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Expert-verified Found in: Page 857 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In Fig. 29-4, a wire forms a semicircle of radius ${\mathbit{R}}{\mathbf{=}}{\mathbf{9}}{\mathbf{.}}{\mathbf{26}}{\mathbf{}}{\mathbf{\text{cm}}}$ and two (radial) straight segments each of length ${\mathbit{L}}{\mathbf{=}}{\mathbf{13}}{\mathbf{.}}{\mathbf{1}}{\mathbf{\text{cm}}}$. The wire carries current ${\mathbf{}}{\mathbit{i}}{\mathbf{=}}{\mathbf{34}}{\mathbf{.}}{\mathbf{8}}{\mathbf{}}{\mathbf{\text{mA}}}$. What are the(a) magnitude and(b) direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C? 1. The magnitude of the magnetic field is $B=1.18×{10}^{-7}\text{T}$
2. The direction of the magnetic field is into the page.
See the step by step solution

## Step 1: Given

1. Radius of circle is $R=9.26\text{cm}=9.26×{10}^{-2}\text{m}$
2. Length of each straight segment is role="math" localid="1663098315135" $L=13.1\text{cm}=13.1×{10}^{-2}\text{m}$
3. Current is $i=34.8\text{mA}=34.8×{10}^{-3}\text{A}$

## Step 2: Determine the formulas

The magnitude of the magnetic field B at centre of circular arc, of radius R, and central angle ${\mathbit{\varphi }}$, carrying a current I is given as follows:

${\mathbit{B}}{\mathbf{=}}\frac{\mathbit{\mu }\mathbit{I}\varphi }{\mathbf{4}\mathbit{\pi }\mathbit{R}}$

For a straight conductor consider the formulas:

${\mathbit{d}}\stackrel{\mathbf{\to }}{\mathbit{B}}{\mathbf{=}}\frac{{\mathbit{\mu }}_{\mathbf{0}}\mathbit{I}\mathbit{d}\stackrel{\mathbf{\to }}{\mathbit{s}}\mathbf{×}\stackrel{\mathbf{^}}{\mathbit{r}}}{\mathbf{4}\mathbit{\pi }{\mathbit{r}}^{\mathbf{2}}}$

Here, ${\mathbit{d}}\overline{\mathbit{B}}$ is the magnetic field through the current carrying conductor, ${\mathbit{d}}\overline{\mathbf{s}}$ length of the element and $\stackrel{\mathbf{^}}{\mathbf{r}}$ is unit vector that is a point from the element to the given point.

Formulae:

${\mathbit{B}}{\mathbf{=}}\frac{\mathbf{\mu I\varphi }}{\mathbf{4}\mathbf{\pi R}}\phantom{\rule{0ex}{0ex}}{\mathbit{B}}{\mathbf{=}}\frac{\mathbf{\mu Ids}\mathbf{}\mathbf{sin\theta }\mathbf{}}{\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}}$

## Step 3: (a) Calculate the magnitude of the net magnetic field at the semicircle’s center of curvature C

To calculate the magnetic field ${B}_{1}$ due to current for left straight segment as follows:

${B}_{1}=\frac{\mu Idssin\theta }{4\pi {r}^{2}}$

$\theta =0°$

So,

${B}_{1}=0$

Calculate the magnetic field ${B}_{2}$ due to current for a right straight segment as follows:

${B}_{2}=\frac{\mu Idssin\theta }{4\pi {r}^{2}}$

Here $\theta =0°$

So

${B}_{2}=0$

Calculate the magnetic field ${B}_{3}$ due to current from circular section as follows

${B}_{3}=\frac{\mu I\varphi }{4\pi R}\phantom{\rule{0ex}{0ex}}{B}_{3}=\frac{\left(4\pi ×{10}^{-7}{\text{N/A}}^{\text{2}}\right)×34.8×{10}^{-3}\text{A}×\pi }{4\pi ×9.26×{10}^{-2}\text{m}}\phantom{\rule{0ex}{0ex}}{B}_{3}=1.18×{10}^{-7}\text{T}$

So, the total magnetic field at the center of semicircle arc is

$B={B}_{1}+{B}_{2}+{B}_{3}\phantom{\rule{0ex}{0ex}}=1.18×{10}^{-7}\text{T}\phantom{\rule{0ex}{0ex}}$

Hence the magnetic field is, $1.18×{10}^{-7}\text{T}$.

## Step 4: (b) Calculate the direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C

Using the right hand rule, direction of magnetic field is into the page ### Want to see more solutions like these? 