26 questions · timed · auto-graded
$\text{R}=\frac{\text{E}}{\text{i}}=\frac{4}{2}=2$
$\text{i}=\frac{\text{L}}{\text{R}}=\frac{1}{2}=0.5$
$\phi_{1}=\text{BA},\phi_{2}=0$
$=\frac{2\times10^{-4}\times\pi(0.1)^2}{2}=\pi\times10^{-5}$
$\text{E}=\frac{\text{d}\phi}{\text{dt}}=\frac{\pi\times10^{-6}}{2}=1.57\times10^{-6}\text{V}$
Total energy stored $=\frac{\text{B}^2\text{V}}{2\mu_0}=\frac{\big(\frac{\mu_0\text{i}}{2\text{r}}\big)^2}{2\mu_0}\text{V}=\frac{\mu_0\text{i}^2}{4\text{r}^2\times2}\text{V}$
$=\frac{4\pi\times10^{-7}\times4^2\times1\times10^{-9}}{4\times(10^{-1})^2\times2}=8\pi\times10^{-14} \text{J}.$
$\text{e}=\text{Bvl}=\frac{\text{B}\times\text{a}\times\omega\times\text{a}}{2}$
$\text{i}=\frac{\text{Ba}^2\omega}{2\text{R}}$
$\text{F}=\text{ilB}=\frac{\text{B}\text{a}^2\omega}{2\text{R}}\times\text{a}\times\text{B}=\frac{\text{B}^2\text{a}^2\omega}{2\text{R}}$ towards right of OA.
$=\text{vBl}\cos\theta$
$\theta$ is angle between normal to plane and $\vec{\text{B}}=90^{\circ}$
$=\text{vBl}\cos90^{\circ}=0.$
$\text{D}\phi=\phi_2=(0.85-0.35) \ \text{weber}=0.5 \ \text{weber}$
$\text{dt}=0.5\text{sec}$
$\text{E}=\frac{\text{d}\phi}{\text{dt}'}=\frac{0.5}{0.5}=1\text{V}$
The induced current is anticlockwise as seen from above.
$\frac{\text{dl}}{\text{dt}}=\frac{\text{lA}}{\text{s}}$
$\text{E}=-\mu\frac{\text{dl}}{\text{dt}}$
$\Rightarrow\text{E}=2.5\times1=2.5\text{V}$
V at a distance $\frac{\text{r}}{2}$
From the centre $=\frac{\text{r}\omega}{2}$
$\text{E}=\text{BlV}$
$\Rightarrow\text{E}=\text{B}\times\text{r}\times\frac{\text{r}\omega}{2}=\frac{1}{2}\text{B}\text{r}^2\omega$
$\phi_2=0$
$\text{E}=\frac{\phi_1-\phi_2}{\text{t}}=\frac{125\pi\times10^{-5}}{5\times10^{-1}}=25\pi\times10^{-4}=7.8\times10^{-3}$
$\int\vec{\text{E}}.\vec{\text{dl}}.$
$\text{vBl}$
$\frac{\text{d}\psi_{\text{B}}}{\text{dt}}$
$\int\vec{\text{E}}.\vec{\text{dl}}=\text{ML}\text{T}^{-3}\text{l}^{-1}=\text{ML}^2\text{l}^{-1}\text{T}^{-3}$
$\text{vBl}=\text{LT}^{-1}\times\text{Ml}^{-1}\text{T}^{-2}\times\text{L}=\text{ML}^2\text{l}^{-1}\text{T}^{-3}$
$\frac{\text{d}\phi_{\text{B}}}{\text{dt}}=\text{Ml}^{-1}\text{T}^{-2}\times\text{L}^2=\text{ML}^2\text{l}^{-1}\text{T}^{-2}$
$=1\times10^{-6}\text{m}^2, \ \text{f}_{\text{cu}}=1.7\times10^{-8}\Omega-\text{m}$$$
$\text{R}=\frac{\text{f}_\text{cu}\text{l}}{\text{A}}=\frac{1.7\times10^{-8}\times100}{1\times10^{-6}}=1.7 \Omega$
$\text{i}=\frac{\text{L}}{\text{R}}=\frac{0.17\times10^{-8}}{1.7}=10^{-2} \text{ sec}=10 \ \text{m} \ \text{sec}.$
$\text{A}=1\text{cm}^2=10^{-4}\text{m}^2$
$\text{T}=1\text{s}$
$\phi=\text{B}.\text{A=}10^{-1}\times10^{-4}=10^{-5}$
$\text{e}=\frac{\text{d}\phi}{\text{dt}}=\frac{10^{-5}}{1}=10^{-5}=10\mu\text{V}$
$\text{e}=\frac{\text{d}\phi}{\text{dt}}=\frac{\text{BA}}{\text{dt}}=\frac{\mu_0\text{i}}{2\pi\text{d}}\times\frac{\text{A}}{\text{dt}}$
$=\frac{4\pi\times10^{-7}\times10}{2\pi\times2\times10^{-1}}\times\frac{10^{-6}}{1\times10^{-1}}=1\times10^{-10}\text{V}$
$\frac{\text{dl}}{\text{dt}}=\frac{\text{ERV}}{\text{L}\Big(\frac{\text{Rx}}{\text{L}}+\text{r}\Big)^2}$
$\mu=\frac{\text{E}}{\frac{\text{di}}{\text{dt}}}=\frac{\text{N}\mu_0\pi\text{a}^2\text{a}'^2}{2(\text{a}^2+\text{x}^2)^{\frac{3}{2}}}$
$\therefore$ The voltmeter will record 1mv.
$\text{E}=\text{Bvl}$
Resistance = r × total length$=\text{r}\times2(\text{l}+\text{vt})=24(\text{l}+\text{vt})$
$\text{i}=\frac{\text{Bvl}}{2\text{r}(\text{l}+\text{vt})}$
$\phi=\text{BA}=\text{Ba}\cos0^{\circ}=\text{BA}$
$\text{e}=\frac{\text{d}\phi}{\text{dt}}=\frac{\text{BA}}{1};\text{i}=\frac {\text{e}}{\text{R}}=\frac{\text{BA}}{\text{R}}$
$\phi=\text{iT}=\frac{\text{BA}}{\text{R}}$
$\therefore$ E = Bv2R
$\therefore$ E = 0
