Irodov 3.354: Biot and Savart from Displacement Current

Problem Irodov 3.354

A point charge $q$ moves at a constant velocity $\vect v$ , slow compared with the speed of light. A circle is drawn perpendicular to $\vect v$ , centred at the point $O$ on the charge’s line of motion, and ahead of the charge; $A$ is a point of the circle, and $\vect r$ is the vector from the charge to $A$ , making the angle $\theta$ with $\vect v$ . By applying the circulation theorem for $\vect H$ to this circle, find $\vect H$ at $A$ in terms of $\vect r$ and $\vect v$ .

The charge q moving with v toward a circle perpendicular to its path, centred at O. The vector r runs from the charge to the point A on the circle, at the angle θ to v. — The charge $q$ moving with $\vect v$ toward a circle perpendicular to its path, centred at $O$ . The vector $\vect r$ runs from the charge to the point $A$ on the circle, at the angle $\theta$ to $\vect v$ .

Given

$q$
$\vect v = \text{const}$ , $v \ll c$
$\vect r$ from $q$ to $A$ , at angle $\theta$ to $\vect v$

Find

\vect H

A

Solution

Step 1Only displacement current threads the circle

The circulation theorem for $\vect H$ counts the conduction current and the displacement current through any surface spanning the contour. The charge never crosses a surface that stays ahead of it, so only the rate of change of the flux of $\vect D$ survives:

\oint \vect H\cdot d\vect l = \frac{d\Phi_D}{dt} \tag{1}

The contour is oriented counterclockwise about $\vect v$ (right-hand rule), so the flux $\Phi_D$ is taken along $\vect v$ .

Step 2The flux of D through the circle

On that sphere $D = q/(4\pi r^2)$ everywhere, so the flux through the cap is $D$ times the cap’s area, $2\pi r^2(1-\cos\theta)$ :

\Phi_D = \frac{q}{4\pi r^2}\cdot 2\pi r^2(1-\cos\theta) = \frac{q}{2}(1-\cos\theta) \tag{2}

The spanning surface S: the cap of the sphere of radius r about the charge, on which D is normal and uniform. H at A is tangent to the circle. — The spanning surface $S$ : the cap of the sphere of radius $r$ about the charge, on which $\vect D$ is normal and uniform. $\vect H$ at $A$ is tangent to the circle.

Step 3How fast the flux grows as the charge approaches

The circle stays fixed while the charge moves. Let $x$ be the charge’s distance from the circle’s plane and $\rho$ the circle’s radius, which is constant. The charge approaches, so $x$ shrinks at the rate $v$ :

\cos\theta = \frac{x}{\sqrt{x^2+\rho^2}}, \qquad \frac{dx}{dt} = -v

Differentiating $\cos\theta$ with respect to $x$ , then using $\sqrt{x^2+\rho^2} = r$ and $\rho = r\sin\theta$ :

\frac{d\cos\theta}{dx} = \frac{\rho^2}{(x^2+\rho^2)^{3/2}} = \frac{\sin^2\theta}{r} \tag{3}

Then the chain rule in (2), with $dx/dt = -v$ , gives the rate at which the flux grows:

\frac{d\Phi_D}{dt} = -\frac{q}{2}\,\frac{d\cos\theta}{dx}\,\frac{dx}{dt} = \frac{q v\sin^2\theta}{2r} \tag{4}

Step 4The magnitude of H

The circulation is then $H$ times the circumference $2\pi\rho = 2\pi r\sin\theta$ . Equating it to (4) through (1):

H\cdot 2\pi r\sin\theta = \frac{q v\sin^2\theta}{2r}

Dividing by $2\pi r\sin\theta$ :

H = \frac{q v\sin\theta}{4\pi r^2} \tag{5}

Step 5The direction, and the vector form

For $q > 0$ the flux (4) grows along $\vect v$ , so $\vect H$ circulates counterclockwise about $\vect v$ . At $A$ that tangent direction is the direction of $\vect v\times\vect r$ , whose magnitude is $v r\sin\theta$ , so (5) is the magnitude of $q\,\vect v\times\vect r/(4\pi r^3)$ . For $q < 0$ both the rate and this vector reverse together.

\vect H = \frac{q\,\vect v\times\vect r}{4\pi r^3}

(6)

Check

Units: $q v/r^2$ is $\text{C}\cdot(\text{m/s})/\text{m}^2 = \text{A/m}$ , the unit of $H$ . Limits: at $\theta \to 0$ the circle shrinks onto the line of motion ahead of the charge and $H \to 0$ , as $\sin\theta$ in (5) says; at $\theta = 90^\circ$ the charge is passing through the circle’s plane and $H = qv/(4\pi r^2)$ is largest, the field abeam of the charge.

Remark

Result (5) is the Biot and Savart law for a current element, with $q\vect v$ standing for $I\,d\vect l$ . Here no current ever threads the circle: the whole magnetic field of the slow charge is produced by the displacement current of its own changing Coulomb field.