Conduction Current and Current Density MCQ Quiz in मल्याळम - Objective Question with Answer for Conduction Current and Current Density - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Current density can be written as:

$\begin{array}{l}\overrightarrow{J}\left(r\right)=\rho \left(r\right)\overrightarrow{V}\left(r\right)\\ =\rho \left(r\right)\left(\overrightarrow{\omega }\times \overrightarrow{r}\right)\end{array}$

$\begin{array}{l}=\rho \left(r\right)\left[\omega {\hat{a}}_z\times r{\hat{a}}_r\right]\\ =\rho \left(r\right)\omega r\hat{a}\varphi \end{array}$

Current is given by

$\begin{array}{l}I\left(r\right)=\underset{r=0}{\overset{r}{\int }}\underset{z=0}{\overset{L}{\int }}\rho \left(r\right)\omega r\hat{a}\varphi .drdz\hat{a}\varphi \\ =\underset{r=0}{\overset{r}{\int }}\underset{z=0}{\overset{L}{\int }}\rho \left(r\right)\omega rdrdz\end{array}$

$\begin{array}{l}=\omega L\underset{r=0}{\overset{r}{\int }}\rho \left(r\right)rdr\\ =\omega L\underset{r=0}{\overset{r}{\int }}{10}^3{r}^{3}dr\end{array}$

$I\left(r\right)=\omega L\times {10}^{-3}\left(\frac{r^4}{4}\right)$

At $r=\frac{R}{2}$

At $I\left(r=\frac{R}{2}\right)=\omega L\times {10}^{-3}\times \frac{R^4}{4\times 16}$ 

= 7.8125 μA

Given,

σ = 7 × 10⁷ s/m; E = 10^-2 v/m

J = σE

= 7 × 10⁷ × 10^-2

= 7 × 10⁵ A/m²

Also, J = ρ_vV_d ; ρ_v = ne

⇒ 7 × 10⁵ = 10²⁹ × 1.6 × 10^-19 V_d

Given, $2\mathrm{a}=0.2\mathrm{m}\Rightarrow \mathrm{a}=0.1\mathrm{m}$

Total current $\mathrm{I}=\int \overrightarrow{\mathrm{J}}.\mathrm{d}\overrightarrow{\mathrm{s}}$

$\begin{array}{l}\Rightarrow \mathrm{I}=\int {\mathrm{J}}_0{\mathrm{e}}^{-10\left[\left(\mathrm{a}-\left|\mathrm{x}\right|\right)+\left(\mathrm{a}-\left|\mathrm{y}\right|\right)\right]}{\hat{\mathrm{a}}}_{\mathrm{z}}\cdot \mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}{\hat{\mathrm{a}}}_{\mathrm{z}}\\ \Rightarrow \mathrm{I}=\underset{\mathrm{y}=-\mathrm{a}}{\overset{\mathrm{a}}{\int }}\underset{\mathrm{x}=-\mathrm{a}}{\overset{\mathrm{a}}{\int }}{\mathrm{J}}_0{\mathrm{e}}^{-10\left(\mathrm{a}-\left|\mathrm{x}\right|\right)}\cdot {\mathrm{e}}^{-10\left(\mathrm{a}-\left|\mathrm{y}\right|\right)}\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}\\ \Rightarrow \mathrm{I}={\mathrm{J}}_0\underset{-\mathrm{a}}{\overset{\mathrm{a}}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}-\left|\mathrm{x}\right|\right)}\mathrm{d}\mathrm{x}\underset{-\mathrm{a}}{\overset{\mathrm{a}}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}-\left|\mathrm{y}\right|\right)}\mathrm{d}\mathrm{y}\end{array}$ 

Evaluating, $\underset{-\mathrm{a}}{\overset{\mathrm{a}}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}-\left|\mathrm{x}\right|\right)}\mathrm{d}\mathrm{x}=\underset{-\mathrm{a}}{\overset{0}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}+\mathrm{x}\right)}\mathrm{d}\mathrm{x}+\underset{0}{\overset{\mathrm{a}}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}-\mathrm{x}\right)}\mathrm{d}\mathrm{x}$ 

$\underset{-\mathrm{a}}{\overset{0}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}+\mathrm{x}\right)}\mathrm{d}\mathrm{x}={\frac{{\mathrm{e}}^{-10\left(\mathrm{a}+\mathrm{x}\right)}}{-10}|}_{-\mathrm{a}}^0=\frac{1-{\mathrm{e}}^{-10\mathrm{a}}}{10}$ 

Similarly,

$\underset{0}{\overset{\mathrm{a}}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}-\mathrm{x}\right)}\mathrm{d}\mathrm{x}={\frac{{\mathrm{e}}^{-10\left(\mathrm{a}-\mathrm{x}\right)}}{10}|}_0^{\mathrm{a}}=\frac{1-{\mathrm{e}}^{-10\mathrm{a}}}{10}$ 

Thus,

$\underset{-\mathrm{a}}{\overset{\mathrm{a}}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}-\left|\mathrm{x}\right|\right)}\mathrm{d}\mathrm{x}=\frac{1-{\mathrm{e}}^{-10\mathrm{a}}}{10}+\frac{1-{\mathrm{e}}^{-10\mathrm{a}}}{10}=\frac{1-{\mathrm{e}}^{-10\mathrm{a}}}{5}$ 

Similarly,

$\underset{-\mathrm{a}}{\overset{\mathrm{a}}{\int }}{\mathrm{e}}^{-10\left(\mathrm{a}-\left|\mathrm{y}\right|\right)}\mathrm{d}\mathrm{x}=\frac{1-{\mathrm{e}}^{-10\mathrm{a}}}{5}$ 

Thus,

$\mathrm{I}={\mathrm{J}}_0\cdot \frac{1-{\mathrm{e}}^{-10\mathrm{a}}}{5}\cdot \frac{1-{\mathrm{e}}^{-10\mathrm{a}}}{5}$