physics-class-12-unit-03-chapter-14-displacement-current-l-11-11-xo4mxdf4xbi

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Ampere's Circuital Law </primary>
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- For surface $S_1 \quad$ $I_{enc} $ = I

- For surface $S_{2} \quad I_{enc}$ = 0 !

- Election plane arrange $S_2$

- $\phi_E  = E A $

- $ =\frac{\sigma}{\epsilon_0} A$
- =$\frac{Q}{\epsilon_0}$

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<footer style = "font-size: 18px">Displacement Current $\rightarrow$ Ampere's Circuital Law $\rightarrow$ <span style = "color:blue"> Ampere's Circuital Law </span> $\rightarrow$ Current $\rightarrow$ Displacement  Current </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Case-1 </primary>
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- r < R

- $\oint \vec{B} \cdot \vec{d l } $ = $B \cdot 2 \pi r $

- $ \Phi_\epsilon=\pi r^2 E $

- $ =\pi r^2 \frac{\sigma}{\epsilon_0} $
- $ =\frac{\pi r^2}{\epsilon_0} \frac{1}{\pi R^2} Q $

- $ =\frac{Q r^2}{\epsilon_0 R^2}$

- $\frac{d\Phi_\epsilon}{d t}=\frac{r^2}{\epsilon_r R^2} \frac{d Q}{d t}$
- =$\frac{r^2}{\epsilon_0 R^2} I $

- $ \sigma=\frac{Q}{A}=\frac{Q}{\pi R^2} $

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<footer style = "font-size: 18px">Displacement  Current $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Case-1 </span> $\rightarrow$ Case-1 $\rightarrow$ Case-2 </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Case-1 </primary>
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- $\oint \vec{B} \cdot \vec{d l} =\mu_0 I_c+\mu_0 \epsilon_0 \frac{d \Phi_{\varepsilon}}{d t} $

- $ =\mu_0 \epsilon_0 \frac{d \Phi_\epsilon}{d t} $

- $B \cdot 2 \pi r \quad  =\mu_0 \epsilon_0 \frac{r^2}{\epsilon_0 R^2} I $

- $B  =\frac{\mu_0 r}{2 \pi R^2} I \quad r < R$

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<footer style = "font-size: 18px">Example-1 $\rightarrow$ Case-1 $\rightarrow$ <span style = "color:blue"> Case-1 </span> $\rightarrow$ Case-2 $\rightarrow$ Case-2 </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Case-2 </primary>
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- r > R

- $ \Phi_E=\pi R^2 \epsilon $

- $=\pi R^2 \frac{\sigma}{\epsilon_0} $
- $=\frac{Q}{\epsilon_0} $

- $ \frac{d \Phi_\epsilon}{d t}$
- =$\frac{1}{\epsilon_0} \frac{d Q}{d t}=\frac{1}{\epsilon_0} I $

- $ \oint \vec{B} \cdot \vec{d l}=\mu_0 I_c+\mu_0 \epsilon_0 \frac{d \Phi_\epsilon}{d t}$

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<footer style = "font-size: 18px">Case-1 $\rightarrow$ Case-1 $\rightarrow$ <span style = "color:blue"> Case-2 </span> $\rightarrow$ Case-2 $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Conduction and Displacement Current </primary>
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- $\oint \vec{B} \cdot \overrightarrow{d L}=\mu_0 I_c+\mu_0 \epsilon_0 \frac{d \Phi_\epsilon}{d t}$

- Conduction Current: $I_c=I$

- Displacement current between $r$ \and R

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<footer style = "font-size: 18px">Magnetic Field $\rightarrow$ Magnetic Field $\rightarrow$ <span style = "color:blue"> Conduction and Displacement Current </span> $\rightarrow$ Conduction and Displacement Current $\rightarrow$ Conduction and Displacement Current </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Conduction and Displacement Current </primary>
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- $\oint \vec{B} \cdot \overrightarrow{d L}=\mu_0 I_c+\mu_0 \epsilon_0 \frac{d \Phi_\epsilon}{d t}$

- $I_c=I$

- $I_{D}$ : Displacement current bstween $r$ \and R

- $ I_D = \epsilon_0 {\frac{d \Phi_\epsilon}{d t}}$

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<footer style = "font-size: 18px">Magnetic Field $\rightarrow$ Conduction and Displacement Current $\rightarrow$ <span style = "color:blue"> Conduction and Displacement Current </span> $\rightarrow$ Conduction and Displacement Current $\rightarrow$ Conduction and Displacement Current </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Conduction and Displacement Current </primary>
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- $I_0  =\epsilon_0 \frac{d \Phi_\epsilon}{d t} $

- $ =\epsilon{\frac{d}{d t}}\left[-E \pi\left(R^2-r^2\right)\right] $

- $ =\epsilon_0 \frac{d}{d t}\left[-\frac{\sigma}{\epsilon_0} \pi\left(R^2-r^2\right)\right] $

- $ =-\frac{d}{d t}\left[\frac{Q}{A} \pi\left(R^2-r^2\right)\right] $

- $ =-\frac{\pi}{\pi R^2}\left(R^2-r^2\right) \frac{d Q}{d t} $

- $ =-I\left(1-\frac{r^2}{R^2}\right)$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-03-chapter-14-displacement-current-l-11_11-xo4mxdf4xbi-16.jpg)

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<footer style = "font-size: 18px">Conduction and Displacement Current $\rightarrow$ Conduction and Displacement Current $\rightarrow$ <span style = "color:blue"> Conduction and Displacement Current </span> $\rightarrow$ Conduction and Displacement Current $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Conduction and Displacement Current </primary>
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- $\oint \vec{B} \cdot \vec{d l}  =\mu_0 I_c+\mu_0 I_D $

- $B \cdot 2 \pi r  =\mu_0 I-\mu_0 I\left(1-\frac{r^2}{R^2}\right) $

- $ =\mu_0 I-\mu_0 I+\frac{\mu_0 I r^2}{R^2} $
- $ =\frac{\mu_0 I r^2}{R^2} $

- $B  =\frac{\mu_0 I r^2}{R^2} \cdot \frac{1}{2 \pi r} $
- $B  =\frac{\mu_0 I r}{2 \pi R^2}$

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### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **A Parallel plates capacitor (are filled) is getting charges and the current at a specific time is 0.45 A. If radius of the plates R = 5cm.**

- **a).** Calculates the total displacement current between the capacitor plates.

- **b).** Calculates the displacement current density.

- **c).** Calculates B as r= 2.5 cm and r = 10cm.

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<footer style = "font-size: 18px">Case-1 $\rightarrow$ Case-2 $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Faradays Law $\rightarrow$ Faradays Law </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Faradays Law </primary>
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- $\oint \vec{E} \cdot \overrightarrow{d l}=-\frac{d \Phi_B}{d t}=-\frac{d}{d t} \int \vec{B} \cdot d \vec{A}$

- **Magnetic field Amperical law ( with $I_c=0$ )**

- $\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 \epsilon_0 \frac{d \Phi_E}{d t}=\mu_0 \epsilon_0 \frac{d}{d t} \int \vec{E} \cdot d \vec{A}$

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<footer style = "font-size: 18px">Case-2 $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Faradays Law </span> $\rightarrow$ Faradays Law $\rightarrow$ Conduction and Displacement Current </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Conduction Current </primary>
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-  E = $E_0$ cos $\omega t$

- $J_c  =\sigma E $

- $ =\sigma E_0 \cos \omega t$

- $J_3  =\epsilon \frac{d E}{d t}$

- $ =-\epsilon \omega E_0 \sin \omega t$

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<footer style = "font-size: 18px">Faradays Law $\rightarrow$ Conduction and Displacement Current $\rightarrow$ <span style = "color:blue"> Conduction Current </span> $\rightarrow$ Displacement current $\rightarrow$ Good Conductor </footer>

### <Success style="font-size:25px"> Displacement Current L-11 </Success>
### <primary style="font-size:24px"> Displacement current </primary>
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- $J_C(max) =\sigma E_0 $

- $J_D(max) =\epsilon \omega E_0 $

- $\left.\frac{J_D}{J_C}\right|_{\operatorname{max}}$

- =$\frac{\epsilon \omega E_0}{\sigma E_0}=\frac{\epsilon \omega}{\sigma} $

- $ =\frac{2 \pi \nu \epsilon}{\sigma} \quad \omega=2 \pi \nu$

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<footer style = "font-size: 18px">Conduction and Displacement Current $\rightarrow$ Conduction Current $\rightarrow$ <span style = "color:blue"> Displacement current </span> $\rightarrow$ Good Conductor $\rightarrow$ Poor Conductor </footer>