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.