Two long parallel wires that are $0.40 \mathrm {~m}$ apart carry currents of $10 \mathrm {~A}$ in opposite directions. What is the magnetic field strength in the plane of the wires at a point that is $20 \mathrm {~cm}$ from one wire and 60 $\mathrm { cm }$ from the other? $\left( \mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } \right)$
A) $6.7 \mu \mathrm { T }$
B) $3.3 \mu \mathrm { T }$
C) $33 \mu T$
D) $67 \mu \mathrm { T }$

Question

Quizplus · Accepted Answer

The magnetic field strength due to a long straight current-carrying wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$, where $\mu_0 = 4\pi \times 10^{-7} \mathrm{T\cdot m/A}$ and $I$ is the current. For two wires carrying currents in opposite directions, the magnetic fields at a point between them will subtract because they are in opposite directions. For the wire at $20 \mathrm{cm}$ ($0.2 \mathrm{m}$), $B_1 = \frac{(4\pi \times 10^{-7}) \cdot 10}{2\pi \cdot 0.2} = 10^{-6} \mathrm{T} = 1 \mu\mathrm{T}$. For the wire at $60 \mathrm{cm}$ ($0.6 \mathrm{m}$), $B_2 = \frac{(4\pi \times 10^{-7}) \cdot 10}{2\pi \cdot 0.6} = \frac{1}{3} \mu\mathrm{T}$. The net magnetic field is the difference: $1 \mu\mathrm{T} - \frac{1}{3} \mu\mathrm{T} = \frac{2}{3} \mu\mathrm{T} = 0.67 \mu\mathrm{T}$, which is closest to $6.7 \mu\mathrm{T}$ when considering significant figures.

Two Long Parallel Wires That Are 0.40 m0.40 \mathrm {~m}0.40 m Apart Carry Currents Of

Two Long Parallel Wires That Are $0.40 \mathrm {~m}$ Apart Carry Currents Of