1 Introduction

In cartesian coordinates with cartesian unit vectors, the approximate magnetic field created by a current $I$ flowing in the direction of $\hat{ϕ}$ in a circle of radius $b$ that is centered on the origin and lies in the $x - - y$ plane for $b ≪ r$ is

$B = \frac{μ _{o}}{4 π} \frac{I π b ^{2}}{r ^{5}} (3 x z \hat{x} + 3 yz \hat{y} + (3 z^{2} - r^{2}) \hat{z})$

where $r^{2} = x^{2} + y^{2} + z^{2}$ . In spherical coordinates with spherical unit vectors, the field is

$B = \frac{μ _{0}}{4 π} \frac{I π b ^{2}}{r ^{3}} (2 cos θ \hat{r} + sin θ \hat{θ})$

With cylindrical unit vectors and spherical coordinates, it was shown in HW #11 that

$B = \frac{μ _{0}}{4 π} \frac{I π b ^{2}}{r ^{3}} (3 cos θ sin θ \hat{s} + (3 cos^{2} θ - 1) \hat{z})$

In the limit $b / r \to 0$ , the current loop is referred to as a perfect dipole. See Vector Potential for the derivation.

1.1 Problem – Magnetic Dipole Transformation from Cartesian to Spherical

Starting with the equation for $B$ above for a magnetic dipole in spherical coordinates with spherical unit vectors, show that

$B = \frac{μ _{o}}{4 π} \frac{I π b ^{2}}{r ^{5}} (3 x z \hat{x} + 3 yz \hat{y} + (3 z^{2} - r^{2}) \hat{z})$

%Answer

%From the equation list on the last page of Griffiths,

%$$\hat{\mathbf{x}} = \sin\theta,\cos\phi,\hat{\mathbf{r}} + \cos\theta,\cos\phi,\hat{\boldsymbol{\theta}} - \sin,\theta,\hat{\boldsymbol{\phi}}$$

%$$\hat{\mathbf{y}} = \sin\theta,\sin\phi,\hat{\mathbf{r}} + \cos\theta,\sin\phi,\hat{\boldsymbol{\theta}} + \cos,\theta,\hat{\boldsymbol{\phi}}$$

%$$\hat{\mathbf{z}} = \cos\theta,\sin\phi,\hat{\mathbf{r}} - \sin,\theta,\hat{\boldsymbol{\theta}}$$

%Recall that one can check these equations by choosing values of (\theta) and (\phi) and verifying that the cartesian unit vector points in the expected direction based on a diagram.

%From the previous problem,

%$$\mathbf{B}=\frac{\pi a^2\mu_o}{4\pi}\left(\frac{3xz}{r^5}\hat{\mathbf{x}}+\frac{3yz}{r^5}\hat{\mathbf{y}}+\frac{3z^2-r^2}{r^5}\hat{\mathbf{z}}\right)$$

%Using the unit vector equations above and the relationships

%$$x=r\sin,\theta,\cos,\phi$$

%$$y=r\sin,\theta,\sin,\phi$$

%$$z=r\cos,\theta$$

%and address each component individually and use (\cos^2\theta+\sin^2\theta=1). The final result should be

%$$\mathbf{B}=\frac{\pi a^2\mu_o}{4\pi}\frac{1}{r^3}\left(2,\cos,\theta\hat{\mathbf{r}}+\sin,\theta,\hat{\boldsymbol {\theta}}\right)$$

%## Using the Magnetic Dipole Equation

%Use the equation for a magnetic dipole to find the magnetic field at the point $y=y_o$ due to a current loop of radius $b$ centered on the $z$-axis and lying in the $z=d$ plane.

%What is the direction of the current required for the dipole equation to apply without modification?

1.2 Magnetic Dipole in Spherical Coordinates

1. Use

$A (r, θ, ϕ) = \frac{μ _{0}}{4 π} \frac{m sin θ}{r ^{2}} \hat{ϕ}$

to compute $\nabla \times A$ using the spherical form of the curl operator. Your answer should be

$B (r, θ, ϕ) = \frac{μ _{0}}{4 π} \frac{m}{r ^{3}} (2 cos θ \hat{r} + sin θ \hat{θ})$

2. Write the equation for $A$ above in cartesian coordinates and compute $\nabla \times A$ using the cartesian form of the curl operator.

1.3 Magnetic Dipole $B$ in Cylindrical Coordinates

In the above $B$ was given in spherical coordinates with cylindrical unit vectors. Find $B$ in cylindrical coordinates with cylindrical unit vectors.

2 Magnetic moment, $m$

For electric dipoles, the potential due to charges $\pm q$ at $z = \pm d$ for $r ≫ d$ is

$V = \frac{1}{4 π ϵ _{o}} \frac{q d}{r ^{2}}$

The generalization of this equation for charges $\pm q$ at $r_{\pm}$ was

$V = \frac{1}{4 π ϵ _{o}} \frac{p \cdot r ^}{r ^{2}}$

where the “electric dipole moment”

$p \equiv q (r_{+} - r_{-})$

was introduced. This equation has the constraints that $r_{+} = r_{-}$ , the center of the line connecting the charges is at the origin and $r_{+} ≪ r$ .

Previously, it was found that the vector potential for a circular current loop of radius $b$ in the $x$ – $y$ plane and centered on the origin was

$A = \frac{μ _{0}}{4 π} \frac{I π b ^{2}}{r ^{2}} sin θ \hat{ϕ}$

If we define the magnetic moment for a circular loop as

$m = I A \hat{n}$

then it can be shown that

$A = \frac{μ _{o}}{4 π} \frac{m \times r ^}{r ^{2}}$

The area $A$ in $m$ is the cross–sectional area of the loop; the normal vector $\hat{n}$ for the circle is the vector that points in the direction of your thumb when you wrap your fingers around the loop.

The equation $m = I A \hat{n}$ actually applies to a loop of any shape provided that the loop is in a plane. The equation for $A$ requires that the maximum extent of the loop is much less than $r$ and that the loop is centered on the origin.

If the loop is not in a plane, then $m = I \int d a \hat{n}$ . See also 5.3.4 of Griffiths. (The integral is over any area bounded by the loop; for a circle, the area could be a disk, which would be the most obvious choice, or a dome, for example. The dome area can be visualized by gluing a circular rubber disk to the circle and then blowing on the disk so as to expand it. Or, think of the shape created when creating soap bubbles.)

%## Example – Computing $\mathbf{B}$ for Circular Loops not in $x$–$y$ plane.

2.1 Computing $m$

A square loop is centered on the origin and lies in the $x$ – $y$ plane. Its sides are parallel to either the $x$ or $y$ axes. It is then rotated around the $x$ –axis by $α$ . Find $m$ .

%## Computing $\mathbf{m}$

%An equilateral triangle has sides of length $b$, is centered on the origin, and lies in the $x$–$y$ plane.

%1. Does $\mathbf{m}$ change if the triangle is rotated around the $z$–axis? %2. Compute $\mathbf{m}$.

3 Coordinate–Free Dipole Field

The equations for $B$ given in the introduction only apply to a circular centered on the origin and lying in the $x - - y$ plane. The “coordinate-free” equation for a loop centered at the origin but with arbitrary orientation and shape is

$B = \frac{μ _{0}}{4 π} \frac{1}{r ^{3}} (3 (m \cdot \hat{r}) \hat{r} - m)$

where $m = I \int d a$ is the magnetic moment and $d a$ is a differential area vector.

This equation can be derived from

$A = \frac{μ _{o}}{4 π} \frac{m \times r ^}{r ^{2}}$

3.1 Using the Coordinate Free Magnetic Dipole Equation

Use the coordinate-free dipole equation to compute the magnetic field created by a magnetic dipole with dipole moment $m$ oriented in the $\hat{z}$ –direction in cartesian coordinates.