The magnitude of the magnetic field that a long and extremely thin current-carrying wire produces at a distance of 3.0 µm from the center of the wire is 2.0 × 10^-3 T.How much current is flowing through the wire? (μ₀ = 4π × 10^-7 T ∙ m/A) A)30 mA B)190 mA C)19 mA D)380 mA

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Quizplus · Accepted Answer

The magnetic field of a long, thin wire is given by:$$B = \frac{\mu_0 I}{2\pi r}$$where:* B is the magnetic field* μ0 is the permeability of free space* I is the current* r is the distance from the wireWe can rearrange this equation to solve for the current:$$I = \frac{2\pi rB}{\mu_0}$$Plugging in the given values, we get:$$I = \frac{2\pi (3.0 	imes 10^{-6} 	ext{ m})(2.0 	imes 10^{-3} 	ext{ T})}{4\pi 	imes 10^{-7} 	ext{ T} \cdot 	ext{m/A}} = 30 	ext{ mA}$$Therefore, the correct answer is A.

The Magnitude of the Magnetic Field That a Long and Extremely