A very long thin wire produces a $\vec { B }$ field of $0.0020 \times 10 ^ { - 4 } \mathrm {~T}$ at a distance of $1.0 \mathrm {~mm}$ from the wire.What is the magnitude of the current? (μ₀ = 4π × 10^-7 T ∙ m/A)

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Quizplus · Accepted Answer

The magnetic field of a long, thin wire is given by:$$B = \frac{\mu_0 I}{2\pi d}$$where:* B is the magnetic field strength* μ0 is the permeability of free space (4π × 10^-7 T ∙ m/A)* I is the current flowing through the wire* d is the distance from the wireWe can rearrange this equation to solve for the current:$$I = \frac{2\pi dB}{\mu_0}$$Plugging in the given values, we get:$$I = \frac{2\pi (1.0 	imes 10^{-3} \mathrm{m})(0.0020 	imes 10^{-4} \mathrm{T})}{4\pi 	imes 10^{-7} \mathrm{T \cdot m/A}} = 1.0 \mathrm{mA}$$Therefore, the magnitude of the current is 1.0 mA.

A Very Long Thin Wire Produces A B⃗\vec { B }B

A Very Long Thin Wire Produces A $\vec { B }$