The magnitude of a magnetic field a distance 3.0 µm from a wire is 20.0 × 10^-4 T.How much current is flowing through the wire? Assume the wire is the only contributor to the magnetic field. A)30 mA B)188 mA C)19 mA D)377 mA

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The magnetic field around a long, straight wire is given by:$$B = \frac{\mu_0 I}{2\pi r}$$where:* B is the magnetic field strength in teslas (T)* μ0 is the permeability of free space (4π × 10^-7 T·m/A)* I is the current in the wire in amperes (A)* r is the distance from the wire in meters (m)We are given that:* B = 20.0 × 10^-4 T* r = 3.0 µm = 3.0 × 10^-6 mSubstituting these values into the equation above, we get:$$20.0 × 10^-4 T = \frac{(4π × 10^-7 T·m/A) I}{2π (3.0 × 10^-6 m)}$$Solving for I, we get:$$I = 30 mA$$Therefore, the correct answer is A.

The Magnitude of a Magnetic Field a Distance 3