Irodov 3.401: The Orbit That Threads the Axis – Amit & Deepali Sinha Foundation

Problem Irodov 3.401

A particle of specific charge $q/m$ is at rest inside a long solenoid of circular cross-section, at the distance $r$ from its axis. The current in the winding is switched on, and the magnetic induction inside the solenoid rises to $B$ . Assuming the particle moves a negligible distance while the current is rising, find the particle’s velocity once the field has reached $B$ , and the radius of curvature of its subsequent path.

The solenoid's cross-section with the particle (the circled glyph) at distance r from the axis; B, drawn out of the page, is the field once the current is on, and q > 0 is drawn. — The solenoid’s cross-section with the particle (the circled glyph) at distance $r$ from the axis; $\vect B$ , drawn out of the page, is the field once the current is on, and $q > 0$ is drawn.

Given

$q/m$
$r$
$B$ after switch-on
no displacement while $B$ rises

Find

the speed

v

and the radius of curvature

\rho

of the path

Solution

Step 1The eddy field at the particle

By symmetry the eddy field is tangent to the circle of radius $r$ about the axis and has the same magnitude $E$ all round it. The circle encloses the uniform field over the area $\pi r^2$ , so Faraday’s law gives, in magnitude,

\oint\vect E\cdot d\vect l = 2\pi r\,E = \pi r^2\,\frac{dB}{dt}

Dividing by $2\pi r$ :

E = \frac{r}{2}\,\frac{dB}{dt} \tag{1}

By Lenz’s rule the field circulates clockwise as seen with the growing $\vect B$ pointing at the viewer.

While B grows out of the page, the eddy field circulates clockwise on the circle of radius r through the particle; E at the particle is tangent to that circle. — While $\vect B$ grows out of the page, the eddy field circulates clockwise on the circle of radius $r$ through the particle; $\vect E$ at the particle is tangent to that circle.

Step 2The kick from the switch-on

The particle stays at the distance $r$ from the axis, so it feels the tangential force $qE$ from (1) all the while the field rises. The magnetic force on the moving particle is perpendicular to its velocity and does not change its speed. Integrating Newton’s law from rest at $B = 0$ to the final field $B$ :

m\,\frac{dv}{dt} = qE = \frac{qr}{2}\,\frac{dB}{dt} \quad\Longrightarrow\quad m v = \frac{qr}{2}\int_0^B dB'

v = \frac{qBr}{2m}

(2)

The velocity is tangent to the circle of radius $r$ about the axis, perpendicular to the line from the axis to the particle.

Step 3The orbit in the steady field

Once the current is steady only the magnetic force acts, and it supplies the centripetal force of a circle of radius $\rho$ . Using (2) for $v$ :

\frac{m v^2}{\rho} = q v B \quad\Longrightarrow\quad \rho = \frac{m v}{qB} = \frac{m}{qB}\cdot\frac{qBr}{2m}

\rho = \frac{r}{2}

(3)

Since the velocity is perpendicular to the radius line from the axis, the centre of this circle lies on that line, halfway between the particle and the axis.

The path after switch-on: a circle of radius ρ = r/2, with v tangent and the magnetic force F toward its centre, halfway to the axis O. The circle passes through the axis. — The path after switch-on: a circle of radius $\rho = r/2$ , with $\vect v$ tangent and the magnetic force $\vect F$ toward its centre, halfway to the axis $O$ . The circle passes through the axis.

Check

Units: $qBr/m$ is $(\text{C}\cdot\text{T}/\text{kg})\cdot\text{m} = \text{m}/\text{s}$ , since $qB/m$ is a frequency. The speed does not depend on how fast the field was switched on, only on its final value, as an impulse argument requires: $\int qE\,dt = (qr/2)\int dB$ for any time profile. A particle on the axis ( $r = 0$ ) feels no eddy field and stays at rest, as (2) gives.

Remark

Every such orbit passes through the axis, whatever $r$ and $q/m$ . This is Busch’s theorem at work: for motion in an axially symmetric magnetic field the combination $m r v_\varphi + q\Phi/2\pi$ is conserved, with $v_\varphi$ the azimuthal velocity and $\Phi$ the flux through the circle of radius $r$ , both signed. It starts at zero, and its magnitude balance $m r v = q\pi r^2 B/2\pi$ is (2) again. On the axis both terms vanish, so only an orbit with the combination equal to zero can ever reach the axis; switching the field on under a particle at rest makes it zero.