A long straight wire (diameter = 2.0 mm) carries a current of 40 A. What is the magnitude of the magnetic field 1.5 mm from the axis of the wire?
A) 3.0 mT
B) 12 mT
C) 5.3 mT
D) 7.4 mT
E) 8.0 mT

Question

Quizplus · Accepted Answer

Using the formula for the magnetic field from a current-carrying wire, we have:
$$B = \frac{\mu_0 I}{2r}$$
where $\mu_0$ is the permeability of free space ($4\pi	imes 10^{-7}$ Tm/A), $I$ is the current, and $r$ is the distance from the axis of the wire. Plugging in the given values, we get:
$$B = \frac{(4\pi	imes 10^{-7}	ext{ Tm/A})(40	ext{ A})}{2(1.5	imes 10^{-3}	ext{ m})} \approx 5.3	ext{ mT}$$
Therefore, the answer is choice C.

A Long Straight Wire (Diameter = 2