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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. 29-44 point ${{\mathbit{P}}}_{{\mathbf{1}}}$ is at distance ${\mathbit{R}}{\mathbf{=}}{\mathbf{}}{\mathbf{13}}{\mathbf{.}}{\mathbf{1}}{\mathbf{}}{\mathbf{\text{cm}}}{\mathbf{}}$ on the perpendicular bisector of a straight wire of length ${\mathbit{L}}{\mathbf{=}}{\mathbf{}}{\mathbf{18}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{cm}}}$. carrying current. (Note that the wire is not long.) What is the magnitude of the magnetic field at ${{\mathbit{P}}}_{{\mathbf{1}}}$ due to i? The magnitude of the magnetic field at ${P}_{1}$ due to i is $5.03×{10}^{-8}\text{T}$

See the step by step solution

## Step 1: Given

1. Current $i=58.2\text{mA}$
2. Distance of point P on the perpendicular bisector of wire $R=13.1\text{cm}=0.131\text{m}$
3. Length of wire $L=18.0\text{cm}=0.18\text{m}$

## Step 2: Determine the formulas:

Use the formula of Biot-Savarts law to find the magnitude of the field at ${{\mathbit{P}}}_{{\mathbf{1}}}$ due to i

Formula:

${\mathbit{d}}{\mathbit{B}}{\mathbf{=}}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}\mathbf{}\mathbf{d}\mathbf{l}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{{\mathbf{r}}^{\mathbf{2}}}$

## Step 3: Calculate the magnitude of the magnetic field at P1 due to i

According to Bio-Savarts law

$dB=\frac{{\mu }_{0}idl}{4\pi }\frac{\mathrm{sin}\theta }{{r}^{2}}$

If r makes an angle $\theta$ with l then

$r=\sqrt{{l}^{2}+{R}^{2}}$

And

$\mathrm{sin}\theta =\frac{R}{R}=\frac{R}{\sqrt{{l}^{2}+{R}^{2}}}\phantom{\rule{0ex}{0ex}}l=\frac{L}{2}\phantom{\rule{0ex}{0ex}}=\frac{0.18\text{m}}{2}\phantom{\rule{0ex}{0ex}}=0.09\text{m}$

Integrating an equation of Bio-Savarts law

$B={\int }_{0}^{0.09}dB\phantom{\rule{0ex}{0ex}}={\int }_{0}^{0.09}\frac{{\mu }_{0}idl}{4\pi }\frac{\mathrm{sin}\theta }{{r}^{2}}\phantom{\rule{0ex}{0ex}}B=\frac{{\mu }_{0}i}{4\pi }{\int }_{0}^{0.09}\frac{\mathrm{sin}\theta dl}{{r}^{2}}\phantom{\rule{0ex}{0ex}}B=\frac{{\mu }_{0}i}{4\pi }{\int }_{0}^{0.09}\frac{R}{\sqrt{{l}^{2}+{R}^{2}}}\frac{dl}{{\left(\sqrt{{l}^{2}+{R}^{2}}\right)}^{2}}\phantom{\rule{0ex}{0ex}}B=\frac{{\mu }_{0}iR}{4\pi }{\int }_{0}^{0.09}\frac{dl}{{\left({l}^{2}+{R}^{2}\right)}^{3/2}}\phantom{\rule{0ex}{0ex}}$

Solve further as:

$B=\frac{{\mu }_{0}iR}{4\pi }\frac{1}{{R}^{2}}{\left|\frac{l}{{\left({l}^{2}+{R}^{2}\right)}^{1/2}}\right|}_{0}^{0.09}\phantom{\rule{0ex}{0ex}}B=\frac{{\mu }_{0}i}{2\pi R}\left[\frac{l}{{\left({l}^{2}+{R}^{2}\right)}^{1/2}}\right]\phantom{\rule{0ex}{0ex}}B=\frac{\left(4\pi ×{10}^{-7}\frac{\text{Tm}}{\text{A}}\right)\left(0.0582\text{A}\right)}{2\pi \left(0.131\text{m}\right)}\left[\frac{0.09\text{m}}{{\left[{\left(0.09\text{m}\right)}^{2}+{\left(0.131\text{m}\right)}^{2}\right]}^{1/2}}\right]\phantom{\rule{0ex}{0ex}}B=5.03×{10}^{-8}\text{T}$

Therefore, the magnitude of the magnetic field at ${P}_{1}$ due to i is $5.03×{10}^{-8}\text{T}$

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