# The Strength of the Tangential and Radial Components of a Dipole Magnetic Field

I am following along well several textbooks (Geophysics) that helps me understand the in-depth physics behind the magnetic field of a dipole magnet. I understand that the basic magnetic potential (when observing a dipole magnet where the one of the poles is far enough to have negligible effect on its partner is:

$$W= -\int_r^{\infty}B \space dr = \frac{u_op}{4 \pi r}$$

Then to find the magnetic potential at point P due to both poles are:

$$W(\theta, r)=\frac{u_om\space \cos \theta}{4 \pi r^2}$$

Now to find the magnetic field strength at point P, I know its the vector addition of both $B_r$ and $B_\theta$ (radial and tangential respectively), and to find both I need to differentiate the potential with respect to r and $\theta$

Now here is finally where I get to ask my question. To find $B_r$ is easy enough by:

$$B_r=\frac{\partial W}{\partial r} = -\frac{2 u_o m\space \cos\theta}{4 \pi r^3}$$

But when I take the potential and differentiate with respect to $\theta$ I should get:

$$B_\theta = \frac{\partial W}{\partial \theta} = -\frac{u_o m\space \sin\theta}{4 \pi r^2}$$

but in ALL the textbooks they multiply $\frac{1}{r}$ to get:

$$B_\theta = \frac{1}{r}\frac{\partial W}{\partial \theta} = -\frac{u_o m\space \sin\theta}{4 \pi r^3}$$

Which is needed to find the total magnetic strength field ($B_r + B_\theta$)

Can anyone please explain where they got that extra $\frac{1}{r}?$

I have a feeling it has something to do with a unit vector, but I can't seem to connect the dots.

.

• hint.- if one can think of gradient of the potential in radial and tangential direction the differential change of position of a point at r at theta and another at same r at theta +d(theta)-we have differential change { r.d(theta)}. – drvrm Mar 6 '16 at 13:40

The simplest way to see why $B_\theta = \frac{1}{r}\frac{\partial W}{\partial \theta}$ and not $B_\theta = \frac{\partial W}{\partial \theta}$ is to imagine what a differential element of length looks like in spherical coordinates.

When you differentiate a field with respect a coordinate, you are effectively dividing a differential change in your field by a differential change in that coordinate (there are some subtleties & caveats regarding that interpretation, but it will give us the intuition you're looking for).

In cartesian coordinates, a differential unit of distance is simply $dx$, $dy$, or $dz$, so differentiating a field $A$ looks like $\frac{dA}{dx}$, $\frac{dA}{dy}$, or $\frac{dA}{dz}$.

In spherical coordinates, it's not so simple. The radial coordinate is simple, because displacement in the radial direction is just $\Delta r$. Therefore a differential element of radius is $dr$, and differentiating $A$ with respect to $r$ looks like $\frac{dA}{dr}$. However, the angular coordinates are more complicated.

A differential unit of distance in the azimuthal direction is given by $r d\theta$. You can see that geometrically in this picture, which shows the two sides of a differential element of area in a polar coordinate system (which is simply a 2D slice of our spherical coordinate system).

Notice that the length of the side of the area element in the azimuthal direction is labeled $r d\theta$. The length of that element is proportional to both the change in angle, and the radius.

Since the differential unit of distance in the azimuthal direction is $r d\theta$, when we differentiate our field $A$ with respect to that coordinate, the result isn't simply $\frac{dA}{d\theta}$, it's $\frac{dA}{r d\theta}$ or $\frac{1}{r} \frac{dA}{d\theta}$.

• Thank you! Visualising this makes it so much better. Thank you! – Uys of Spades Mar 6 '16 at 23:32

The magnetic field is the gradient of magnetic potential. The gradient operation is defined in Cartesian coordinates $(x, y, z)$ as $$\vec{\nabla} u = {\partial u \over \partial x} \mathbf{\hat{x}} + {\partial u \over \partial y} \mathbf{\hat{y}} + {\partial u \over \partial z} \mathbf{\hat{z}}$$ In order to work in polar coordinates $(r, \theta, \phi)$, you need to express derivates with respect to $(x, y, z)$ in terms of derivatives with respect to $(r, \theta, \phi)$ and unit vectors $\mathbf{\hat{x}}, \mathbf{\hat{y}}, \mathbf{\hat{z}}$ in terms of $\mathbf{\hat{r}}, \mathbf{\hat{\theta}}, \mathbf{\hat{\phi}}$. For example: $${\partial f \over \partial x} = {\partial f \over \partial r}{\partial r \over \partial x} + {\partial f \over \partial \theta}{\partial \theta \over \partial x} + {\partial f \over \partial \phi}{\partial \phi \over \partial x}$$ When you do this for all coordinates, you will see that $$(\vec{\nabla} u)_\theta = {1 \over \theta}{\partial u \over \partial \theta}$$ Getting to this point using this method will take some work though.

• So I would need to refreshen up on my polar coordinates, haven't touched them in ages. Although I still don't understand; deriving everything from the beginning we stay with Cartesian coordinates even when finding the potential of point P, where and when does the switch over take place? – Uys of Spades Mar 6 '16 at 13:49