We know that a straight conductor creates a circular magnetic field. The direction of the magnetic field, `vec B`, at a given point is simply the tangent to that circle. The *magnitude* is given by Ampere’s law:

`|vec B| = mu_0I/(2pir)`,

where `vec B` is the magnetic field in teslas (T), `mu_0` is the permeability of free space (it has a constant value of 4π × 10^{−7} T·m/A), and `r` is the distance from the conductor in metres (m).

For a coiled conductor (a solenoid), we use the equation

`|vec B| = mu_0(NI)/(L)`,

where `vec B`, `mu_0`, and `I` are the same as before, and where `N` is the number of loops (unitless; always an integer; counted number, therefore `oo` significant digits; if you are solving for it, *round up*, never down) and `L` is the length of the solenoid in metres (m).

Find the magnitude of the magnetic field 3.2 cm from a straight conductor with 0.75 A of current.

All we have to do is plug everything into the equation:

`|vec B| = (4pixx10^-7\ "T·m/A")(0.75\ "A")/(2pi(0.032\ "m")) ~~ 4.6875xx10^-6\ "T"`.

Therefore, the magnitude of the magnetic field is 4.7 × 10^{−6} T.