# Magnetic field strength dependent on distance from source

1. Aug 17, 2011

### babemagnetics

I am trying to find an equation that tells the strength of a magnetic field a given distance away from the source. It would be very helpful if all terms are defined, since the internet is notorious for not saying what variables mean.

Gracias

2. Aug 17, 2011

### vanhees71

In leading order the magnetostatic field far away from the sources (permanent magnets or currents) goes like $1/r^3$, where $r$ is the distance from the source. The multipole expansion for magnetic fields always starts with the dipole term since there are no magnetic monopoles (at least nobody could find any yet). I think, the corresponding Wikipedia article is a quite good source for first orientation:

http://en.wikipedia.org/wiki/Dipole#Vector_form

3. Aug 17, 2011

### babemagnetics

First of all, thank you. Second, on the Wikipedia page, I don't see the equation you provided; is the strength of a magnetic field really as simple as just multiplying the magnetic field at the source by 1 divided by the cube of the distance you are from the magnet? Also, what are the units that $1/r^3$ is made to work with please (Tesla? Gauss?)?

Gracias

4. Aug 18, 2011

### Philip Wood

It depends what you mean by 'source'. Replies so far assume you mean a current loop. In that case, for a plane loop carrying current I around the edge of a 'hole' of area A, the field on the axis of the loop, and a long way, x, from the loop is given by
$$B = \frac{\mu_{0} I A}{2\pi x^3}$$

IA is called the magnetic moment, M, of the loop.

B comes out in tesla, if other quantities are entered in SI units.

Formulae for the field at points off-axis, a long way from the loop can also be given. But if you're closer to to the loop than, say, ten times its radius (or side length or other characteristic dimension) these formulae become increasingly inaccurate. More accurate ones are complicated and depend on the shape of the loop, not just its area.

Another approach is to regard a current element (a short length of conductor carrying a current) as a source. There is a famous formula for the field due to a current element, usually called the Biot-Savart Rule. It's an inverse square law, with the complication of an angular dependence. It's in all e-m texts. The field at a point in the vicinity of a circuit (that is a current loop of any shape) can be found by integrating up the fields at that point from all the current elements in the circuit, remembering that the fields are vectors.

Last edited: Aug 18, 2011
5. Aug 18, 2011

### babemagnetics

So the 1/r^3 equation is no good? That guy has a PhD. If 1/r^3 is correct, then would someone please tell me whether it's in Gauss or Tesla? Also, how far is "magnetostatic field far away from the sources (permanent magnets or currents)"? And I am assuming r is in meters?

Gracias once again

Last edited: Aug 18, 2011
6. Aug 19, 2011

### vanhees71

Philip Wood's reply is correct. It's the dipole approximation for a current loop, and he's given the magnitude of the magnetic field (formerly called magnetic induction) in SI units. So, as he said, his result gives you the magnetic field in Tesla, if you plug in the current in Ampere, the length in meter, and the area in $\mathrm{m}^2$.

7. Aug 19, 2011

### Philip Wood

No, what vanHees77 said is correct (at large distances from the current loop). The formula I gave was also, please note, an inverse cube law - for for the simple special case of the field along the axis of the loop (that is along a line normal to the plane of the loop, passing through the centre of the loop. It included the 'constants', because I thought you wanted to know these. A more general formula for the magnitude of the field at a distance along way, r, from the centre of the dipole, and along a 'radius' at angle $\theta$ to the axis is

$$B = \frac{\mu_{0}AI \sqrt{1 + cos^2(\theta)}}{4\pi r^3}$$

If you need to know the direction of the field at (r, $\theta$), you'll need a vector equation or scalar equations for both the radial and tangential field components.

All these equations give [B in tesla, as long as what you put into the formulae is SI.

Just to clear up a possible confusion generated by my last post... I said that the Biot-Savart law gave an inverse square law for the field due to a current element. What I might have added, is that if you apply it to a current loop, adding up the contributions from all the elements into which you split the loop, you get just exactly the inverse cube formula I've just quoted (at points a long way from the loop).

Last edited: Aug 19, 2011
8. Aug 24, 2011

### babemagnetics

Using 1/r^3, do you multiply 1/r^3 by the magnetic field source strength, or do you subtract 1/r^3 from the magnetic field source strength?

Gracias

9. Aug 24, 2011

### Born2bwire

You multiply. The field falls off as 1/r^3. This is, as already stated by the others, a rough approximation since we have no knowledge of what your magnet is like. But it is generally a good guess that most magnetic sources are primarily dipoles.

10. Aug 24, 2011

### babemagnetics

Gracias sir

11. Nov 17, 2012

### downtownjapan

Can anyone add any information on calculating how the field changes when the distance is very, very small? For example 0.000075 meters? I am assuming the 1/r^3 formula does not apply at this distance.
I have a coil 0.00004 meters in diameter producing a magnetic field of 31.69 tesla. Can anyone suggest how I can measure the field strength at a point 0.000075 meters away from the center of the coil?
(I hope it is okay to post a question here and not an answer. I am trying to calculate the same type of thing as the original poster. Apologies if this is not the correct thing to do!)

12. Nov 17, 2012

### Philip Wood

That's a VERY small coil and a VERY large field. Is this a real set-up or a theoretical problem?

Do you mean measure the field strength, or do you mean predict what it will be?

If the coil is a circular loop then it's easy to predict the flux density at any point on its axis. If you can't find the relevant formula, please ask. For off-axis points, it's very much harder!

13. Nov 19, 2012

### Philip Wood

Downtownjapan. Have you received the information you wanted?

14. Nov 19, 2012

### Staff: Mentor

Looks problematic as real setup. As a current loop and with 31T field strength in the center, this would require ~1000A.
Assuming ~(40µm)^2 cross-section for the current loop (note that this does not fit within the described diameter), this corresponds to a current density of 600kA/mm^2 or 6*10^7 A/cm^2. I found values up to ~10^7A/cm^2 for superconductors - but those are at zero magnetic field!

Bigger current loops have a better scaling - the cross-section increases with diameter squared, while the distance term drops with diameter.

15. Nov 20, 2012

### tris_d

Can you point some reference about that equation?

16. Nov 20, 2012

### Staff: Mentor

17. Nov 20, 2012

### tris_d

$$B = \frac{\mu_{0} I A}{2\pi x^3}$$

I don't see this equation was give explicitly. What is it supposed to be describing:

a.) Field at Center of Current Loop

b.) Field on Axis of Current Loop

c.) Something else

I could also not find it anywhere on Wikipedia, but I found something which confused me even more:

http://en.wikipedia.org/wiki/Biot–Savart_law

Electric currents (along closed curve):

1.)
$$\mathbf{B} = \frac{\mu_0}{4\pi} \int_C \frac{I d\mathbf{l} \times \mathbf{r}}{|\mathbf{r}|^3}$$

- where r is the full displacement vector from the wire element to the point at which the field is being computed and r̂ is the unit vector of r. Using this the equation can be equivalently written:

2.)
$$\mathbf{B} = \frac{\mu_0}{4\pi}\int_C \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|\mathbf{r}|^2}$$

I think the first equation can not be without unit vector, and that it is CIRCLE integral and should be written like this:

1.)
$$\mathbf{B} = \frac{\mu_0}{4\pi} \oint_C \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|\mathbf{r}|^3}$$

where top 'r' is unit vector which they missed to denote, and the second equation is LINE integral and should be written like this:

2.)
$$\mathbf{B} = \frac{\mu_0}{4\pi}\int_L \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|\mathbf{r}|^2}$$

Now, if I got that right, the question is why would circle integral be 1/r^3 instead of 1/r^2?

Last edited: Nov 20, 2012
18. Feb 21, 2013

### mrv11

I have a practical question to add...

Using a calculator found here: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html#c3

I can calculate the magnetic field within a given solenoid. To verify this I have a measured current flowing through the wire and a compass. I align the solenoid at a right angle to the earth's magnetic field and bring the solenoid closer to the compass until it is deflected 45 degrees.

Therefore using vector analysis, I know the field produced by my solenoid at this distance to be equal to the field of the earth at my location (roughly 4.0 x 10-5 Tesla).

To verify the calculated magnetic field strength in my solenoid do I use the a re-arranged version of
B=(μ0NIA)/(2∏x3) ?
N= number of turns

I cannot imagine Bsolenoid=Bat compassr3 would work... The dimensional analysis would leave me with improper units!

Thank you in advance!

19. Feb 22, 2013

### Staff: Mentor

Last edited: Feb 22, 2013
20. Aug 1, 2013

### mrv11

Thank you all for the help on this little project!

I have to apologize about the delay for my thanks. School became overwhelming so I had to put this on hold. I reviewed the info you all provided, along with a great text a professor lent me. While not one of those individual sources described exactly my situation, I was able to synthesize it all into a decent model to describe my data.

Thanks again!

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