A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.
A) 1.6 μT
B) 4.9 μT
C) 1.0 μT
D) 9.8 μT
E) 15 μT

Question

Quizplus · Accepted Answer

The magnetic field at the center of the semicircle can be calculated using the formula $B=\frac{\mu_0}{4\pi}\frac{2I}{R}$, where $I$ is the current, $R$ is the radius of the semicircle, and $\mu_0$ is the permeability of free space. The radius of the semicircle can be found using the formula $C=\pi R$, where $C$ is the circumference of the semicircle and is equal to half the length of the wire, i.e. $C=\frac{3.0	ext{ m}}{2}=\frac{3}{2}	ext{ m}$. Solving for $R$ gives $R=\frac{C}{\pi}=\frac{3}{2\pi}	ext{ m}$. Substituting the values in the expression for magnetic field, we get $B=\frac{\mu_0}{4\pi}\frac{2(15	ext{ A})}{\frac{3}{2\pi}	ext{ m}}\approx4.9\mu	ext{T}$. Therefore, the correct answer is B.

A Segment of Wire of Total Length 3