What is the energy density in the magnetic field 25 cm from a long straight wire carrying a current of 12 A? (μ₀ = 4π × 10^-7 T • m/A) A) 7.3 × 10^-5 J/m³ B) 3.7 × 10^-5 J/m³ C) 3.6 × 10^-4 J/m³ D) 1.2 × 10^-4 J/m³ E) The density cannot be determined without knowing the volume.

Question

Quizplus · Accepted Answer

The energy density $u$ in a magnetic field is given by the formula $u = \frac{B^2}{2\mu_0}$, where $B$ is the magnetic field strength and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}$). The magnetic field $B$ around a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$, where $I$ is the current and $r$ is the distance from the wire. Substituting $I = 12 \, \text{A}$ and $r = 0.25 \, \text{m}$, and then calculating $u$, results in the energy density value corresponding to choice B.

What Is the Energy Density in the Magnetic Field 25