Physics M.C.Q (1 Marks)

Force on $\mathrm{BC}=16 \mathrm{\,N}$ (left)

so force on $\mathrm{ABC}=\sqrt{(12)^{2}+(16)^{2}}=20 \mathrm{\,N}$

Hence a change in flux through the loop is necessary for emf induction, which

happens only in case of options $C$ and $D.$

Hence correct answers are options $\mathrm{C}$ and $\mathrm{D}$.

$A =10 \mathrm{cm} \times 10 \mathrm{\,cm} $

$=100 \mathrm{\,cm}^{2}=10^{-2} \mathrm{\,m}^{2}$

Initial magnetic flux linked with the loop

$\phi_{1} =\mathrm{B}_{1} \mathrm{A} \cos \phi $

$=0.1 \times 10^{-2} \times \cos 45^{\circ}$

$=\frac{0.1 \times 10^{-2} \times 1}{\sqrt{2}}=\frac{10^{-3}}{\sqrt{2}} \mathrm{\,Wb}$

Final magnetic flux linked with loop

$\phi_{2}=0 \mathrm{Wb} \quad\left(\because B_{2}=0\right)$

The induced emf in the loop is

$\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\left(\phi_{2}-\phi_{1}\right)}{\mathrm{t}}$

$=-\frac{\left(0-\frac{10^{-3}}{\sqrt{2}}\right)}{0.7}=\frac{10^{-3}}{0.7 \times \sqrt{2}}=10^{-3} \mathrm{\,volt}$

The induced current in the loop is

$I=\frac{e}{R}=\frac{10^{-3} V}{1 \Omega}=10^{-3} A=1.0 \mathrm{\,mA}.$

${{\rm{V}}_{{\rm{PS}}}} = {\rm{ir}} = \frac{{{\rm{B}}\ell {\rm{v}}}}{{\rm{r}}} \cdot \frac{{\rm{r}}}{6} = \frac{{{\rm{Fr}}}}{{6{\rm{B}}\ell }}\left\{ {{\rm{v}} = \frac{{{\rm{Fr}}}}{{{{\rm{B}}^2}{\ell ^2}}}} \right\}$

$e = \frac{1}{2}B{l^2}\omega $

where $l$ is the length of each spoke i.e., the radius of the

wheel $l = 0.4\,{\rm{m}}$

All the induced cells are connected in parallel as shown in the adjoining figure. Therefore induced emf between the rim and the centre of the wheel is

$e = \frac{1}{2}B{l^2}\omega $

$=\frac{1}{2} \times 0.4 \times 10^{-4} \times(0.4)^{2} \times \frac{2 \pi \times 180}{60}$

$=6 \times 10^{-5} \mathrm{V}$

Given $: \ell=1\,m , B =5 \times 10^{-3}\,Wb / m ^2$

$f =\frac{1800}{60}=30 \text { rotations } / sec$

in one rotation, the moving rod of the metal races a circle of radius $r=\ell$

$\therefore$ Area swept in one rotation $=\pi r ^2$

$\frac{ d \phi}{ dt }=\frac{ d }{ dt }( BA )= B \cdot \frac{ dA }{ dt }=\frac{ B \pi r ^2}{ T }$

$= B f \pi r ^2=\left(5 \times 10^{-3}\right) \times 3.14 \times 30 \times 1=0.471 \,V$

$\therefore \;i = 1\left( {1 - {e^{\frac{{ - 5 \times 2}}{{10}}}}} \right) = (1 - {e^{ - 1}})amp$

Since magnetic field at the centre of circular coil carrying current is given by $B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi Ni}}{r}$

$\therefore \;N.\frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi Ni}}{r}.\pi {r^2} = Li \Rightarrow L = \frac{{{\mu _0}{N^2}\pi r}}{2}$

Hence self inductance of a coil

$ = \frac{{4\pi \times {{10}^{ - 7}} \times 500 \times 500 \times \pi \times 0.05}}{2} = 25\;mH$

$\Rightarrow NB\frac{{dA}}{{dt}} = \frac{{Ldi}}{{dt}}$

$ \Rightarrow \frac{{1 \times 1 \times 5}}{{{{10}^{ - 3}}}} = L \times \left( {\frac{{2 - 1}}{{2 \times {{10}^{ - 3}}}}} \right) $

$\Rightarrow L = 10H$

$\Rightarrow 20 = L \times \frac{{(18 - 2)}}{{0.05}}$

$ \Rightarrow L = 62.5\;mH$

since current decreases so $\frac{{di}}{{dt}}$

is negative, hence $e = - 5 \times \left( { - 2} \right) = + 10\,V$

From the graph it is clear that for time $t_{1},$ current does not change with time, i.e. $\frac{d i}{d t}=0 .$ so, induced emf is zero. Hence graph $(a)$ and $(b)$ are wrong.

from time interval $\left(t_{1}\right) \rightarrow\left(t_{2}\right),(d i / d t)$ is negative and constant. Hence, choise $(C)$ is correct.

Choice$(D)$ is wrong because in the graph the induced emf is never negative and is not constant.

Now $ \mathrm{L}=\frac{\mathrm{N} \phi}{\mathrm{i}}$

$\mathrm{N} \phi=\mathrm{Li}=(2.0)(50)=10 \mathrm{\,Wb}$

Rate of change of current is constant for one period at a positive value and is constant at negative value for the second time period.

Therefore emf is a constant positive value for first half and constant negative value for second half. Option $B$ is correct.

bigger loop and $\mathrm{cw}$

$0<\mathrm{t}<\mathrm{T} / 4 \rightarrow \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}} \rightarrow+\mathrm{ve} $ and constant

$\mathrm{T} / 4<\mathrm{t}<\mathrm{T} / 2 \rightarrow \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}=0$

$\frac{\mathrm{T}}{2}<\mathrm{t}<3 \frac{\mathrm{T}}{4} \rightarrow \frac{\mathrm{dI}}{\mathrm{dt}} \rightarrow-\mathrm{ve} $ and constant

$\mathrm{L}=\frac{\mu_{0} \mathrm{N}^{2} \pi \mathrm{r}^{2}}{\ell}$

${\rm{L}} = \frac{{{\mu _0}{{\rm{N}}^2}4{\pi ^2}{{\rm{r}}^2}}}{{4\pi \ell }} = \frac{{{\mu _0}\ell _\omega ^2}}{{4\pi \ell }}$

$\mathrm{V}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{in}}}{\mathrm{I}_{\mathrm{P}}}=\frac{1000}{8}=125$

$\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}}$

$\Rightarrow \mathrm{N}_{\mathrm{S}}=\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}} \times \mathrm{N}_{\mathrm{P}}=\frac{500}{125} \times 100=400$

$\mathrm{V}_{\mathrm{A}}-5-15+5=\mathrm{V}_{\mathrm{B}}$

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=15$ $\mathrm{volt}$

${\mathrm{L}=\frac{\mathrm{e}}{\mathrm{di} / \mathrm{dt}}}$

$P=e \times i$

or $\quad \mathrm{e}=\frac{\mathrm{P}}{\mathrm{i}}$

$\therefore \mathrm{L}=\frac{\mathrm{P}}{\mathrm{i}(\mathrm{di} / \mathrm{dt})}$ ($\because $ Power and curect rate is same )

or $\mathrm{L} \propto \frac{1}{\mathrm{i}}$

$\therefore \frac{i_{1}}{i_{2}}=\frac{L_{2}}{L_{1}}=\frac{1}{4}$

Final current, $I_{2}=0 A$

Change in current, $d l=I_{1}-I_{2}=5 A$

Time taken for the charge, $t=0.1 s$

Average emf, $e=200 \mathrm{V}$

For self inductance $(L)$ of the coil, we have the relation for average emf as$:$

$e=L \frac{d i}{d t}$

$L=\frac{e}{\frac{d i}{d t}}$

$=\frac{200}{\frac{5}{0.1}}=4 H$

Hence the self inductance of the coil is $4 H$

$ \varepsilon =150 \mathrm{\,V} $

As $|\varepsilon| =\mathrm{L}\left|\frac{\mathrm{d} |}{\mathrm{dt}}\right|$

$\therefore L = \frac{{|\varepsilon |}}{{\left| {d/dt} \right|}} = \frac{{150}}{{25}} = 6\,{\rm{H}}$

Final current, when the switch is opened is zero.

$\therefore \frac{{{\rm{dI}}}}{{{\rm{dt}}}} = \frac{{0 - 2}}{{2 \times {{10}^{ - 3}}}} =  - 1 \times {10^3}{\mkern 1mu} {\rm{A}}{{\rm{s}}^{ - 1}}$

As $\varepsilon=-L \frac{\mathrm{d} I}{\mathrm{dt}}$

$=-10\left(-1 \times 10^{3}\right)=10^{4} \mathrm{\,V}$

$ = \frac{{4\pi \times {{10}^{ - 7}} \times 2000 \times 300 \times 1.2 \times {{10}^{ - 3}}}}{{0.30}} \times \frac{{|2 - ( - 2)|}}{{0.25}}$

$ = 48.2 \times {10^{ - 3}}V = 48\;mV$

$ \Rightarrow \frac{{d{i_1}}}{{dt}} = 4\;A/\sec .$

$\Rightarrow 9 \times {10^{ - 5}} = M \times 3$

$ \Rightarrow M = 3 \times {10^{ - 5}}H$

magnetic flux of the secondary coil

$I_{2}=B_{1} A_{2}=\frac{\mu_{0} I_{1} \pi r^{2}}{2 R}$

mutual inductance

$\frac{\mu_{0} \pi r^{2}}{2 R}$

$4=2 \times \Delta I \Rightarrow \Delta I=2 A m p$

$\Rightarrow \phi_{\mathrm{M}}=\left(\frac{\mathrm{I}_{\mathrm{N}}}{\mathrm{I}_{\mathrm{M}}}\right) \phi_{\mathrm{N}}=\frac{3}{2} \times 1.8 \times 10^{-3}$

$=2.7 \times 10^{-3} \mathrm{\,Wb}$

$=M \frac{d}{d t}\left(I_{0} \sin \omega t\right)=M I_{0} \omega \cos \omega t$

$\therefore \quad \varepsilon_{\max }=0.005 \times 10 \times 100 \pi \times 1=5 \pi$

Considering it to be uniform, magnetic flux passing through secondary coil is

$\phi_{2}=\mathrm{BA}$

$=\frac{\mu_{0} i_{1}}{2 R_{1}}\left(\pi R^{2}\right)$

$M=\frac{\phi_{2}}{i_{1}}$

$M=\frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}}$

$M \propto \frac{R_{2}^{2}}{R_{1}}$

$M=K \sqrt{L_{1} L_{2}}$

or, $\quad M=1 \sqrt{2 \times 10^{-3} \times 8 \times 10^{-3}} \quad(\because K=1)$

$=4 \times 10^{-3}=4 \mathrm{mH}$

$\qquad=100 \times 10^{-3} \mathrm{V}=0.1 \mathrm{V}$

$e=m \frac{\Delta i}{\Delta t}$

$e=m \cdot\left(\frac{i_{2}-i_{1}}{t_{2}-t_{1}}\right)$

$0.1=m\left(\frac{10-0}{0.1-0}\right)$

$m=\frac{0.1}{100}=10^{-3} H$

$m=1 m H$

${\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\mathrm{M} \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}} $

or  $\left| {\rm{e}} \right| = {\rm{M}}\frac{{{\rm{dI}}}}{{{\rm{dt}}}}$

$\therefore $ $|\mathrm{e}|=5 \times \frac{10}{5 \times 10^{-4}}=1 \times 10^{5} \mathrm{\,V}$