The drawing shows two long, thin wires that carry currents in the positive z direction. Both wires are parallel to the z axis. The 50-A wire is in the x-z plane and is 5 m from the z axis. The 40-A wire is in the y-z plane and is 4 m from the z axis. What is the magnitude of the magnetic field at the origin?

Question

Quizplus · Accepted Answer

The magnetic field at the origin due to a current-carrying wire is given by:

$B=\frac{\mu_0 I}{4\pi r}$,

where $I$ is the current, $r$ is the distance from the wire, and $\mu_0$ is the permeability of free space.

For the 50-A wire, $I=50$ A and $r=5$ m, so the magnetic field at the origin due to this wire is:

$B_1=\frac{\mu_0 (50	ext{ A})}{4\pi (5	ext{ m})}=1.2566	imes10^{-5}	ext{ T}$

For the 40-A wire, $I=40$ A and $r=4$ m, so the magnetic field at the origin due to this wire is:

$B_2=\frac{\mu_0 (40	ext{ A})}{4\pi (4	ext{ m})}=1.2566	imes10^{-5}	ext{ T}$

Since the two wires are parallel to each other, the magnetic fields they produce will add up. Thus, the total magnetic field at the origin is:

$B_{	ext{total}}=B_1+B_2=2.513	imes10^{-5}	ext{ T}$

Rounding to one significant figure, the answer is choice C: 3 × 10^(-5) T.

The Drawing Shows Two Long, Thin Wires That Carry Currents