A very long thin wire produces a magnetic field of 0.0050 × 10^-4 T at a distance of 3.0 mm. from the central axis of the wire. What is the magnitude of the current in the wire? (μ₀ = 4π × 10^-7 T ∙ m/A) A) 7.5 mA B) 1.7 mA C) 3300 mA D) 24,000 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 (4π × 10^-7 T ∙ m/A)* I is the current in the wire* r is the distance from the central axis of 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:$$I = \frac{2\pi (3.0 	imes 10^{-3} 	ext{ m})(0.0050 	imes 10^{-4} 	ext{ T})}{4\pi 	imes 10^{-7} 	ext{ T} \cdot 	ext{m/A}} = 7.5 	imes 10^{-3} 	ext{ A} = 7.5 	ext{ mA}$$Therefore, the magnitude of the current in the wire is 7.5 mA.

A Very Long Thin Wire Produces a Magnetic Field of 0.0050