Q6P - Fundamentals Of Physics

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?

Step-by-Step Solution

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 1: Given

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

Step 2: 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.

Step 3: Determining the value of the integral for the dashed path ∮ B ⇀ × d s ⇀

The current for the dashed region can be written as,

$$\begin{gathered} i_{d, enc}=i_d\times \frac{area of dashed loop}{area of total plate}, \\ {i}_{d, enc}=i_d\times \frac{H\times W}{L^2}, \end{gathered}$$

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

$\oint \stackrel{\rightharpoonup }{B}\times d\stackrel{\rightharpoonup }{s}={\mu }_0{i}_{d, enc ,}$

Using the above-derived equation, it can be written as,

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

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