Physics 3 Marks Question

$\text{H}=\frac{\text{B}}{\mu_0}$

$\frac{\text{E}_0}{\text{H}_0}=\frac{\frac{\text{B}_0}{(\mu_0\in_0\text{C})}}{\frac{\text{B}_0}{\mu_0}}=\frac{1}{\in_0\text{C}}$

$=\frac{1}{8.85\times10^{-12}\times3\times10^8}$

$=376.6\Omega=377\Omega$

$\text{Dimension}\frac{1}{\in_0\text{C}}=\frac{1}{\big[\text{LT}^{-1}\big]\big[\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^2\big]}$

$=\frac{1}{\text{M}^{-1}\text{L}^{-2}\text{T}^{3}\text{A}^2}=\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-2}=[\text{R}]$

$\text{B}_0=200\mu\text{T}$

$\text{E}_0=\text{C}\times\text{B}_0$

$\text{E}_0=200\times10^{-6}\times3\times10^{8}$

$\text{E}_0=6\times10^4$

Average energy density $=\frac{1}{2\mu_0}\text{B}_0^2=\frac{\big(200\times10^{-6}\big)^2}{2\times4\pi\times10^{-7}}$

$=\frac{4\times10^{-8}}{8\pi\times10^{-7}}=\frac{1}{20\pi}=0.0159=0.016$ 

We know, $\text{I}=\frac{1}{2}\in_0\text{E}^2_0\text{C}$

$\text{E}^2_0=\frac{2\text{I}}{\in_0\text{C}}$ or $\text{E}_0=\sqrt{\frac{2\text{I}}{\in_0\text{C}}}$

$\text{E}_0=\sqrt{\frac{2\times2.5\times10^{14}}{8.85\times10^{-12}\times3\times10^8}}$

$\text{E}_0=0.4339\times10^9$

$\text{E}_0=4.33\times10^8\text{N/c}$

$\text{B}_0=\mu_0\in_0\text{CE}_0$

$\text{B}_0=4\times3.14\times10^{-7}\times8.854\times10^{-12}\times3\times10^8\times4.33\times10^8$

$\text{B}_0=1.44\text{T}$

$\phi=\text{E.A.}=\frac{\text{Q}}{\in_0\text{A}}\frac{\text{A}}{2}=\frac{\text{Q}}{\in_02}$

$\text{i}_0=\in_0\frac{\text{d}\phi_\text{E}}{\text{dt}}=\in_0\frac{\text{d}}{\text{dt}}\Big(\frac{\text{Q}}{\in_02}\Big)$

$=\frac{1}{2}\Big(\frac{\text{dQ}}{\text{dt}}\Big)$

$=\frac{1}{2}\frac{\text{d}}{\text{dt}}\Big(\text{ECe}^{\frac{-\text{t}}{\text{RC}}}\Big)$

$=\frac{1}{2}\text{EC}-\frac{1}{\text{RC}}\text{e}^{\frac{-\text{t}}{\text{RC}}}$

$=\frac{-\text{E}}{2\text{R}}\text{e}^{\frac{-\text{td}}{\text{R}\text{E}_0\lambda}}$

$\phi=\text{E.A.}=\frac{\text{Q}}{\in_0\text{A}}\frac{\text{A}}{2}=\frac{\text{Q}}{\in_02}$

$\text{i}_0=\in_0\frac{\text{d}\phi_\text{E}}{\text{dt}}=\in_0\frac{\text{d}}{\text{dt}}\Big(\frac{\text{Q}}{\in_02}\Big)$

$=\frac{1}{2}\Big(\frac{\text{dQ}}{\text{dt}}\Big)$

$\phi_\text{E}=\text{EA}=\frac{\text{KqA}}{\text{x}^2}$

$\text{l}_\text{d}=\in_0\frac{\text{d}\phi\text{E}}{\text{dt}}=\in_0\frac{\text{d}}{\text{dt}}\frac{\text{kqA}}{\text{x}^2}=\in_0\text{KqA}=\frac{\text{d}}{\text{dt}}\text{x}^{-2}$

$=\in_0\times\frac{1}{4\pi\in_0}\times\text{q}\times\text{A}\times-2\times\text{x}^{-3}\times\frac{\text{dx}}{\text{dt}}=\frac{\text{qAv}}{2\pi\text{x}^3}$

We know, $\text{B}_0=\mu_0\in_0\text{CE}_0$

Putting the values,

$\text{B}_0=4\pi\times10^{-7}\times8.85\times10^{-12}\times3\times10^8\times810$

$\text{B}_0=27010.9\times10^{-10}$

$\text{B}_0=2.7\times10^{-6}\text{T}$

$\text{B}_0=2.7\mu\text{T}$