The magnitude of the $\vec { B }$ 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.What is the current through the wire? (μ₀ = 4π × 10^-7 T ∙ m/A)

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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 strength* μ0 is the permeability of free space (4π × 10^-7 T m/A)* I is the current through the wire* r is the distance from the wireWe are given that B = 2.0 × 10^-3 T and r = 3.0 µm = 3.0 × 10^-6 m. Substituting these values into the equation above, we get:$$2.0 	imes 10^{-3} T = \frac{4\pi 	imes 10^{-7} T m/A 	imes I}{2\pi 	imes 3.0 	imes 10^{-6} m}$$Solving for I, we get:$$I = \frac{2.0 	imes 10^{-3} T 	imes 3.0 	imes 10^{-6} m}{4\pi 	imes 10^{-7} T m/A} = 30 mA$$Therefore, the current through the wire is 30 mA.

The Magnitude of The B⃗\vec { B }B Field That a Long and Extremely Thin Current-Carrying Wire Produces

The Magnitude of The $\vec { B }$ Field That a Long and Extremely Thin Current-Carrying Wire Produces