Physics M.C.Q (1 Marks)

$=(2 \hat{k} \times(2 \hat{i}+3 \hat{j}+\hat{k})) \cdot(\hat{i}+2 \hat{j})$

$=(-6 \hat{i}+4 \hat{j}) \cdot(\hat{i}+2 \hat{j})$

$=-6+8=2 V$

$=12$ volt

$B=\frac{\mu_{0} I}{2 \pi r}\left(\text { using Ampere's law, } \int B . d l=\mu_{0} I\right)$

Now, induced emf due to moving rod is $e=B l v=\frac{\mu_{0} I}{2 \pi r} \times l v=\frac{\mu_{0} v I l}{2 \pi r}$

Hence the workdone is zero.

Option $D$ is correct.

$=B .4 x \frac{d x}{d t}$

$e=4 B x v$

$I=\frac{e}{R}=\frac{4 B x v}{4 x r}=\frac{B v}{r}$

or $V=(2)(4)=8$ volt

Now energy stored per unit time

$=$power $=V i$

$=(8)(2)$

$=16 J / s$

Therefore $B$ will be negatively charged and $A$ will be positively charged.

$\varepsilon=\left(\mathrm{B}_{\mathrm{H}} \tan \delta\right) \ell \mathrm{v}$

$\varepsilon=3 \times 10^{-4} \times \frac{4}{3} \times 0.25 \times 10 \times 10^{-2}$

$\varepsilon=10 \times 10^{-6} \mathrm{v}=10\, \mu \mathrm{V}$

$(\overrightarrow {\rm{v}}  \times \overrightarrow {\rm{B}} ) = \left| {\begin{array}{*{20}{c}}
{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\
2&3&1\\
1&2&0
\end{array}} \right| =  - 2\widehat {\rm{i}} + \widehat {\rm{j}} + \widehat {\rm{k}}$

$\overrightarrow \ell   = 2\widehat {\rm{k}}\,{\rm{m}}$

$\therefore $ $\mathrm{e}=(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}) \cdot(2 \hat{\mathrm{k}})=2 \mathrm{\,volt}$

Now $\frac{f_{n}}{f_{n+1}}=\frac{n}{n+1}=\frac{252}{336}$

$\Rightarrow n=3 \quad \therefore f_{1}=\frac{252}{3}=84 \mathrm{Hz}$

current $1=\frac{10-\mathrm{e}}{\mathrm{R}}=\frac{10-0.125 \mathrm{\,V}}{2}$

Force $B1\ell $

$=0.5\left(\frac{10-0.125 \mathrm{V}}{2}\right) 0.25=0.5 \mathrm{\,N}$ (given)

Solving $\mathrm{V}=16 \mathrm{\,m} / \mathrm{\,s}$

$I_{1}=I_{2}=\frac{I}{2}$

$I=2 \frac{B \ell v}{3 R}$

$\mathrm{I}_{1}=\mathrm{I}_{2}=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}$

as $\mathrm{e}=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ (in magnitude)

$\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+10 \mathrm{t}+5\right) \times 10^{-3} \mathrm{\,Wb} \sec ^{-1}$

$=(10 t+10) 10^{-3}$ $\mathrm{\,volt}$

$\therefore \mathrm{e}=(10 \mathrm{x} 5+10) \times 10^{-3}=0.06 \mathrm{\,volt}$

$=0.0048 \mathrm{\,V} $

 = $B\,v\,(3\ell )$
= $3\,B\,v\ell $

$\mathrm{P}=\frac{\mathrm{E}^{2}}{\mathrm{R}}=\frac{\mathrm{B}^{2} \ell^{2} \mathrm{v}^{2}}{\mathrm{R}}$

$P^{\prime}=\frac{B^{2} \ell^{2}(2 v)^{2}}{R}=4 \,P$

$ = 5\hat j \cdot (\hat i \times (3\hat i + 4\hat j + 5\hat k)) = 10\,\,V$

In case $(iii)$ $\mathrm{V}_{2}=\mathrm{B}$ $. bv$ $\quad(\mathrm{b}=2 \mathrm{\,cm})$

$\therefore \frac{\mathrm{V}_{1}}{\mathrm{V}_{2}}=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{Bbv}}=\frac{\ell}{\mathrm{b}}=\frac{8}{2}=4$

Induced current $=0$

As shown in figure $(b)$ rod will cut horizontal component of earth's magnetic field.

Now $\tan \theta=\frac{B}{B_{H}} \Rightarrow B_{H}=B \cot \theta$

induced $\mathrm{emf}$ ${\rm{e}} = {{\rm{B}}_{\rm{H}}}{\rm{v}}\ell  = {\rm{Bv}}\ell \cot \theta $

$\mathrm{B}_{\mathrm{v}}$ only and $\mathrm{B}_{\mathrm{v}}=\mathrm{B} \sin \delta=\mathrm{B} \sin 30^{\circ}$

$=\frac{B}{2}=\frac{1}{2}$

$10^{4}$ Gauss $=1 \mathrm{\,T}$

$\mathrm{e}=\frac{1}{2} \mathrm{B}_{\mathrm{v}} \omega \ell^{2}$

$=\frac{1}{2} \times \frac{1}{2} \times \frac{2 \pi \times 120}{60} \times\left(\frac{1}{2}\right)^{2}$

$=\frac{\pi}{4} \mathrm{\,volt}$

$e=2 \times 1 \times 4 \times \frac{1}{2}=4 \mathrm{\,V}$

$=4 \times 2 \times 1 \times \frac{1}{2}=4$ $\mathrm{volt}$

$I=\frac{B V \ell}{R}$

$I=\frac{1 \times 5 \times 10^{-2} \times 2 \times 10^{-2}}{10}$

$\mathrm{I}=0.1 \mathrm{\,mA}$

$=\theta=60^{\circ} $

$=1 \times 20 \times 10^{-2} \times 10 \times 10^{-2} \times \sqrt{3} / 2 $

$=17 \times 10^{-3} \mathrm{\,V} $

$E = Bv 1 \sin 60^{\circ}$

As we know to take that component of length vector which is perpendicular to the velocity vector

$=1 \times 0.2 \times 0.1 \times \sqrt{3} / 2$

$=17.32 \times 10^{-3}\,V$

$V_{A}>V_{B}$

$\therefore q_{A}$ is positive and $q B$ negative $q=C V=C=(B v l)$

$=\left(20 \times 10^{-6}\right)(0.5)(0.1)$

$=0.2 \times 10^{-6} C=0.2 \mu C$

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

$\left(F_{m}\right)=\frac{B^{2} v l^{2}}{R}=\frac{(0.15)^{2} \times(2) \times(0.5)^{2}}{3}$

$=3.75 \times 10^{-3} N$

$I=2 \frac{B \ell v}{3 R}$

$I_{1}=I_{2}=\frac{B \ell v}{3 R}$

Rod will cut $B_H$

$\mathrm{e}=\mathrm{B}_{\mathrm{H}} v \ell$

$\mathrm{e}=0.3 \times 10^{-4} \times 5 \times 20$

$\mathrm{e}=3\, \mathrm{m}$ $volt$

so $\mathrm{f}=\frac{\mathrm{f}}{\ell}=\mathrm{I} \mathrm{B} \sin \theta$

$f=1 \times 3 \times 10^{-5}$

$=3 \times 10^{-5} \,\mathrm{N} / \mathrm{m}$

Wing spanof jet plane, $l=25 \mathrm{m}$ Earth's magnetic field strength, $\mathrm{B}=5.0 \times 10^{-4} \mathrm{T}$

Angle of dip, $\delta=30^{\circ}$

Vertical component of Earth's magnetic field,

$\mathrm{Bv}=\mathrm{B} \sin \delta$

$=5 \times 10^{-4} \sin 30^{\circ}$

$=2.5 \times 10^{-4} \mathrm{T}$

Voltage difference between the ends of the wing can be calculated as:

$e=\left(B_{V}\right) \times l \times v$

$=2.5 \times 10^{-4} \times 25 \times 500$

$=3.125 \mathrm{V}$

Hence, the voltage difference developed between the ends of the wings is $3.125 \mathrm{V}$

Area swept out in one rotation $=\pi\left(\frac{\ell}{2}\right)^{2}=\left(\frac{\pi \ell^{2}}{4}\right)$

Angular velocity $=\omega$ $rad$ $s^{-1}$

Frequency of revolution $=\frac{\omega}{2 \pi}$

Area swept out per second $=\frac{\pi \ell^{2}}{4}\left(\frac{\omega}{2 \pi}\right)=\frac{\ell^{2} \omega}{8}$

Magnetic induction $=\mathrm{B}$

Rate of change of magnetic flux $=\left(\frac{\mathrm{B} \ell^{2} \omega}{8}\right)$

Magnitude of induced $\mathrm{emf}$ $=\frac{\mathrm{B} \ell^{2} \omega}{8}$

Magnitude of induced $\mathrm{emf}$ between the axis and the other end is also $\left(\frac{\mathrm{B} \ell^{2} \omega}{8}\right) .$ These two $\mathrm{emf's}$ are in opposite directions. Hence, the potential difference between the two ends of the rod is zero.

$F=\frac{\ell^{2} B^{2} v}{R} ;$ where $V=2 \,m / \sec$

$B=.15 \,T$

$C=50 \mathrm{cm}=.5$ $\mathrm{meter}$

$F=3.75 \times 10^{-3} \,N$

Hence total charge induced in the conducting loop depend upon the total change in magnetic flux and resistance.