Irodov 3.40: Where the Dipole Field Turns Sideways – Amit & Deepali Sinha Foundation

Problem Irodov 3.40

A point dipole of moment $\vect{p}$ sits at the origin and points along the positive $z$ axis. A point $S$ has position vector $\vect{r}$ , of length $r$ , making the angle $\theta$ with the positive $z$ axis. Find the component $E_z$ of the electric field at $S$ along the $z$ axis and the component $E_\perp$ perpendicular to the $z$ axis (in the plane containing the $z$ axis and $\vect{r}$ ). At which points is the field $\vect{E}$ perpendicular to $\vect{p}$ ?

The point dipole p at the origin, pointing up the z axis, and the point S at position r, its polar angle θ measured from the z axis. — The point dipole $\vect{p}$ at the origin, pointing up the $z$ axis, and the point $S$ at position $\vect{r}$ , its polar angle $\theta$ measured from the $z$ axis.

Given

$\vect{p}$ along $+z$ at the origin
$S$ at distance $r$ , polar angle $\theta$

Find

E_z

and

E_\perp

S

; the points where

\vect{E} \perp \vect{p}

Solution

Step 1The dipole’s potential

At distances large compared with the dipole’s size, the potential is that of the moment alone. With $\vect{p}\cdot\vect{r} = pr\cos\theta$ ,

\varphi = \frac{1}{4\pi\varepsilon_0}\frac{\vect{p}\cdot\vect{r}}{r^3} = \frac{p}{4\pi\varepsilon_0}\frac{\cos\theta}{r^2} \tag{1}

Step 2The field along r and along growing theta

The field is minus the gradient of (1). In polar coordinates in the plane of $z$ and $\vect{r}$ , the gradient has a radial part $\partial\varphi/\partial r$ and a transverse part $(1/r)\,\partial\varphi/\partial\theta$ , the latter pointing in the direction of growing $\theta$ (away from the $z$ axis above the equator, toward it below).

E_r = -\frac{\partial\varphi}{\partial r} = \frac{p}{4\pi\varepsilon_0}\frac{2\cos\theta}{r^3} \tag{2}

E_\theta = -\frac{1}{r}\frac{\partial\varphi}{\partial\theta} = \frac{p}{4\pi\varepsilon_0}\frac{\sin\theta}{r^3} \tag{3}

The field at S split along r (E_r) and along the direction of growing θ (E_θ). — The field at $S$ split along $\vect{r}$ ( $E_r$ ) and along the direction of growing $\theta$ ( $E_\theta$ ).

The same field split along the z axis (E_z) and across it (E_⊥). — The same field split along the $z$ axis ( $E_z$ ) and across it ( $E_\perp$ ).

Step 3Projecting onto the z axis and across it

The unit vector along $\vect{r}$ has components $\cos\theta$ along $z$ and $\sin\theta$ across it; the unit vector of growing $\theta$ has $-\sin\theta$ along $z$ and $\cos\theta$ across it. Projecting $E_r$ and $E_\theta$ from (2) and (3) onto $z$ :

\begin{aligned} E_z &= E_r\cos\theta - E_\theta\sin\theta\\ &= \frac{p}{4\pi\varepsilon_0 r^3}\left(2\cos^2\theta - \sin^2\theta\right) \end{aligned}

Writing $\sin^2\theta = 1 - \cos^2\theta$ in the bracket,

E_z = \frac{p}{4\pi\varepsilon_0}\,\frac{3\cos^2\theta - 1}{r^3}

(4)

Across the axis, in the same way,

\begin{aligned} E_\perp &= E_r\sin\theta + E_\theta\cos\theta\\ &= \frac{p}{4\pi\varepsilon_0 r^3}\left(2\sin\theta\cos\theta + \sin\theta\cos\theta\right) \end{aligned}

E_\perp = \frac{p}{4\pi\varepsilon_0}\,\frac{3\sin\theta\cos\theta}{r^3}

(5)

Step 4Where the field is perpendicular to p

$\vect{E}$ is perpendicular to $\vect{p}$ exactly where its component along $\vect{p}$ , which is $E_z$ , vanishes. Setting (4) to zero:

\begin{aligned} 3\cos^2\theta - 1 &= 0\\ \cos\theta &= \pm\frac{1}{\sqrt{3}} \end{aligned}

The two roots describe one cone about the $z$ axis with its vertex at the dipole, opening upward ( $\theta_1$ ) and downward ( $\theta_2 = 180^\circ - \theta_1$ ). $\vect{E}$ is perpendicular to $\vect{p}$ at every point of its lateral surface:

\begin{gathered}\theta_1 = \arccos\frac{1}{\sqrt{3}} = 54.7^\circ\\ \theta_2 = 125.3^\circ\end{gathered}

(6)

Field lines of the point dipole and the cone (gray dashes): its upper nappe at θ_1 = 54.7° from the z axis, its lower at θ_2 = 125.3°. Wherever a line crosses the cone it runs horizontally, as E does at the dot. — Field lines of the point dipole and the cone (gray dashes): its upper nappe at $\theta_1 = 54.7^\circ$ from the $z$ axis, its lower at $\theta_2 = 125.3^\circ$ . Wherever a line crosses the cone it runs horizontally, as $\vect{E}$ does at the dot.

On the cone only $E_\perp$ survives, so the field there is all perpendicular. With $\cos\theta = 1/\sqrt{3}$ and $\sin\theta = \sqrt{2/3}$ in (5):

3\sin\theta\cos\theta = 3\sqrt{\frac{2}{3}}\,\frac{1}{\sqrt{3}} = \sqrt{2}

E = E_\perp = \frac{\sqrt{2}\,p}{4\pi\varepsilon_0 r^3}

(7)

On the lower nappe, $\cos\theta = -1/\sqrt{3}$ , the same magnitude holds with $E_\perp$ negative: there the field points toward the axis.

Check

On the axis, $\theta = 0$ : (4) gives $E_z = 2p/(4\pi\varepsilon_0 r^3)$ and (5) gives zero, the familiar axial field along $\vect{p}$ . On the equator, $\theta = 90^\circ$ : $E_z = -p/(4\pi\varepsilon_0 r^3)$ and $E_\perp = 0$ , half the axial strength and antiparallel to $\vect{p}$ . Between these the sign of $E_z$ must change, and (4) changes it exactly once in $0 < \theta < 90^\circ$ , at $54.7^\circ$ .

Remark

Squaring and adding the two boxes gives the strength at any point,

E = \frac{p}{4\pi\varepsilon_0 r^3}\sqrt{1 + 3\cos^2\theta}

which falls from twice the equatorial value on the axis to $\sqrt{2}$ times it on the cone. The cone angle $\cos^2\theta = 1/3$ is the same magic angle that appears wherever a dipole interacts through $3\cos^2\theta - 1$ : the interaction energy of two parallel dipoles whose separation lies along it is zero, which is why nuclear magnetic resonance spins samples about an axis at $54.7^\circ$ to the field.