As shown in the figure, an insulated wire is bent into a circular loop of radius 6.0 cm and has two long straight sections. The loop is in the xy-plane, with the center at the origin. The straight sections are parallel to the z-axis. The wire carries a current of 8.0 A. What is the magnitude of the magnetic field at the origin? (μ₀ = 4π × 10^-7 T ∙ m/A) A) 75 µT B) 81 µT C) 88 µT D) 110 µT E) 120 µT

Question

Quizplus · Accepted Answer

The magnetic field at a point on the loop due to a current element of length dl is given by:

$\vec{dB}$ $= \frac{\mu_{0}}{4\pi} \frac{I\vec{dl}	imes\vec{r}}{r^2}$

where $\vec{r}$ is the position vector pointing from the current element to the point where we want to find the magnetic field, and $	imes$ denotes the vector cross product.

For the two straight sections of the wire, the magnetic field at the origin is horizontal and has the same magnitude and direction for both sections. Therefore, we only need to consider one section.

Let X be a point on the section that is 6.0 cm away from the origin, and let Y be a point on the section that is 6.0 cm away from X, in the direction of the current. The magnitude of $\vec{dl}$ is the distance between X and Y, which is $dl$ = 2.0π(0.06 m) = 0.377 m. The vector $\vec{dl}$ points in the direction of the current.

The position vector $\vec{r}$ pointing from the current element to the origin has magnitude $r$ = 6.0 cm = 0.06 m, and direction opposite to $\vec{dl}$. Therefore, $\vec{dl}	imes\vec{r}$ points in the vertical direction.

The magnitude of the magnetic field at the origin due to the section is:

$|\vec{dB}|$ $= \frac{\mu_{0}}{4\pi} \frac{I dl}{r^2}$

$|\vec{dB}|$ $= \frac{4\pi	imes10^{-7}	extrm{ T}\cdot	extrm{m}/	extrm{A}}{4\pi} \frac{8.0	extrm{ A} \cdot 0.377	extrm{ m}}{(0.06	extrm{ m})^2}$

$|\vec{dB}|$ = 0.088 T = 88 µT

Since the magnitude of the magnetic field due to each section is the same and both fields point in the same direction, the total magnetic field at the origin is twice the magnitude of the field due to one section, which gives:

$|\vec{B}|$ = 2$|\vec{dB}|$ = 0.176 T = 176 µT

which is closest to choice C, 88 µT.

As Shown in the Figure, an Insulated Wire Is Bent