# 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:

$\left|\overrightarrow{B}\right|=(4\pi \times {10}^{-7}\text{}\text{T\u22c5m/A})\frac{0.75\text{}\text{A}}{2\pi (0.032\text{}\text{m})}\approx 4.6875\times {10}^{-6}\text{}\text{T}\text{.}$

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