91Ó°ÊÓ

Figure 29.33

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

The formula is as follows:

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0{i}_{enc}$

Where,

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

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

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

According to Ampere’s law,

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0{i}_{enc}$

From the given figure, it can interpret that,

For path (a),

role="math" localid="1663002871558" $$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0\left(\text{3}i-i\right) \\ âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i \end{gathered}$$

Hence, the path of (a) is $âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i$.

For path (b),

$$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0\left(\text{2}i-i\right) \\ âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i \end{gathered}$$

Hence, the path of (b) is $âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i$.

For path (c),

$$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0\left(\text{3}i+i\right) \\ âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}=\text{4}{\mathrm{μ}}_{0}i \end{gathered}$$

Hence, the path of (c) is $âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}=\text{4}{\mathrm{μ}}_{0}i$.

For path (d),

$$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0\left(\text{3}i+i\right) \\ âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}=\text{4}{\mathrm{μ}}_{0}i \end{gathered}$$

Hence, the path of (d) is $âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}=\text{4}{\mathrm{μ}}_{0}i$.

For path (e),

$$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0\left(\text{3}i-i\right) \\ âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}=\text{2}{\mathrm{μ}}_{0}i \end{gathered}$$

Hence, the path of (e) is$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}=\text{2}{\mathrm{μ}}_{0}i$.

Hence, the ranking of the paths according to the value of $âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}$, 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.