Consider a long, straight, thick wire of radius 1 cm that carries a steady 0.9-A current. The value of the magnetic field 0.5 cm from the center of the wire is

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The magnetic field inside a long, straight wire is given by:$$B = \frac{\mu_0 I}{2\pi r}$$where:* B is the magnetic field in teslas (T)* \mu_0 is the permeability of free space (4π × 10^-7 T·m/A)* I is the current in amperes (A)* r is the distance from the center of the wire in meters (m)In this problem, we have:* I = 0.9 A* r = 0.5 cm = 0.005 mPlugging these values into the equation, we get:$$B = \frac{(4\pi 	imes 10^{-7} 	ext{ T$\cdot$m/A})(0.9 	ext{ A})}{2\pi (0.005 	ext{ m})}$$$$B = 9.0 	imes 10^{-6} 	ext{ T}$$Therefore, the correct answer is A.

Consider a Long, Straight, Thick Wire of Radius 1 Cm 9.0×10−69.0 \times 10 ^ { - 6 }9.0×10−6

Consider a Long, Straight, Thick Wire of Radius 1 Cm $9.0 \times 10 ^ { - 6 }$