ and $a$ positive constants.

The given field profile $B = B_0 - ar^2$ against the distance from the axis, falling to zero at $r = (B_0/a)^{1/2}$ .

Given

$B(r)$ , axially symmetric
case $B = B_0 - ar^2$ , $B_0 > 0$ , $a > 0$

Find

the orbit radius

r_0

, in general and for this case

Solution

Step 1The condition as an equation for

r_0

On an orbit of radius $r_0$ the betatron condition asks $B(r_0) = \langle B\rangle/2$ , with $\langle B\rangle$ the flux through the orbit divided by its area. The flux is the sum over rings of radius $r$ and width $dr$ :

\langle B\rangle = \frac{\Phi}{\pi r_0^2} = \frac{1}{\pi r_0^2}\int_0^{r_0} B(r)\,2\pi r\,dr = \frac{2}{r_0^2}\int_0^{r_0} B(r)\,r\,dr

Putting this into $B(r_0) = \langle B\rangle/2$ and multiplying by $r_0^2$ :

r_0^2\,B(r_0) = \int_0^{r_0} B(r)\,r\,dr

(1)

The orbit radius is the root of this equation.

Step 2The parabolic profile

For $B = B_0 - ar^2$ the integral on the right of (1) is elementary:

\int_0^{r_0} (B_0 - ar^2)\,r\,dr = \frac{B_0 r_0^2}{2} - \frac{a r_0^4}{4}

Setting it equal to $r_0^2 B(r_0) = r_0^2(B_0 - ar_0^2)$ and dividing by $r_0^2$ :

\begin{aligned} \frac{B_0}{2} - \frac{a r_0^2}{4} &= B_0 - a r_0^2, \\ \frac{3}{4}\,a r_0^2 &= \frac{B_0}{2} \end{aligned}

r_0 = \sqrt{\frac{2B_0}{3a}}

(2)

Step 3The orbit as a crossing

The mean field inside a circle of radius $r$ follows from the same integral, $\langle B\rangle = B_0 - ar^2/2$ , so half of it falls more slowly than $B$ itself and starts lower. The two curves cross exactly once, at the orbit.

The field B on a circle of radius r and half the mean field inside it, ( B)/2, against r, in the units B_0 = a = 1. They cross at the orbit radius r_0. — The field $B$ on a circle of radius $r$ and half the mean field inside it, $\langle B\rangle/2$ , against $r$ , in the units $B_0 = a = 1$ . They cross at the orbit radius $r_0$ .

Check

Units: $B_0/a$ is an induction over an induction per square metre, an area, so $r_0$ is a length. At the orbit $B(r_0) = B_0 - 2B_0/3 = B_0/3$ , while the mean field inside it is $B_0 - ar_0^2/2 = B_0 - B_0/3 = 2B_0/3$ : exactly twice the field on the orbit, as the condition demands. The orbit also lies inside the region of positive field, since $r_0^2 = (2/3)B_0/a$ is less than $B_0/a$ .

Remark

The betatron condition places the orbit; whether the orbit holds when an electron strays from it is a separate question, answered by how steeply the field falls with $r$ . Weak focusing needs the field index $-(r/B)\,dB/dr$ to lie between 0 and 1, and for this profile it is $2ar_0^2/B(r_0) = 4$ at the orbit: the radius is right, but the profile falls too steeply to hold a straying electron on it.