Detailed solution not required.
Detailed solution not required.
$d\vec{B} = \frac{\mu_{0}}{4\pi} \frac{i d\vec{l} \times \vec{r}}{r^{3}}$
$B = \frac{\mu_{0}N}{l}i$
$= 4\pi \times 10^{-7} \times \frac{10}{10^{-2}} \times 5$
$= 2\pi \times 10^{-3}\text{ Tesla}$
$B = \frac{\mu_{0}}{4\pi} \frac{2i}{d}$
$\frac{B_{1}}{B_{2}} = \frac{d_{2}}{d_{1}} \implies \frac{40}{10}$
$B_{2} = \frac{B_{1}}{4} = \frac{0.04}{4} = 0.01\text{ T}$
At mid point, magnetic field due to both the wires are equal & opposite.
So $B_{net} = 0$
$r = \frac{mv}{qB} = \frac{\sqrt{2mqV}}{qB}$
$\frac{R_{1}}{R_{2}} = \sqrt{\frac{m_{x}}{m_{y}}}$ (since $q, V, B \implies \text{Same}$)
$\frac{m_{x}}{m_{y}} = \left(\frac{R_{1}}{R_{2}}\right)^{2}$
$F = qvB\sin\theta$
if $\theta = 0^{\circ}$, $F = 0$
Magnetic force on charge will be zero hence path will be straight line.
Detailed solution not required.
$F_{CD} = F_{CB}$
$\frac{\mu_{0}}{4\pi} \frac{2 \times 15 \times 5}{9} \times l = \frac{\mu_{0}}{4\pi} \frac{2 \times 5 \times 10}{y} \times l$
$y = 6$
$d = 9 + 6 = 15$
Detailed solution not required.
$W_{1} = nW_{2}$
$MB(1 - \cos 90^{\circ}) = nMB(1 - \cos 60^{\circ})$
$n = 2$
$\vec{\tau} = \vec{M} \times \vec{B}$
$= 50\hat{i} \times (0.5\hat{i} + 3\hat{j})$
$= 150\hat{k}\text{ N.m}$
Induced charge $q = \frac{\Delta\phi}{R}$
Detailed solution not required.
$e = \frac{-d}{dt}(3t^{2} + 4t + 9)$
$= -(6t + 4)$
At $t=2\text{s}$, $|e| = 16\text{ V}$
Detailed solution not required.
$V_{A} - 1 \times 5 + 15 + 5 \times 10^{-3}(10^{3}) = V_{B}$
$V_{A} - V_{B} = 15\text{ Volt}$
$L = \frac{e}{\frac{di}{dt}} = \frac{5}{\frac{3-2}{1 \times 10^{-3}}} = 5 \times 10^{-3}\text{ H}$
$L = \mu_{0} \frac{N^{2}A}{l}$
$M = \frac{\mu_{0}N_{1}N_{2}A}{l}$
If there are no losses then $P_{in} = P_{out}$
$U = \frac{1}{2}LI^{2}$
$= \frac{1}{2} \times 100 \times 10^{-3} \times 10^{2} = 5\text{ J}$
$T = \frac{2\pi}{\omega} = \frac{2\pi}{200\pi}$
$t = \frac{T}{4} \implies \frac{1}{4 \times 100}\text{ sec}$
$T = \frac{1}{f} = \frac{1}{50}$
$t = \frac{T}{4} = \frac{1}{4 \times 50} = 5 \times 10^{-3}\text{ sec}$
$I_{0} = I_{rms}\sqrt{2} = 10\sqrt{2}$
Since $R = 0$, $P_{ave} = 0$
$\omega_{0} = \frac{1}{\sqrt{LC}}$
Resonance frequency does not depend on R.
$\tan\phi = \frac{X_{L}}{R} = \frac{\omega L}{R}$
$= \frac{2\pi \times 200 \times \frac{1}{\pi}}{300}$
$\phi = \tan^{-1}\left(\frac{4}{3}\right)$
$\tan\phi = \frac{X_{C}}{R} = \sqrt{3}$
$\phi = \frac{\pi}{3}$
$Z = \frac{V}{i} = \frac{100}{0.5} = 200~\Omega$
$V_{L} = V_{C} = 100\text{ V}$
This is the condition of resonance.
$\therefore V_{R} = 200\text{ V}$
$P = i^{2}R = \frac{V^{2}}{Z^{2}}R = \frac{V^{2}R}{(R^{2} + \omega^{2}L^{2})}$
$\tan 60^{\circ} = \frac{X_{L}}{R}$ and $\tan 60^{\circ} = \frac{X_{C}}{R}$
$X_{L} = X_{C}$ $\implies Z = R$
$P_{ave} = \frac{V^{2}}{R} = \frac{200 \times 200}{100} = 400\text{ W}$
$R = \frac{P}{i_{rms}^{2}} = \frac{240}{16} = 15\Omega$
$Z = \frac{V}{i} = \frac{100}{4} = 25\Omega$
$X_{L} = \sqrt{Z^{2} - R^{2}} = 20\Omega$
$2\pi fL = 20 \implies L = \frac{1}{5\pi}\text{ H}$
Detailed solution not required.
$X_{C} = \frac{1}{2\pi fC}$
Detailed solution not required.
Detailed solution not required.
$\frac{E_{0}}{B_{0}} = C$
$K = \frac{2\pi}{\lambda}$
$\omega = 2\pi\nu$
$V = \frac{C}{\sqrt{\mu_{r}\epsilon_{r}}} = \frac{3 \times 10^{8}}{\sqrt{1.3 \times 2.14}}$
$\phi = 90^{\circ}$
$P_{ave} = V_{rms}I_{rms}\cos 90^{\circ} = 0$
$V_{rms} = \frac{V_{0}}{\sqrt{2}} = \frac{120}{\sqrt{2}}$
Detailed solution not required.