91Ó°ÊÓ

$$\begin{gathered} L=12cm=0.12m, \\ W=4.0cm=0.04m, \\ H=2.0cm=0.02m, \\ {i}_d=0.75A. \end{gathered}$$

Using the Ampere-Maxwell law, the integral for the given dashed path can be written. Using the relationship between the displacement current and displacement current that is encircled by the integration loop, the required integral can be found.

The formula is as follows:

$âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}={\mathrm{μ}}_0{E}_0\frac{dE}{dt}+{\mathrm{μ}}_0{i}_d$

where,

B = magnetic induction,

E = electric field,

A = area,

i= current.

The current for the dashed region can be written as,

$$\begin{gathered} i_{d,enc}=i_d×\frac{areaofdashedloop}{areaoftotalplate}, \\ {i}_{d,enc}=i_d×\frac{H×W}{L^2}, \end{gathered}$$

From Ampere–Maxwell law, the integral can be written as,

role="math" localid="1663131778344" $âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}={\mathrm{μ}}_0{i}_{d,enc,}$

Usingtheabove-derived equation, it can be written as,

role="math" localid="1663132268088" $$\begin{gathered} âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}={\mathrm{μ}}_0{i}_d×\frac{H×W}{L^2}, \\ âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}=\left(4\mathrm{Ï€}×{10}^{-7}\right)×0.75×\frac{0.02×0.04}{0.{12}^2}, \\ âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}=\left(4\mathrm{Ï€}×{10}^{-7}\right)×0.75×\frac{0.02×0.04}{0.{12}^2}, \\ âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}=5.23×{10}^{-8}T.m, \\ âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}=52nT.m, \end{gathered}$$

Hence, the value of the integral for the dashed path$âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}$is $âˆ\circledR \stackrel{â‡\P Ä}{B}×d\stackrel{â‡\P Ä}{s}=52nT.m$