Question

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25 cm, what is the maximum current the wire can carry?

Answer 1

We use the Ampère's law:
\[B=μI/2πr\]Therefore, we have that
\[I=B(2πr)/μ\]Which gives us
\[I=\frac{ 0.55∗10^{−4}∗2∗3.14∗0.25 }{ 4∗3.14∗10^{−7} }\]\[I=68.75A\]