Ampere’s law

We know that a straight conductor creates a circular magnetic field. The direction of the magnetic field, $\overrightarrow{B}\text{,}$ at a given point is simply the tangent to that circle. The magnitude is given by Ampere’s law:

$\left|\overrightarrow{B}\right|={\mu }_0\frac{I}{2\pi r}\text{,}$

where $\overrightarrow{B}$ 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

$\left|\overrightarrow{B}\right|={\mu }_0\frac{NI}{L}\text{,}$

where $\overrightarrow{B}\text{,}$ ${\mu }_0\text{,}$ and $I$ are the same as before, and where $N$ is the number of loops (unitless; always an integer; counted number, therefore $\infty$ significant digits; if you are solving for it, round up, never down) and $L$ is the length of the solenoid in metres (m).

Example

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:

$\left|\overrightarrow{B}\right|=(4\pi \times {10}^{-7}\text{T⋅m/A})\frac{0.75\text{A}}{2\pi (0.032\text{m})}\approx 4.6875\times {10}^{-6}\text{T}\text{.}$

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