91Ó°ÊÓ

• Figure showing four circular Amperian loops.
• The current is uniform across the wire’s circular cross-section.

Using Ampere’s law, Rank theloops according to the magnitude offrom the corresponding values of the magnitude of enclosed currents.

Formulae are as follows:

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

Where,

role="math" localid="1663001431872" $d\stackrel{→}{s}=$is the infinitesimal segment of the integration path,

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

$i=$is the enclosed electric current by the path.

According to Ampere’s law,

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

So,

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

From the given figure, it interprets that,

role="math" localid="1663001760102" $$\begin{gathered} \left|i_{enc,a}\right|=4\text{A} \\ \left|i_{enc,b}\right|=5\text{A} \\ \left|i_{enc,c}\right|=0\text{A} \\ \left|i_{enc,d}\right|=3\text{A} \end{gathered}$$

Hence, the ranking of Amperian loops according to the magnitude of $âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}$around each is $b>a>d>c$.

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