The magnetic field due to the current in a long, straight wire is $8.0 \mu T$ at a distance of $4.0 \mathrm {~cm}$ from the center of the wire. What is the current in the wire? $\left( \mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } \right)$

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The magnetic field $ B $ due to a long, straight wire is given by $ B = \frac{\mu_0 I}{2\pi r} $. Solving for $ I $, we have $ I = \frac{B \cdot 2\pi r}{\mu_0} $. Substituting $ B = 8.0 \times 10^{-6} \, \text{T} $, $ r = 0.04 \, \text{m} $, and $ \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} $, we find $ I = 1.6 \, \text{A} $.

The Magnetic Field Due to the Current in a Long 8.0μT8.0 \mu T8.0μT

The Magnetic Field Due to the Current in a Long $8.0 \mu T$