Irodov 3.147: A Current Without Any Wire – Amit & Deepali Sinha Foundation

Problem Irodov 3.147

A long cylinder of radius $a = 1.0$ cm carries a charge spread uniformly over its surface and moves along its own axis at the constant speed $v = 10$ m/s. The electric field at the surface of the cylinder is $E = 0.9$ kV/cm. Find the convection current, the current produced by the mechanical motion of the charge.

A length of the long cylinder of radius a, uniformly charged over its surface (drawn for a positive charge), moving along its axis with velocity v. — A length of the long cylinder of radius $a$ , uniformly charged over its surface (drawn for a positive charge), moving along its axis with velocity $\vect{v}$ .

A current is the charge carried across a fixed cross-section per unit time. Here no charge moves relative to the cylinder; the charge is carried along by the cylinder itself, so the current is the charge per unit length times the speed. The field at the surface tells us that charge.

Given

$a = 1.0$ cm
$v = 10$ m/s
$E = 0.9$ kV/cm at the surface
air outside (assumed)

Find

the convection current

I

Solution

Step 1The charge per unit length from the field

Let the cylinder carry $\lambda$ per unit length; it is surrounded by air (assumed), with permittivity taken as $1$ . Far from the ends, by symmetry the field is radial and the same all round the axis, and Gauss’s theorem on a coaxial cylinder of length $\ell$ just outside the surface gives

E\cdot 2\pi a\ell = \frac{\lambda\ell}{\varepsilon_0}

Solving for $\lambda$ :

\lambda = 2\pi\varepsilon_0 aE \tag{1}

Step 2The charge carried past a fixed plane

In a time $dt$ the cylinder moves on by $v\,dt$ , so the charge on that length crosses a fixed plane normal to the axis. Dividing it by $dt$ gives the current, into which (1) then goes:

\begin{aligned} dq &= \lambda v\,dt\\ I &= \frac{dq}{dt} = \lambda v \end{aligned}

Numerically, with $a = 0.010$ m, $E = 0.9$ kV/cm $= 9 \times 10^4$ V/m and $v = 10$ m/s:

I = 2\pi \times 8.85 \times 10^{-12} \times 0.010 \times 9 \times 10^4 \times 10\ \text{A} = 5.0 \times 10^{-7}\ \text{A}

I = 2\pi\varepsilon_0 aEv \approx 0.5\ \mu\text{A}

(2)

For a positive charge the current points along $\vect{v}$ , for a negative one against it.

Check

Units: $\varepsilon_0 E$ is a surface charge density, C/m $^2$ ; times the radius $a$ it is a charge per unit length, C/m; times a speed it is C/s, an ampere. A second route to the number: the field just outside fixes the surface density, $\sigma = \varepsilon_0 E$ , and each second a band of surface $2\pi a$ wide and $v$ long passes the plane:

\begin{gathered} \sigma = 8.0 \times 10^{-7}\ \text{C/m}^2\\ 2\pi a\,v = 0.63\ \text{m}^2/\text{s} \end{gathered}

Their product is $5.0 \times 10^{-7}$ A again.

Remark

The current is tiny, and it cannot be made much larger: the field at the surface is capped by the breakdown of air, about $30$ kV/cm, which at this radius and speed allows only about $17~\mu$ A. That is why the first demonstration that moving charge makes a magnetic field just as a current in a wire does (Rowland, 1876) needed a charged disc spun at high speed and a very delicate magnetometer.