Magnetic Fields — Invisible Forces That Shape the Universe
Magnetic Fields
Magnetism is one of the fundamental forces of nature. Every magnet — from the ones on your refrigerator to the Earth's core — generates an invisible field that can push, pull, and deflect without physical contact. Magnetic fields are described by the vector quantity B, measured in teslas. Unlike electric charges, magnetic poles always come in pairs: every magnet has a north pole and a south pole, and cutting a magnet in half simply produces two smaller magnets, each with both poles.
The magnetic field of a bar magnet forms characteristic closed loops running from the north pole outside the magnet to the south pole, and then continuing inside the magnet from south back to north. This is fundamentally different from electric field lines, which begin and end on charges. The mathematical statement of this fact is Gauss's law for magnetism: the divergence of B is zero, or equivalently, the total magnetic flux through any closed surface is zero. There are no magnetic monopoles.
Magnetic field strength varies enormously in nature. The Earth's field at the surface is about 25 to 65 microteslas — weak enough that a compass needle barely responds. A refrigerator magnet produces about 5 milliteslas. Medical MRI machines use superconducting magnets generating 1.5 to 7 teslas. Neutron stars called magnetars have surface fields exceeding 10 to the 11th teslas — the strongest magnetic fields in the known universe.
In this lesson you will learn how to quantify magnetic fields, calculate the field produced by simple current configurations, and use the right-hand rule to determine field directions. The mathematics involves vectors, cross products, and inverse-square laws — the same tools used throughout physics and engineering.
The magnetic field $\vec{B}$ is a vector field measured in teslas (T). It describes the force environment created by magnets, currents, and changing electric fields.
$$1 \text{ T} = 1 \frac{\text{kg}}{\text{A·s}^2} = 1 \frac{\text{Wb}}{\text{m}^2}$$
A wire carrying current $I$ produces a circular magnetic field at perpendicular distance $r$:
$$B = \frac{\mu_0 I}{2\pi r}$$
where $\mu_0 = 4\pi \times 10^{-7}$ T·m/A is the permeability of free space.
The net magnetic flux through any closed surface is always zero:
$$\oint \vec{B} \cdot d\vec{A} = 0$$
This means there are no magnetic monopoles — magnetic field lines always form closed loops.
A long straight wire carries a current of $I = 10$ A. Find the magnetic field at a distance of $r = 5$ cm from the wire.
Step 1: Write down the formula for the magnetic field of a long straight wire.
$$B = \frac{\mu_0 I}{2\pi r}$$
Step 2: Convert units: $r = 5$ cm $= 0.05$ m.
Step 3: Substitute values.
$$B = \frac{(4\pi \times 10^{-7})(10)}{2\pi (0.05)}$$
Step 4: Simplify.
$$B = \frac{4\pi \times 10^{-6}}{0.1\pi} = \frac{4 \times 10^{-6}}{0.1} = 4 \times 10^{-5} \text{ T} = 40 \text{ μT}$$
This is comparable to the Earth's magnetic field — everyday currents produce surprisingly weak fields.
Two parallel wires are 20 cm apart. Wire 1 carries 5 A and wire 2 carries 8 A in the same direction. Find the magnetic field at the midpoint between them.
Step 1: At the midpoint, $r = 0.10$ m from each wire.
Step 2: Field from wire 1: $B_1 = \frac{\mu_0 (5)}{2\pi (0.10)} = \frac{4\pi \times 10^{-7} \times 5}{0.2\pi} = 10 \times 10^{-6} = 10$ μT
Step 3: Field from wire 2: $B_2 = \frac{\mu_0 (8)}{2\pi (0.10)} = 16$ μT
Step 4: By the right-hand rule, at the midpoint the fields from parallel currents point in opposite directions.
$$B_{\text{net}} = |B_2 - B_1| = 16 - 10 = 6 \text{ μT}$$
At what distance from a wire carrying 20 A does the field equal 100 μT?
Step 1: Solve for $r$: $B = \frac{\mu_0 I}{2\pi r} \implies r = \frac{\mu_0 I}{2\pi B}$
Step 2: Substitute: $r = \frac{(4\pi \times 10^{-7})(20)}{2\pi (100 \times 10^{-6})}$
Step 3: $r = \frac{8\pi \times 10^{-6}}{200\pi \times 10^{-6}} = \frac{8}{200} = 0.04$ m $= 4$ cm
Practice Problems
Show Answer Key
1. $B = \mu_0(15)/(2\pi \times 0.03) = 100$ μT
2. $r = \mu_0(25)/(2\pi \times 50\times10^{-6}) = 0.10$ m $= 10$ cm
3. $I = 2\pi r B/\mu_0 = 2\pi(0.02)(50\times10^{-6})/(4\pi\times10^{-7}) = 5$ A
4. Both fields point the same way at the midpoint (opposite currents). $B_{\text{net}} = 2 \times \mu_0(6)/(2\pi \times 0.05) = 48$ μT
5. $B = \mu_0(500)/(2\pi \times 15) = 6.67$ μT
6. $B \propto I/r$. Doubling both leaves $B$ unchanged.
7. $r = \mu_0(1)/(2\pi \times 10^{-6}) = 0.20$ m $= 20$ cm
8. $B = (4\pi\times10^{-7})(3)/(2\times0.10) = 18.8$ μT