Question

A capacitor with square plates of edge length L is being discharged by a current of 0.75 A. Figure 32-29 is a head-on view of one of the plates from inside the capacitor. A dashed rectangular path is shown. If L = 12cm, W = 4.0 cm , and H = 2.0 cm , what is the value $\mathbf{\oint }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}\mathbf{\times }\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{s}}$of around the dashed path?

Short answer

The value of the integral for the dashed path $\oint \stackrel{\rightharpoonup }{\mathrm{B}}\times d\stackrel{\rightharpoonup }{\mathrm{s}}$ is$\oint \stackrel{\rightharpoonup }{\mathrm{B}}\times d\stackrel{\rightharpoonup }{\mathrm{s}}=52nT.M.$

Step-by-step solution

Given

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

Determining the concept

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:

$\oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}={\mu }_0{E}_0\frac{dE}{dt}+{\mu }_0{i}_d$

where,

B = magnetic induction,

E = electric field,

A = area,

i= current.

Determining the value of the integral for the dashed path ∮B⇀×ds⇀

The current for the dashed region can be written as,

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

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

role="math" localid="1663131778344" $\oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}={\mu }_0{i}_{d,enc,}$

Usingtheabove-derived equation, it can be written as,

role="math" localid="1663132268088" $$\begin{gathered} \oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}={\mu }_0{i}_d\times \frac{H\times W}{L^2}, \\ \oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}=\left(4\mathrm{\pi }\times {10}^{-7}\right)\times 0.75\times \frac{0.02\times 0.04}{0.{12}^2}, \\ \oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}=\left(4\mathrm{\pi }\times {10}^{-7}\right)\times 0.75\times \frac{0.02\times 0.04}{0.{12}^2}, \\ \oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}=5.23\times {10}^{-8}T.m, \\ \oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}=52nT.m, \end{gathered}$$

Hence, the value of the integral for the dashed path$\oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}$is $\oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}=52nT.m$