# What is the magnetic field due to an electromagnet at a point that is not along its axis?

For e.g. at a point below(or perpendicular) to a cuboidal/cylindrical core electromagnet? I am confused whether the magnetic field for an iron-core electromagnet is limited to the axis passing through it because if that is the case how can it be used for the various applications that requires applying force on other metals? Please specify the method to calculate the field,if any. I am a beginner so kindly pardon any mistakes.

For analyzing the magnetic field from a coil at some distance, the fall back position is the biot savart law. Here's the simple example alluded to in the question: -

And, for the magnitude of the fields off-centre I've found that the only way to practically do this is go back to the basic biot savart law: -

A magnet can be treated as a magnetic dipole and the magnetic field can be calculated in similar form to the calculations used for an electric dipole. By convention, the vector is taken to point away from the magnet at the N pole and toward the magnet at the S pole. If you imagine the magnet in the center of a clock face, with N pointing to 12, the vectors rotate through $360^\circ$ as you move around from 12 to 6 on the clock and diminish to a minimum of strength at 3 and 9 on the clock, where the magnitude will be half what it was at the same distance on the clock face at 12 and 6. The field will vary in strength $\propto \tfrac{1}{r^3}$.

If you want to measure the actual field, you might first figure out the strength of the Earth's magnetic field (tangent to the surface) at your location. In the US, it's generally about $2\times 10^{-5}$ Tesla, though in Maine it's about 25% weaker and in Florida it's about 50% stronger. You can then take a sensitive compass and make some measurements outside, away from any metal structures (or metal table materials.) Set the compass so that it is undisturbed and settles down to "pointing north." Now, approach the compass with your magnet, where you approach along a line that is perpendicular to the N-S line the compass is currently describing, holding your magnet with N (or S) pointing straight at the center of the compass. When the compass needle rotates exactly $70^\circ$ away from its original direction (and towards or away from) your magnet, measure out the distance from the center of the compass to the center of your magnet. At this point:

$$\vec{B}_{magnet} = \vec{B}_{earth}\cdot \textrm{tan}\left(70^\circ\right)$$

If you are in the US where $\vec{B}_{earth}\approx 2\times 10^{-5}\:\textrm{T}$, then you have now determined the distance where $\vec{B}_{magnet}\approx 5.5\times 10^{-5}\:\textrm{T}$. Using this distance, you can now compute the magnetic dipole moment of your magnet as:

$$\mu = \frac{4\: \pi\cdot B_{magnet}\cdot r^3}{2\:\mu_0}$$

If you stay in SI units, the answer results in magnetic dipole moment units of $\textrm{A}\cdot \textrm{m}^2$ (or, equivalently, Joules per Tesla.)

You can do these experiments with various magnets and/or electromagnets, if you prefer. Once you know it's magnetic dipole moment, you can sit down and map out the field on paper, if you like.

If you are in a different place in the world, you should be able to find a reasonable value to use for $\vec{B}_{earth}$ near your location. Just follow the above equations using your own value, instead.

A common bar magnet weighing a couple of ounces might yield $3\:\textrm{A}\cdot \textrm{m}^2$, for example.