Two long, straight wires, each carrying an electric current of 5.0 A are kept parallel to each other at a separation of 2.5 cm. Find the magnitude of the magnetic force experienced by 10 cm of a wire

Solution:

The field at the site of one wire due to the other is   $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi d}}=\frac{\left(2\times {10}^{-7}{\mathrm{TmA}}^{-1}\right)(5.0\mathrm{A})}{2.5\times {10}^{-2}\mathrm{m}}=4.0\times {10}^{-5}\mathrm{T}$
The force experienced by 10 cm of this wire due to the other is
$\begin{array}{l}\mathrm{F}=\mathrm{ilB}\\ =(5.0\mathrm{A})\left(10\times {10}^{-2}\mathrm{m}\right)\left(4.0\times {10}^{-5}\mathrm{T}\right)=2.0\times {10}^{-5}\mathrm{N}.\end{array}$