Field of a household cable and a compass

Problem

A household electrical cable carries a current $i = 10$ A. Estimate the magnitude of the magnetic field it produces at a distance of $15$ cm and compare it with Earth’s magnetic field ($B_T \approx 5\cdot 10^{-5}$ T). Could it disturb a compass placed next to it?

Solution

Treating the cable as an infinite straight wire: $B = \frac{\mu_0\,i}{2\pi r} = \frac{2\cdot 10^{-7}\cdot 10}{0{,}15}$ $\ev{B \approx 1{,}3\cdot 10^{-5}\;\text{T}}$ This value is about a quarter of Earth’s magnetic field: the disturbance to a compass is small but not entirely negligible at close range, while at larger distances (the field goes as $1/r$) it quickly becomes irrelevant. It should also be noted that in a real cable the two conductors carry opposite currents very close together, whose fields almost completely cancel, further reducing the effect.

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Topics: Magnetism Concepts: Magnetic field Skills: Fermi estimation