# 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)}. Mar 6, 2016 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! Mar 6, 2016 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? Mar 6, 2016 at 13:49