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Expert-verified Found in: Page 856 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Shows four identical currents i and five Amperian paths (a through e) encircling them. Rank the paths according to the value of ${\mathbf{\oint }}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{.}}{\mathbit{d}}\stackrel{\mathbf{\to }}{\mathbf{s}}$taken in the directions shown, most positive first. The ranking of the paths according to the value of $\oint \stackrel{\to }{B}.d\stackrel{\to }{s}$, the most positive first is $\text{d > a = e > b > c}$.

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See the step by step solution

Figure 29.33

## Step 2: Determining the concept.

Using Ampere’s law, find the values ${\mathbf{\oint }}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{.}}{\mathbit{d}}\stackrel{\mathbf{\to }}{\mathbf{s}}$along each path. Comparing them, rank the paths according to the value ${\mathbf{\oint }}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{.}}{\mathbit{d}}\stackrel{\mathbf{\to }}{\mathbf{s}}$.

The formula is as follows:

$\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}{i}_{enc}$

Where,

$\stackrel{\to }{B}$ = is the magnetic field,$d\stackrel{\to }{s}$= is the infinitesimal segment of the integration path,

${\mu }_{0}$= is the empty's permeability,

${i}_{enc}$ = is the enclosed electric current by the path.

## Step 3: Determining the path according to the value of ∮B→.ds→.

According to Ampere’s law,

$\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}{i}_{enc}$

From the given figure, it can interpret that,

For path (a),

role="math" localid="1663002871558" $\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}\left(\text{3}i-i\right)\phantom{\rule{0ex}{0ex}}\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}i$

Hence, the path of (a) is $\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}i$.

## Step 4: Determining the path (b).

For path (b),

$\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}\left(\text{2}i-i\right)\phantom{\rule{0ex}{0ex}}\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}i$

Hence, the path of (b) is $\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}i$.

## Step 5: Determining the path (c).

For path (c),

$\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}\left(\text{3}i+i\right)\phantom{\rule{0ex}{0ex}}\oint \stackrel{\to }{B}.d\stackrel{\to }{s}=\text{4}{\mu }_{0}i$

Hence, the path of (c) is $\oint \stackrel{\to }{B}.d\stackrel{\to }{s}=\text{4}{\mu }_{0}i$.

## Step 6: Determining the path (d),

For path (d),

$\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}\left(\text{3}i+i\right)\phantom{\rule{0ex}{0ex}}\oint \stackrel{\to }{B}.d\stackrel{\to }{s}=\text{4}{\mu }_{0}i$

Hence, the path of (d) is $\oint \stackrel{\to }{B}.d\stackrel{\to }{s}=\text{4}{\mu }_{0}i$.

## Step 7: Determining the path (e).

For path (e),

$\oint \stackrel{\to }{B}.d\stackrel{\to }{s}={\mu }_{0}\left(\text{3}i-i\right)\phantom{\rule{0ex}{0ex}}\oint \stackrel{\to }{B}.d\stackrel{\to }{s}=\text{2}{\mu }_{0}i$

Hence, the path of (e) is $\oint \stackrel{\to }{B}.d\stackrel{\to }{s}=\text{2}{\mu }_{0}i$.

Hence, the ranking of the paths according to the value of $\oint \stackrel{\to }{B}.d\stackrel{\to }{s}$, the most positive first is $\text{d > a = e > b > c}$.

Ampere’s law gives the relation between magnetic flux and current enclosed by the loop.

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