Each of two parallel wires separated by 6.0 mm carries a 40-A current. These two currents are in the same direction. Determine the magnitude of the magnetic field at a point that is 5.0 mm from each of the wires.
A) 2.6 mT
B) zero
C) 1.9 mT
D) 1.6 mT
E) 3.2 mT

Question

Quizplus · Accepted Answer

The magnetic field due to a single long straight wire carrying current I at a distance r from the wire is given by the formula $B = \frac{\mu_0 I}{2\pi r}$, where $\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}$ is the permeability of free space. For two wires carrying currents in the same direction, the magnetic fields at a point equidistant from both wires will add up.Given that each wire carries a current $I = 40 \, \text{A}$ and the point of interest is $r = 5.0 \, \text{mm} = 5.0 \times 10^{-3} \, \text{m}$ from each wire, the magnetic field due to one wire at this point is:$$B = \frac{(4\pi \times 10^{-7}) \times 40}{2\pi \times 5.0 \times 10^{-3}} = \frac{4 \times 10^{-7} \times 40}{2 \times 5.0 \times 10^{-3}} = \frac{16 \times 10^{-7}}{10^{-2}} = 1.6 \times 10^{-3} \, \text{T} = 1.6 \, \text{mT}$$Since there are two wires and their magnetic fields add up at the point of interest, the total magnetic field is:$$B_{\text{total}} = 2 \times 1.6 \, \text{mT} = 3.2 \, \text{mT}$$Therefore, the correct answer is E) 3.2 mT.

Each of Two Parallel Wires Separated by 6