A long wire carrying a 2.0-A current is placed along the y-axis.What is the magnitude of the $\vec { B }$ field at a point that is 0.60 m from the origin along the x-axis? (μ₀ = 4π × 10^-7 T ∙ m/A)

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Quizplus · Accepted Answer

The magnitude of the magnetic field ($\vec{B}$) created by a long straight current-carrying wire is given by the formula $B = \frac{\mu_0 I}{2\pi r}$, where $\mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}$, $I = 2.0 \, \text{A}$, and $r = 0.60 \, \text{m}$. Plugging these values into the formula gives $B = \frac{(4\pi \times 10^{-7}) \times 2.0}{2\pi \times 0.60} = \frac{4 \times 10^{-7}}{0.60}$, which calculates to approximately $0.67 \times 10^{-6} \, \text{T}$ or $0.67 \, \mu\text{T}$.

A Long Wire Carrying a 2 B⃗\vec { B }B Field at a Point That Is 0

A Long Wire Carrying a 2 $\vec { B }$ Field at a Point That Is 0