Question

Shows four identical currents iand five Amperian paths (athrough e) encircling them. Rank the paths according to the value of $\mathbf{\oint }\overrightarrow{\mathbf{B}}\mathbf{.}\mathit{d}\overrightarrow{\mathbf{s}}$taken in the directions shown, most positive first.

Short answer

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

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Step-by-step solution

Given

Figure 29.33

Determining the concept.

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

The formula is as follows:

$\oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0{i}_{enc}$

Where,

$\overrightarrow{B}$= is the magnetic field,$d\overrightarrow{s}$=istheinfinitesimal segment of the integration path,

${\mu }_0$= is the empty's permeability,

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

Determining the path according to the value of ∮B→.ds→.

According to Ampere’s law,

$\oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0{i}_{enc}$

From the given figure, it can interpret that,

For path (a),

role="math" localid="1663002871558" $$\begin{gathered} \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0\left(\text{3}i-i\right) \\ \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_{0}i \end{gathered}$$

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

Determining the path (b).

For path (b),

$$\begin{gathered} \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0\left(\text{2}i-i\right) \\ \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_{0}i \end{gathered}$$

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

Determining the path (c).

For path (c),

$$\begin{gathered} \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0\left(\text{3}i+i\right) \\ \oint \overrightarrow{B}.d\overrightarrow{s}=\text{4}{\mu }_{0}i \end{gathered}$$

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

Determining the path (d),

For path (d),

$$\begin{gathered} \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0\left(\text{3}i+i\right) \\ \oint \overrightarrow{B}.d\overrightarrow{s}=\text{4}{\mu }_{0}i \end{gathered}$$

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

Determining the path (e).

For path (e),

$$\begin{gathered} \oint \overrightarrow{B}.d\overrightarrow{s}={\mu }_0\left(\text{3}i-i\right) \\ \oint \overrightarrow{B}.d\overrightarrow{s}=\text{2}{\mu }_{0}i \end{gathered}$$

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

Hence, the ranking of the paths according to the value of $\oint \overrightarrow{B}.d\overrightarrow{s}$, 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.