Physics M.C.Q (1 Marks)

$=3 \times 10^{-3} \mathrm{\,H} $

$\varepsilon  =  - M\frac{{dI}}{{dt}} =  - 3 \times {10^{ - 3}}\left[ {\frac{{ - 1 - 1}}{{0.25}}} \right] = 24\,{\rm{mV}}$

$emf$ induced per unit turn $ = \frac{E}{N} = \frac{{MI}}{t}$

$ = 2\pi \times 50 \times 300 \times 4 \times {10^{ - 2}} \times (25 \times {10^{ - 2}} \times 10 \times {10^{ - 2}})$

$ = 30\;\pi \;volt$

$ = 2 \times {(3.14)^2} \times \frac{{1800}}{{60}} \times 4000 \times 0.5 \times {10^{ - 4}} \times {(7 \times {10^{ - 2}})^2}$

$ = 0.58\;V$

$ = 2 \times 3.14 \times 1000 \times 5000 \times 0.2 \times 0.25$= $157 \,kV$

$\frac{\varepsilon_{2}}{\varepsilon_{1}}=\frac{r_{1}}{r_{2}}=\frac{b}{2 b}=\frac{1}{2}$

$\mathrm{E}_{\max }=\mathrm{N} \mathrm{AB} \omega$

Hence the flux through the coil is $\phi=\vec{B} . \vec{A}=B A \cos \theta$

Hence the induced emf $=$ Rate of change of flux $=-\frac{d \phi}{d t}=B A \sin \theta \frac{d \theta}{d t}=B A \omega \sin \theta$

Hence it is minimum when $\theta=0^{\circ}, 180^{\circ}$

Which means when the plane of coil is at right angle to the magnetic field.

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

$\Longrightarrow|E|(2 \pi r)=\frac{d B}{d t} \cdot \pi a^{2}$

$\Longrightarrow|E| \propto \frac{1}{r}$

$=-N \frac{d}{d t}(B A \cos \omega t)=N B A \omega \sin \omega t $

$ \Rightarrow e_{\max }=N B A \omega $

$P=I^{2} R \ldots(2)$

From on $( 1)$ and $( 2)$

$\frac{L}{2 R}=\frac{U}{A P} \Rightarrow \frac{L}{R}=\frac{2 U}{\Pi}$ mplies $\tau=\frac{2 U}{P}$

${L_P} = \frac{{{L_1}{L_2}}}{{{L_1} + {L_2}}} = 2.4H$ ..... $(ii)$

On solving $(i)$ and $(ii)$ $L_1$ $L_2$ = 24 ..... $(iii)$

Also ${({L_1} - {L_2})^2} = {({L_1} + {L_2})^2} - 4{L_1}{L_2}$

$ \Rightarrow {({L_1} - {L_2})^2} = {(10)^2} - 4 \times 24 = 4$$ \Rightarrow {L_1} - {L_2} = 2H$

Thus, emf produced in first coil is negative and the emf produced in second coil is positive so, net inductance is

$L = {L_1} + {L_2} = L + L$

$ \Rightarrow L = - \frac{\phi }{i} + \frac{\phi }{i} = 0$

$=\frac{1}{2} \times(2) \times(1)^{2}=1 \mathrm{\,J}$

Energy developed in resistance

$=1^{2} \mathrm{RT}=1^{2} \times 10 \times 10=100 \mathrm{\,J}$

Hence the required ratio is $\frac{1}{100}$

$R_{e q}=\frac{R}{3}+R=\frac{4 R}{3}=\frac{4(220)^{2}}{300}$

$\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(220)^{2} \times 300}{4(220)^{2}}=\frac{300}{4}=75 \mathrm{\,W}$

that across $L_{2}$ i.e. $\mathcal{E}_{1}=\mathcal{E}_{2}$

Or $-L_{1} \frac{d i_{1}}{d t}=-L_{2} \frac{d I_{2}}{d t}$

Or $L_{1} \int d i_{1}=L_{2} \int d i_{2}$

Or $L_{2} i_{2}=L_{1} i_{1}$

or $\frac{i_{1}}{i_{2}}=\frac{L_{2}}{L_{1}}$

Given, $\phi==\left(2 t^{2}+4 t+6\right) \,m W b$

$\mathrm{As}, \mathrm{\varepsilon}=\frac{\mathrm{d} \phi}{\mathrm{dt}}$

$=\frac{d}{d t}\left(2 t^{2}+4 t+6\right) \times 10^{-3} \mathrm{\,Wb} \mathrm{s}^{-1}$

$=(4 t+4) \times 10^{-3} \,V$

At $\quad t=4 \,s$

$ \varepsilon =(4 \times 4+4) \times 10^{-3} \mathrm{\,V}$

$=20 \times 10^{-3} \mathrm{\,V}=0.02 \mathrm{\,V} $

Here, all the inductances are connected in parallel. 

Hence, the equivalent inductance between $\mathrm{A}$ and $\mathrm{B}$ is

$\frac{1}{L_{A B}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{4}{4}=1$

or $L_{A B}=1 \,H$

and in parallel connection

$\frac{L_{1} L_{2}}{\left(L_{1}+L_{2}\right)}=2.4 \mathrm{\,H}$         ........$(ii)$

Substituting the value of $\left(L_{1}+L_{2}\right)$ from $(i)$ into $(ii),$ we get

$\mathrm{L}_{1} \mathrm{L}_{2}=(2.4)\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right)=(2.4) \times 10=24$

$\left(L_{1}-L_{2}\right)^{2}=\left(L_{1}+L_{2}\right)^{2}-4 L_{1} L_{2}$

$\mathrm{L}_{1}-\mathrm{L}_{2}=\left[(10)^{2}-4 \times 24\right]^{1 / 2}=2 \mathrm{\,H}$        .......$(iii)$

Solving $(i)$ and $(iii),$ we get

$\mathrm{L}_{1}=6 \mathrm{\,H}, \mathrm{L}_{2}=4 \mathrm{\,H}$

==> $0.37{i_0} = {i_0}{e^{ - Rt/L}}$

==> $0.37 = \frac{1}{e} = {e^{ - Rt/L}}$

==> $t = \frac{L}{R}$

$\therefore \,\,I = {I_0}\left[ {1 - {e^{\frac{{ - R}}{L} \times \frac{L}{R}}}} \right] = {I_0}\,(1 - {e^{ - 1}}) = {I_0}\,\left( {1 - \frac{1}{e}} \right)$.

$ = {I_0}\,\left( {1 - \frac{1}{{2.718}}} \right) = 0.63{I_0} = 63\% $ of ${I_0}$

$\tau_{1}=\frac{L}{2 R}$

Time contant for decay

$\tau_{2}=\frac{L}{3 R}$

$\frac{\tau_{1}}{\tau_{2}}=\frac{3}{2}$

$U=\frac{1}{2} L I^{2} \Rightarrow 32=\frac{1}{2} \times L \times(4)^{2}$

$L=4 H$

$P=I^{2} R \Rightarrow 320=(4)^{2} \times R$

$R=20 \Omega$

$\tau=\frac{L}{R}=\frac{4}{20}=0.25$

$P=I^{2} R \ldots(2)$

From on $(1)$ and $(2)$

$\frac{L}{2 R}=\frac{U}{A P} \Rightarrow \frac{L}{R}=\frac{2 U}{\Pi}$ mplies$\tau$ $=\frac{2 U}{P}$

$\mathrm{P}_{\text {heater }}=\left(\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{H}}+\mathrm{R}_{\mathrm{b}}}\right)^{2} \mathrm{R}_{\mathrm{H}}$

$\mathrm{I}_{0}=\mathrm{E} / \mathrm{R}=\frac{100}{100}=1 \mathrm{\,Amp}$

$\mathrm{t}=1 \mathrm{\,m} . \mathrm{s.}=10^{-3} \mathrm{\,sec}$

$\alpha=\mathrm{L} / \mathrm{R}=\frac{100 \times 10^{-3}}{100}$

$I=1 \times e-\frac{10^{-3}}{10^{-3}} \Rightarrow \frac{1}{e}$

Energy stored in $\mathrm{L}=\frac{1}{2} \mathrm{Li}_{0}^{2}=\frac{1}{2} \mathrm{L}\left(\frac{\varepsilon^{2}}{\mathrm{R}_{1}^{2}}\right)$

$=$ heat produced in $\mathrm{R}_{2}$ during discharge.

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{net}}}=\frac{10}{6+8}=\frac{10}{14}=\frac{5}{7} \mathrm{\,A}$

Rate $=\frac{\mathrm{d} \mathrm{U}}{\mathrm{dt}}=\mathrm{LI}\left(\frac{\mathrm{dI}}{\mathrm{dt}}\right)$

At $t=0, I=0$                   $\therefore$ Rate $=0$

At $\mathrm{t}=\infty, \mathrm{I}=\mathrm{I}_{0}$ but $\frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}=0,$ therefore, rate $=0$

$=\frac{1}{\ell}\left[\frac{1}{2}\left(\frac{\mu_{0} \mu_{\mathrm{r}} \mathrm{N}^{2} \mathrm{A}}{\ell}\right) \mathrm{I}^{2}\right]$

$=\frac{1}{2} \mu_{0} \mu_{r}\left(\frac{\mathrm{N}}{\ell}\right)^{2} \mathrm{AI}^{2}=\frac{1}{2} \mu_{0} \mu_{\mathrm{r}} \mathrm{n}^{2} \mathrm{AI}^{2}$

$=\frac{1}{2} \times 4 \pi \times 10^{-7} \times 1000(5000)^{2}\left(10 \times 10^{-4}\right)(1)^{2}$

$=15.7 \mathrm{\,J} / \mathrm{m}$

$I_{N}=\frac{10}{6}$

$t=\infty$ $L$ will be shorted

$I_{N}=\frac{10}{2}=5$

$\mathrm{R}_{\mathrm{N}}=2 \mathrm{R}$

$\left(\tau_{\mathrm{N}}\right)_{1}=\frac{\mathrm{L}}{2 \mathrm{R}}$

$\frac{\left(\tau_{\mathrm{N}}\right)_{1}}{\left(\tau_{\mathrm{N}}\right)_{2}}=\frac{3 \mathrm{R}}{2 \mathrm{R}}$

Decay

$\mathrm{R}_{\mathrm{N}}=3 \mathrm{R}$

$\left(\tau_{\mathrm{N}}\right)_{2}=\frac{\mathrm{I}}{3 \mathrm{R}}$

$\therefore R=\left(\frac{E}{i}\right)=\left(\frac{4}{2}\right)=2 \Omega$

$i=\frac{L}{R}=\frac{1}{2}=0.5 A$