# q. about magnetic fields **URL:** \ **Category:** Factual Questions\ **Created:** [July 23, 2013, 4:05pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302 "2013-07-23T16:05:48Z")\ **Posts on this page:** 16\ **Page:** 1
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [July 23, 2013, 4:05pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/1 "2013-07-23T16:05:48Z")
It occurs to me that a magnetic field can be very long range but not particularly intense (like the Earth’s magnetic field), or very intense but only over a short distance (like a MRI scanner)\*. So what’s the proper way to categorize the “strength” of a magnetic field? \*And yes, I know it could be both or neither; that’s not my point ---
**Author:** ![friedo](https://avatars.discourse-cdn.com/v4/letter/f/8edcca/32.png) [@friedo](https://boards.straightdope.com/u/friedo)\ **Post date:** [July 23, 2013, 4:46pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/2 "2013-07-23T16:46:54Z")
Magnetic field strength is measured in [teslas](http://en.wikipedia.org/wiki/Tesla_%28unit%29). An MRI machine magnet is anywhere from 0.5 to 3 teslas, depending on the machine. ETA: Sometimes you’ll also see field strength measured in gauss units. One guass is 1/10,000 of a tesla. ---
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [July 23, 2013, 4:55pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/3 "2013-07-23T16:55:11Z")
I think the OP’s question is actually asking about the energy in the field. As **friedo** says, the actual strength is in Teslas, and that measures the field strength(what the OP is terming intensity) at a point. But if you want to measure the whole field and compare big and weak versus small and strong and in some way equate the two, you mean the energy. ETA - on reflection, the footnote in the OP means the question is possibly better answered by **friedo** ’s answer. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [July 23, 2013, 5:44pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/4 "2013-07-23T17:44:08Z")
If you want to take the size of the field into account, then the energy technique is probably a good one. There’s an energy _density_ associated with magnetic fields, given by the square of the field strength (in tesla, if you’re in SI units) divided by 2\*[µ[sub]0[/sub].](http://en.wikipedia.org/wiki/Vacuum_permeability) Take this energy density, multiply it by the volume over which the field occurs (or, more accurately, integrate over the volume), and that gives you the amount of energy required to produce that field. The stronger a field is, or the more volume over which it extends, the more energy it will take to create. ---
**Author:** ![Pasta](https://avatars.discourse-cdn.com/v4/letter/p/ecccb3/32.png) [@Pasta](https://boards.straightdope.com/u/Pasta)\ **Post date:** [July 23, 2013, 7:07pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/5 "2013-07-23T19:07:21Z")
And if you’re just interested in making comparisons and not in the actual stored energy, you can leave out the various multiplicative factors and just integrate (field[sup]2[/sup]) over all of space. Order of magnitude estimates for MRI machine and Earth: MRI: assume uniform 3 T field inside a 3 meter long tube with diameter 1 meter. (3 T)[sup]2[/sup]\*(3 m)_pi_(0.5 m)[sup]2[/sup] = 20 T[sup]2[/sup] m[sup]3[/sup]. Earth: assume about 0.0001 T inside a sphere of radius, oh let’s say, 3 earth diameters. This yields 10[sup]14[/sup] T[sup]2[/sup] m[sup]3[/sup]. As **MikeS** said, if you do divide this by 2µ[sub]0[/sub], you get the energy stored in the field (since I’ve already multiplied by volume here)… MRI: 10[sup]7[/sup] joules Earth: 10[sup]20[/sup] joules ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [July 23, 2013, 7:56pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/6 "2013-07-23T19:56:47Z")
> [@Lumpy](#): > > It occurs to me that a magnetic field can be very long range but not particularly intense (like the Earth’s magnetic field), or very intense but only over a short distance (like a MRI scanner)\*. So what’s the proper way to categorize the “strength” of a magnetic field? > > \*And yes, I know it could be both or neither; that’s not my point I think it’s important to point out that if we’re looking at the magnetic field **outside** a magnet it decreases with distance. The Earth’s magnetic field has a long reach, but the strength does decrease with distance. According to [this calculator](http://www.ngdc.noaa.gov/geomag-web/#igrfwmm) from the National Geophysical Data Center the total field strength where I am is: 50914.8 nT at sea level 44781.1 nT at 300 km above sl 39558.9 nT at 600 km above sl If we look at magnets that are somewhat more comparable than a MRI scanner and a planet, like different sizes of bar magnets of different materials, it’s been my experience that you’re right. The influence of a strong neodymium magnet seems to decrease faster than that of a simple old iron bar magnet. Now the field strength of a magnet can’t be described by the simple inverse square law of a the electrical field of a point electrical charge, but I seem to recall that there is still a relationship between field strength and **distance from the center of the magnet** , even if it’s not a simple one. Now say you have a 10 cm bar magnet and a 1 cm neodymium magnet, at the surface of the magnet the neodymium magnet is a lot stronger. But if you measure the strength at 5 cm from the surface you’ve only doubled the distance from the center of the bar magnet, while you’re 6 times further from the center of the neodymium magnet. Now the Earth is a very large magnet, so you need to be a long way from the surface before the field strength decreases much, but it does decrease. And the short range of the MRI’s field isn’t a result of it being strong but as a result of the way it’s created, with the strong uniform field being **inside** the magnet. ---
**Author:** ![Pasta](https://avatars.discourse-cdn.com/v4/letter/p/ecccb3/32.png) [@Pasta](https://boards.straightdope.com/u/Pasta)\ **Post date:** [July 23, 2013, 9:21pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/7 "2013-07-23T21:21:06Z")
> [@naita](#): > > …10 cm bar magnet…1 cm neodymium magnet…Earth… All of these are well described as [dipoles](http://en.wikipedia.org/wiki/Magnetic_dipole) at distances a few times larger than the physical extent of the source. The field of a dipole drops off as the cube of the distance. You are right that near to these sources things will be more complicated, but not overly so until you get into the physical source itself. > [@](#): > > Now the Earth is a very large magnet, so you need to be a long way from the surface before the field strength decreases much, but it does decrease. The source of the Earth’s field is in the core, so one is already well away from it at the surface of the Earth. Thus, the cube-law works well: ((6400 km + 0 km) / (6400 km + 300 km))[sup]3[/sup] \* (0.51 gauss) = **0.44 gauss** at 300 km ((6400 km + 0 km) / (6400 km + 600 km))[sup]3[/sup] \* (0.51 gauss) = **0.39 gauss** at 600 km in agreement with your numbers. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 24, 2013, 1:42am UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/8 "2013-07-24T01:42:07Z")
One can also make a magnet where the dipole moment is exactly zero, in which case the long-range field will be due to the quadropole moment, and will die off even faster. Or cancel that out too, leaving the octopole moment, which dies off faster yet, and so on. The simple way to do this is to just place two identical magnets next to each other, in the opposite orientation. To the OP, it really depends on why you’re comparing them. The total energy stored in the field is the most plausible thing you might wonder about which depends both on the strength of the field and its physical extent, but one could construct many other quantities which depend on both of those factors, and come up with situations where one might wonder about them. And just as a side note, because I think it’s a fun fact: The magnetic field of a magnetar can have an energy density tens of millions times greater than that of lead. That’s as in, the _total_ energy density of lead, including the rest energy. ---
**Author:** ![rbroome](https://avatars.discourse-cdn.com/v4/letter/r/838e76/32.png) [@rbroome](https://boards.straightdope.com/u/rbroome)\ **Post date:** [July 24, 2013, 1:48am UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/9 "2013-07-24T01:48:26Z")
> [@Chronos](#): > > And just as a side note, because I think it’s a fun fact: The magnetic field of a magnetar can have an energy density tens of millions times greater than that of lead. That’s as in, the _total_ energy density of lead, including the rest energy. Whew! you mean that the energy density of the magnetic field per unit volume is hugely greater than the energy density per unit volume of lead, as in e=mc\*\*2 energy? Convert the lead to energy and the field has much more energy? that is mind boggling. ---
**Author:** ![j\_sum1](https://avatars.discourse-cdn.com/v4/letter/j/8baadc/32.png) [@j\_sum1](https://boards.straightdope.com/u/j_sum1)\ **Post date:** [July 24, 2013, 2:00am UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/10 "2013-07-24T02:00:46Z")
Ok. I’d like some clarification. I understand that a gravitational field exhibits an inverse square law. This is of course an approximation treating the mass as a point source. Which is ok. I had assumed that a magnetic field would behave similarly – notwithstanding the fact that it is generated by a dipole. Pasta now tells me that it is an inverse cube law. > [@Pasta](#): > > The field of a dipole drops off as the cube of the distance. Colour me surprised but also ok. An explanation would be welcome. **Chronos** chimes in with quadrapole moments and octopole moments and I am well out of my depth. Now my question. I have often heard it said that the weak force and the strong force operate at extremely short distances. Do these follow some different law from the inverse square and inverse cube previously discussed? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 24, 2013, 2:37am UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/11 "2013-07-24T02:37:09Z")
All monopoles, of any sort, have fields that fall off as the inverse square. Gravitational and electric monopoles both behave this way, and if we ever managed to find magnetic monopoles, they’d do the same. Likewise, all dipoles of any sort have fields that fall off as the inverse cube. There’s no such thing as a gravitational dipole (since it would require negative mass to construct one), but electric and magnetic dipoles are both inverse cube (and gravitational ones would be, if they could exist). And so on for quadrupoles (which can exist for gravity, electricity, and magnetism alike) and every other multipole. This is _not_ related to how the nuclear forces fall off very quickly with distance. Those have (at least approximately) exponential falloff, not any polynomial falloff of any degree. This is due, basically, to the fact that the force carriers for those forces are massive, as opposed to the massless photon and graviton. And I won’t even get into the whole bit about the strong force as opposed to the color force, here. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 24, 2013, 2:38am UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/12 "2013-07-24T02:38:10Z")
Oh, and **rbroome** , yes, that’s exactly what I’m saying, and yes, it is indeed mindboggling. ---
**Author:** ![JWT\_Kottekoe](https://avatars.discourse-cdn.com/v4/letter/j/3ec8ea/32.png) [@JWT\_Kottekoe](https://boards.straightdope.com/u/JWT_Kottekoe)\ **Post date:** [July 24, 2013, 4:32am UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/13 "2013-07-24T04:32:05Z")
As a follow up to Chronos’ answer, here is where the inverse cube comes from depending on your knowledge of differential calculus: A dipole is two monopoles of opposite sign placed closed together, for example an electron placed next to a positron, or the north end of a magnet in close proximity to the south end. Let’s call the distance between the monopoles “a”. The monopole field falls off like the square of the distance. The dipole field falls off like the difference between the two inverse square law fields of the monopoles. We are interested in the dependence of the field at distances much larger than the separation “a”. 1. Calculus answer: Compute the difference by taking the derivative of the field of a monopole and multiply by the separation. The derivative of 1/R^2 is -2/R^3. The dipole field will be smaller by a factor of roughly 2a/R (separation divided by distance away). 2. Non calculus answer: take my word for it. If a field goes down like the square of the distance, the difference between two such fields will go down like the cube of the distance. Note that I have left out a discussion of the angular dependence. It depends on your angle away from the line connecting the two monopoles. ---
**Author:** ![j\_sum1](https://avatars.discourse-cdn.com/v4/letter/j/8baadc/32.png) [@j\_sum1](https://boards.straightdope.com/u/j_sum1)\ **Post date:** [July 24, 2013, 5:09am UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/14 "2013-07-24T05:09:59Z")
> [@JWT\_Kottekoe](#): > > 1. Calculus answer: Compute the difference by taking the derivative of the field of a monopole and multiply by the separation. The derivative of 1/R^2 is -2/R^3. The dipole field will be smaller by a factor of roughly 2a/R (separation divided by distance away). > … > Note that I have left out a discussion of the angular dependence. It depends on your angle away from the line connecting the two monopoles. Got that. Both the calculus and the approximation that results from distance\>\>separation. I will admit that I was thinking primarily of a monopole situation and not thinking about the other pole. Am I correct in assuming that inverse square would be a reasonable first approximation for positions very close to the pole? (In asking this I realise that the answer for an electric dipole made from two separated charges may differ from a bar magnet which is composed of a large number of dipoles at the atomic level.) ---
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [July 24, 2013, 7:51pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/15 "2013-07-24T19:51:16Z")
> [@JWT\_Kottekoe](#): > > If a field goes down like the square of the distance, the difference between two such fields will go down like the cube of the distance. The way tidal pull (the difference in felt gravity by the near and far sides of a body) falls off as the cube of the distance? ---
**Author:** ![Pasta](https://avatars.discourse-cdn.com/v4/letter/p/ecccb3/32.png) [@Pasta](https://boards.straightdope.com/u/Pasta)\ **Post date:** [July 24, 2013, 8:09pm UTC](https://boards.straightdope.com/t/q-about-magnetic-fields/664302/16 "2013-07-24T20:09:51Z")
> [@Lumpy](#): > > The way tidal pull (the difference in felt gravity by the near and far sides of a body) falls off as the cube of the distance? Yes. Both of these involve differentiation of the underlying monopole force law, so both end up with a cube.