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) = (\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= (2 \pi r B)/(\mu _0) = (2 \pi (0.12 m)(0.5 \cdot 10^(-4)T))/( 4 \pi \cdot 10^(-7) NA^(-2))= 30 A`