Question

If an electric wire is allowed to produce a magnetic field no larger than that of the earth (0.50×10−4t) at a distance of 12 cm from the wire, what is the maximum current the wire can carry

Answer 1

The magnetic field generated by a wire carrying a current I is:
$B(r) = \frac{\mu_0 I}{2 \pi r}$
where r is the distance at which the magnetic field is measured, and $\mu_0 = 4 \pi \cdot 10^{-7} NA^{-2}$ is the magnetic permeability in vacuum.

The problem says that the magnetic field at a distance r=12 cm=0.12 m from the wire must be no larger than $B=0.5 \cdot 10^{-4}T$. Substituting these values, we can find the maximum value of the current I that the wire can carry:
$I= \frac{2 \pi r B}{\mu _0} = \frac{2 \pi (0.12 m)(0.5 \cdot 10^{-4}T)}{ 4 \pi \cdot 10^{-7} NA^{-2}}= 30 A$