A very long thin wire produces a magnetic 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)

Question

A very long thin wire produces a magnetic 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)

Quizplus · Accepted Answer

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 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 are given that:* B = 0.0020 × 10^-4 T* r = 1.0 mm = 1.0 × 10^-3 mSubstituting these values into the equation, we get:$$0.0020 	imes 10^{-4} \mathrm{T} = \frac{(4\pi 	imes 10^{-7} \mathrm{T} \cdot \mathrm{m}/\mathrm{A})I}{2\pi (1.0 	imes 10^{-3} \mathrm{m})}$$Solving for I, we get:$$I = \frac{(0.0020 	imes 10^{-4} \mathrm{T})(2\pi (1.0 	imes 10^{-3} \mathrm{m}))}{4\pi 	imes 10^{-7} \mathrm{T} \cdot \mathrm{m}/\mathrm{A}} = 1.0 \mathrm{mA}$$Therefore, the magnitude of the current is 1.0 mA.

A Very Long Thin Wire Produces a Magnetic Field Of 0.0020