Physics

The aim of using displacement current was to allow currrent ot flow inspite of a change in the field. This was not possible in the case of actual current since it lead to some complex situations, due to which scientists had to find a new solution for generating an electric field in such cases.

Kindly go through the solution

-> $v=\omega \sqrt[]{A^2-x^2}\Rightarrow \frac{v^2}{A^2{\omega }^2}+\frac{x^2}{A^2}=1$ Path is ellipse.

Maxwell simply wanted to generate a magnetic field using electric current, which displacement current had the capability to offer. Since both current flows were able to produce an electric field, both of them could be simultaneously used for using in the equation.

Let the spring is in extension state and   $a_B>a_A,{\overrightarrow{a}}_{AB}=-a_A\widehat{i}-\left(-a_B\widehat{i}\right)=\left(-a_A+a_B\right)\widehat{i}$

Hence we can say that block moves away from block B in the frame of B

F – kx = Ma_B           . (1)

kx = Ma                  . (2)

$\Rightarrow a_B=\frac{F}{M}-a$

$\widehat{r}-\widehat{i}=2cos\theta \widehat{n}......\left(1\right)$

$\Rightarrow \widehat{r}=\widehat{i}-2\left(\widehat{i}.\widehat{n}\right)\widehat{n}$

$\Rightarrow \overrightarrow{b}=\overrightarrow{a}-2\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{c}$

The key difference between these two terms is that displacement current is deprived of electrons which flow in the case of an actual current. It's role is to help generate magnetic field which is produced when the medium changes between two different bodies.

  $N=mgcos30°+qEsin30°$

$a=\frac{mgsin\theta -qEcos\theta -\mu N}{m}=2.30m/s^2$

$S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow t=\sqrt[]{\frac{2\mathcal{l}}{a}}=1.31sec$             

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

-> $Z=\sqrt[]{R^2+{\left(X_L-X_C\right)}^2}$

$Z=\sqrt[]{{\left(120\right)}^2+{\left(10-100\right)}^2}=150\Omega$

$\omega =\frac{1}{\sqrt[]{LC}}=\frac{1}{\sqrt[]{10^{-1}*10^{-4}}}=\sqrt[]{10^5}$

$?\omega =2\pi f$

$\Rightarrow f=\frac{10^3}{2\pi \sqrt[]{10}}=50Hz$

$A=A_0{e}^{-\lambda t_1}$     [Radio active decay law]

$\frac{A}{5}=A_0{e}^{-\lambda \left(t_2-t_1\right)}$

$\Rightarrow Averagelife=\frac{1}{\lambda }=\frac{\left(t_2-t_1\right)}{\mathcal{l}n5}$