Question

A long wire carries a current of 4.00 A. The energy stored in the magnetic field inside the volume of \(1\,\text{mm}^{3}\) at a distance of 10 cm from the wire is given by:

• A. \(2.55\times10^{-14}\,\text{J}\)
• B. \(5.10\times10^{-14}\,\text{J}\)
• C. \(7.65\times10^{-14}\,\text{J}\)
• D. Zero J

Hint:

Get B with the long-wire formula, then energy is (B squared over 2 mu-zero) times the volume.

Answer 1

The Correct Option is A

Step 1: Find the magnetic field at 10 cm from a long straight wire:

\[B=\frac{\mu_0 I}{2\pi r}=\frac{(4\pi\times10^{-7})(4.00)}{2\pi(0.10)}=8.0\times10^{-6}\,\text{T}.\]

Step 2: Compute the magnetic energy density:

\[u=\frac{B^{2}}{2\mu_0}=\frac{(8.0\times10^{-6})^{2}}{2(4\pi\times10^{-7})}=\frac{6.4\times10^{-11}}{2.513\times10^{-6}}=2.55\times10^{-5}\,\text{J/m}^{3}.\]

Step 3: Multiply by the volume \(V=1\,\text{mm}^{3}=1\times10^{-9}\,\text{m}^{3}\):

\[U=uV=(2.55\times10^{-5})(1\times10^{-9})=2.55\times10^{-14}\,\text{J}.\]

Step 4: This matches option (A).

\[\boxed{U\approx2.55\times10^{-14}\,\text{J}}\]