# Inverse square law and magnetic fields

1. Oct 6, 2007

### kamikaze762

Ok this may be a silly question, but I have trouble grasping the function of the inverse square law and how it relates to the intensity of a magnetic field.

It seems to me that if we take the same calculations of I2 = (I1 * D1^2) / D2^2 we come up with different real world answers based on which distance measurement we use.

Is the only unit we can use here the meter?

Example:

D1 = 1m^2 = 1m

D1 = 1000mm^2 = 1,000,000 or 1000m

They are the same distance, yet they calculate differently with the square product.

A re-wording I've seen of this function is that if you double the distance, you cut the intensity by 1/4. Must we know the strength of the source or can the source intensity be calculated by knowing a set intensity at a set distance?

The problem occurs when I reach 1m and try to go below it. If we set D2 at 0.00000001m the square product shows the field essentially approaching infinity, which is certainly not what is hapenning.

It then follows that below 1m, the field becomes infinitly stronger, which is wrong.

Any clarity would be much appreciated.

Last edited: Oct 6, 2007
2. Oct 6, 2007

### rcgldr

I2 = I1 * (D1 / D2)^2
It doesn't matter what the units are, it's the square of the ratio of the distances.

3. Oct 6, 2007

### Zorodius

they're the same, you just need to keep track of the units. The m or mm squares also, not just the number that goes along with it. (1,000 mm)² is 1,000,000 square-millimeters. To convert this into square meters, multiply by (1 meter / 1,000 millimeter) (=1) twice. The result is one square meter, the same as (1m)^2.

The magnitude of an inverse-square law does tend towards infinity as r tends towards zero. The repulsive force between two electrons, say, becomes enormous as they are brought very close to one another. You might wonder why you aren't crushed to the surface of the earth with enormous force. The answer is that you aren't zero distance away from the center of the earth, but rather, you are separated from it by a distance equal to the radius of the earth, some 6,400 kilometers. (And if you went underground, only the parts of the earth at a lesser radius would attract you. So the relevant mass of the earth would be less as your R got smaller.)

4. Oct 6, 2007

### kamikaze762

That was quite an oversight I made about the square and ratio. Thanks for explaining.

If this applies to magnets, does that mean that the theoretical or mathematical source is actually below the surface?

Basically I'm trying to figure this out:

A magnet weighing 5g levitates above the surface of another magnet at 50mm.

The polar region of the repeling magnet is a 100mm perfect cube, placing the mathematical source magnetism 50mm below its surface.

Assuming perfect conditions and assuming no interference from the other pole (monopole), could we get a good approximation of the surface force simply by checking weight and dimensions?

for example: 5g * 9.8m/s^2 = 49 mN

F1 = 49mN

We know that gravitational and magnetic forces are equal at levitation.

So levitation occurs: 49mN @ 100mm (above true source?)

F2 = surface force

F2 = F1 *(D1/D2)^2

F2 = 49mN * (100mm/50mm)^2

F2 = 196mN

If this can be correctly related to "force", is this a reasonable appoximation without knowing other variables?

Last edited: Oct 6, 2007
5. Oct 7, 2007

### rbj

the inverse-square law speaks to ratios of intensity of field or radiation or whatever given ratios of distance. when you compare a ratio of like-dimensioned physical quantity, it only makes sense to compare such ratio of that quantity expressed in the same units. we say that "the distance from New York to San Francisco is 31 times farther than the distance from New York to Philadelphia" because 2906 miles is 31 x 93 3/4 miles. we don't say that "the distance from New York to San Francisco is 49890 times farther than the distance from New York to Philadelphia" because 2906 miles is 4676754 meters and that is 49890 x 93 3/4 miles. be careful to compare apples to apples.

the inverse-square relationship says that when you increase the distance between two interacting objects by a factor of r, the intensity of interaction is affected inversely (meaning "reciprocal") by the square of that factor or bya factor of 1/r2. here is what Wikipedia sez about it (and why inverse-square makes fundamental sense in a 3-dimensional universe):

Let the total power radiated from a point source, e.g., an omnidirectional isotropic antenna, be P. At large distances from the source (compared to the size of the source), this power is distributed over larger and larger spherical surfaces as the distance from the source increases. Since the surface area of a sphere of radius r is $A = 4 \pi r^2 \$, then intensity I of radiation at distance r is

$$I = \frac{P}{A} = \frac{P}{4 \pi r^2}. \,$$

$$I \propto \frac{1}{r^2} \,$$

$$\frac{I_1} {I_2 } = \frac{{r_2}^2}{{r_1}^2} \,$$

$$I_1 = I_{2} \cdot {r_{2}^2} \cdot \frac{1}{{r_1}^2} \,$$

The energy or intensity decreases by a factor of 1/4 as the distance r is doubled, or measured in dB it would decrease by 6.02 dB. This is the fundamental reason why intensity of radiation, whether it is electromagnetic or acoustic radiation, follows the inverse-square behavior, at least in the ideal 3 dimensional context (propagation in 2 dimensions would follow a just an inverse-proportional distance behaviour and propagation in one dimension, the plane wave, remains constant in amplitude even as distance from the source changes).

Last edited: Oct 7, 2007