A very long wire generates a magnetic field of 0.0050 × 10^-4T at a distance of 3.0 mm.What is the magnitude of the current?

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Quizplus · Accepted Answer

The magnetic field of a long wire is given by:$$B = \frac{\mu_0 I}{2\pi r}$$where:* B is the magnetic field in teslas (T)* μ0 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 wire in meters (m)We can rearrange this equation to solve for the current:$$I = \frac{2\pi rB}{\mu_0}$$Plugging in the given values, we get:$$I = \frac{2\pi (3.0 	imes 10^{-3} 	ext{ m})(0.0050 	imes 10^{-4} 	ext{ T})}{4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m}/	ext{A}} = 7.5 	imes 10^{-3} 	ext{ A} = 7.5 	ext{ mA}$$Therefore, the magnitude of the current is 7.5 mA.

A Very Long Wire Generates a Magnetic Field of 0