Physics M.C.Q (1 Marks)

$ \Rightarrow |e| = A\alpha $

$\therefore \phi \propto {r^2} \Rightarrow \phi = k{r^2}$      ( $k = constant$)

$\therefore e = \frac{{d\phi }}{{dt}} = k.2r\frac{{dr}}{{dt}}$

From $0 -1$, $r$ is constant,   $\frac{{dr}}{{dt}} = 0$ hence, $e = 0$

From $1 -2$, $r = \alpha t,$  $\frac{{dr}}{{dt}} = \alpha $ hence $e \propto r$==> $e \propto t$ 

From $2 -3$, again $r$ is constant,   $\frac{{dr}}{{dt}} = 0$ hence $e = 0$

$ = - \frac{{NBA}}{R}(\cos {180^o} - \cos {0^o})$ ==> $B = \frac{{QR}}{{2NA}}$

$A=\pi R^{2}=\pi\left(R_{0}+t\right)^{2}, \frac{d A}{d t}=2 \pi\left(R_{0}+t\right)$

therefore, emf induced in the loop is

$e=B \frac{d A}{d t}=2 \pi\left(R_{0}+t\right) B$

since area is increasing, induced emf is in anticlockwise sense to oppose the chang in flux.

$\delta t=\frac{\phi_{i}-\phi_{f}}{e}=\frac{B A-0}{e}=\frac{20 \times 10^{-2}}{10}$

$=20 \times 10^{-3} \mathrm{sec}=20 \mathrm{ms}$

$\Rightarrow \Delta q=\frac{\Delta(f l u x)}{R}=\frac{2 B A}{R}$

$\phi=\overrightarrow{ B } \cdot \overrightarrow{ A }= BA \cos 90^{\circ}=0$

By Lenz's law this field is to oppose the movement of the magnet towards the ring.

Therefore the $N$-pole should be facing the ring, as only then will the induced field by the ring will oppose the movement of the approaching magnet. Therefore, the north pole faces the ring and the magnet moves towards it and the south pole faces the ring and the magnet moves away from it.

$E=\frac{-d \phi}{d t}$

$E=\frac{-d}{d t}(B A)=-B \frac{d}{d t}(A)-A \frac{d}{d t}(B)$

either magnetic field or area must be changed w.r.t. time to generate emf, and here the magnetic field and plane of ring are parallel thus no emf induced in either situation.

${\mathrm{E}=-\mathrm{d} \phi / \mathrm{dt}} $

${=-\frac{\mathrm{d}}{\mathrm{dt}}\left(3 \mathrm{t}^{2}+2 \mathrm{t}+3\right) \times 10^{-3}}$

(because given flux is in $\mathrm{mWb}$ ). Thus

$E=(-6 t-2) \times 10^{-3}$

at $t=2$ $sec,$

${\mathrm{E}=(-6 \times 2-2) \times 10^{-3}} $

${=-14 \mathrm{\,mV}}$

Frequency $v=\frac{1}{T}=\frac{\omega}{2 \pi}$

$\mathrm{So}, \in=\frac{B \times \pi l^{2} \times \omega}{2 \pi}$

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

$=+\frac{800 \times 0.05 \times 4 \times 10^{-5}}{0.1}=1600 \times 10^{-5}=0.016 \mathrm{\,V}$

$\left(\frac{d \varphi}{d t}\right)_{\text {relative velocity2v }}=2\left(\frac{d \varphi}{d t}\right)_{\text {1case }}=2 e$

$A=20 \mathrm{\,cm}^{2}=20 \times 10^{-4} \mathrm{\,m}^{2}$

$\mathrm{N}=400, \mathrm{I}_{1}=2 \mathrm{A}, \mathrm{I}_{2}=0, \mathrm{dt}=10^{-3} \mathrm{\,s}$

As $\varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{BAN})}{\mathrm{dt}}$

$=\frac{\mu_{0} \mathrm{Nd} \mathrm{I} \mathrm{AN}}{\ell \mathrm{dt}} \quad\left(\because \mathrm{B}=\frac{\mu_{0} \mathrm{NI}}{\ell}\right)$

$=\frac{\mu_{0} \mathrm{N}\left(\mathrm{I}_{1}-\mathrm{I}_{2}\right) \mathrm{AN}}{(\mathrm{dt}}$

$=\frac{4 \pi \times 10^{-7} \times(400)^{2} \times 2 \times 20 \times 10^{-4}}{20 \times 10^{-2} \times 10^{-3}}=4 \mathrm{\,V}$

$\frac{d r}{d t}=r_0 \Rightarrow \phi=B \pi r^2$

$\varepsilon=\frac{-d \phi}{d t}=-B \pi(2 r) \frac{d r}{d t}$

$\varepsilon=-2 \pi B r r_0$

$\phi=a t(T-t)=a T t-a t^2$

$\varepsilon=-\frac{d \phi}{d t}=-(a T-2 a t)$

$P=-\frac{\varepsilon^2}{R}=\frac{(a T-2 a t)^2}{R}$

$H=\int \limits_0^T P d t=\int \limits_0^T \frac{(a T-2 a t)^2}{R}$

$=\frac{a^2 T^3}{3 R}$

$\varepsilon=n B \pi r^2 \frac{d B}{d t}$

$\varepsilon^{\prime}=4 n B \pi\left(\frac{r}{2}\right)^2 \frac{d B}{d t}$

$\varepsilon^{\prime}=n B \pi r^2 \frac{d B}{d t}=\varepsilon$

$P^{\prime}=\frac{\left(\varepsilon^{\prime}\right)^2}{R}=\frac{\varepsilon^2}{R}$

Intensity of bulb remains the same because source is $D C$, so steady state current will be independent of the inductance of the inductor for $D C$ circuit,

$i_{\text {stoady }}=\frac{E_{\text {source }}}{R_{\text {bubb }}}$

Taking on element as thrown

we write $d \varepsilon=B l v \quad(\varepsilon H F)$

Now $l=d x v=w x$

$\int d \varepsilon=B w \int_{0}^{r} x d x$

$\varepsilon=\frac{B w r^{2}}{2}$

$ = Blv = {B_H}lv$ $ = 3 \times {10^{ - 5}} \times 2 \times 50 = 30 \times {10^{ - 3}}volt = 3\,millivolt$
.

By Fleming's right hand rule, end $A$ becomes positively charged.

Magnetic force acting on the wire ${F_m} = Bil = B\left( {\frac{{Bvl}}{R}} \right)\;l$

$ \Rightarrow {F_m} = \frac{{{B^2}v{l^2}}}{R}$ External force needed to move the rod with constant velocity

$({F_m}) = \frac{{{B^2}v{l^2}}}{R} = \frac{{{{(0.15)}^2} \times (2) \times {{(0.5)}^2}}}{3}$$ = 3.75 \times {10^{ - 3}}N$

For part $AO$ ; ${e_{OA}} = {e_O} - {e_A} = \frac{1}{2}B{l^2}\omega $

For part $OC$; ${e_{OC}} = {e_O} - {e_C} = \frac{1}{2}B{(3l)^2}\omega $
$\therefore {e_A} - {e_C} = 4\;B{l^2}\omega $