Two long straight parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What is the magnitude of the resulting magnetic field at a point that is 15 cm from the wire carrying the 30-A current and 25 cm from the other wire?
A) 51 μT
B) 33 μT
C) 72 μT
D) 64 μT
E) 46 μT

Question

Quizplus · Accepted Answer

To find the magnitude of the resulting magnetic field, we can use the formula:
$B = \dfrac{\mu_0 I_1 I_2}{2\pi r}$
Where $\mu_0$ is the permeability of free space, $I_1$ and $I_2$ are the currents in the wires, and $r$ is the distance from the point to each wire. Plugging in the values, we get:
$B = \dfrac{(4\pi 	imes 10^{-7} T\cdot m/A)(30 A)(40 A)}{2\pi (0.15 m + 0.25 m)}$
Simplifying, we get:
$B \approx 33 \mu T$
So the answer is choice B.

Two Long Straight Parallel Wires Separated by a Distance of 20