Answer to Question #220236 in Electricity and Magnetism for Vlad

Question

At what distance from the central axis of a long straight thin wire carrying a current of 5.0 A is the magnitude of the magnetic field due to the wire equal to the strength of the Earth's magnetic field of about 5.0 × 10^-5 T? (μ0 = 4π × 10^-7 T ∙ m/A)

Answer 1

$B =\frac{ μ₀I}{2πr}$

$B = 5.0 × 10^{-5} T$

$I = 5.0A$

$μ₀ = 4π×10^{-7}T.m/A$

$5.0 × 10^{-5}= \frac{4π × 10^{-7} T ∙ m/A × 5.0}{2πr}$

$5.0 × 10^{-5} × 2πr = {4π×10 ^ {−7} T∙m/A×5.0}$

$r = \frac{4π × 10^{-7}m}{2π×10^{-5}}$

$r = 2cm$/ $0.02m$