JEE physics MCQ

$\tau=N I \times A \times B$

$105=500 \times 3 \times 10^{-4} \times \frac{1}{2} \times B$

$B=20$

$N =1$

For $ABCD$

$\overrightarrow{ M }_{1}= abI \hat{ K }$

For $DEFA$

$\overrightarrow{ M }_{2}= abI \hat{ j }$

$\overrightarrow{ M }=\overrightarrow{ M }_{1}+\overrightarrow{ M }_{2}$

$=\operatorname{ab} I (\hat{ k }+\hat{j})$

$=\operatorname{ab} I \sqrt{2}\left(\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{ k }}{\sqrt{2}}\right)$

$\mathrm{I} \alpha=-\mathrm{MB} \sin \theta$

for small $\theta$

$\alpha=-\frac{\mathrm{MB}}{\mathrm{I}} \theta$

$\omega=\sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}=\sqrt{\frac{(\mathrm{I})\left(\pi \mathrm{R}^{2}\right) \mathrm{B}}{\left(\frac{\mathrm{mR}^{2}}{2}\right)}}$

$\omega=\sqrt{\frac{2 \mathrm{I} \pi \mathrm{B}}{\mathrm{m}}}$

$\therefore \mathrm{T}=\frac{2 \pi}{\omega}=\sqrt{\frac{2 \pi \mathrm{m}}{\mathrm{IB}}}$

$\therefore \quad \mathrm{I}_{g} \mathrm{G}=\left(\mathrm{I}_{0}-\mathrm{I}_{g}\right) \mathrm{R}_{\mathrm{A}}$

$\therefore \quad R_{A}=\left(\frac{I_{g}}{I_{0}-I_{g}}\right) G$

When galvanometer is used as a voltmeter, resistance is used in series with galvanometer.

$\mathrm{I}_{g}\left(\mathrm{G}+\mathrm{R}_{\mathrm{v}}\right)=\mathrm{V}=\mathrm{GI}_{0}\left(\text { given } \mathrm{V}=\mathrm{GI}_{0}\right)$

$\therefore R_{v}=\frac{\left(I_{0}-I_{g}\right) G}{I_{g}}$

$\therefore \quad {{\text{R}}_{\text{A}}}{{\text{R}}_{\text{v}}} = {{\text{G}}^2}\& \frac{{{{\text{R}}_{\text{A}}}}}{{{{\text{R}}_{\text{v}}}}} = {\left( {\frac{{{{\text{I}}_g}}}{{{{\text{I}}_0} - {{\text{I}}_g}}}} \right)^2}$

$\mathrm{v}_{1}=\frac{2}{10^{-3}}=100+\mathrm{R}_{1}$

$1900=\mathrm{R}_{1}$

$\mathrm{V}_{2}=\frac{10}{10^{-3}}=\left(2000+\mathrm{R}_{2}\right)$

$\mathrm{R}_{2}=8000$

$\mathrm{V}_{3}=\frac{20}{10^{-3}}=10 \times 10^{3}+\mathrm{R}_{3}=10 \times 10^{3} \mathrm{R}_{3}$

$\mathrm{C} \theta=\mathrm{i} \mathrm{N} \mathrm{A} \mathrm{B}$

$10^{-6} \times \frac{\pi}{180}=10^{-3} \times 10^{-4} \times 175 \times B$

$B=10^{-3}$ $Tesla$

$(0.002)\left(R_{G}\right)=(0.5-0.002)\left(r_{s}\right)$

$\Rightarrow r_{s} \approx 0.2\, \Omega$

$\mathrm{S}=5000 \,\Omega$

$\mathrm{I}_{\mathrm{g}}=4 \times 10^{-3}$

$\mathrm{V}=\mathrm{i}_{g}(\mathrm{G}+\mathrm{S})$

$V=4 \times 10^{-3}(50+5000)$

$=4 \times 10^{-3}(5050)=20.2$ $volt$

${{\text{i}}_{{g_0}}} = 4 \times {10^{ - 4}} \times 25 = {10^{ - 2}}{\mkern 1mu} {\text{A}}$

$\mathrm{V}_{0}=2.5\, \mathrm{V}$

$R_{0}+R=\frac{V_{0}}{i_{0}}=\frac{2.5}{10^{-2}}=250$

$\Rightarrow \quad \mathrm{R}=200\, \Omega$

$\mathrm{N}_{\mathrm{L}}=\mathrm{N}_{\mathrm{g}}=\mathrm{N}=30$

$\mathrm{FOM}=\frac{1}{\phi}=0.005 \,\mathrm{A/Div}$

Current sentivity $=\mathrm{CS}=\left(\frac{1}{0.005}\right)=\frac{\phi}{1}$

$I{g_{\max }} = 0.005 \times 30$

$=15 \times 10^{-2}=0.15$

$15=0.15[20+\mathrm{R}]$

$100=20+R$

$R=80$.

${{{\text{m}}^\prime } = \frac{{I4{\ell ^2}}}{\pi }}$      ${{\text{r}} = \frac{{2\ell }}{\pi }}$

${{{\text{m}}^\prime } = \frac{{I4{\ell ^2}}}{\pi }}$      ${\pi {r^2} = \frac{{\pi 4{\ell ^2}}}{{{\pi ^2}}} = \frac{{4{\ell ^2}}}{\pi }}$  

${{{\text{m}}^\prime } = \frac{4}{\pi }\,{\text{m}}}$

$\tau=\mathrm{NI} \times \mathrm{A} \times \mathrm{B} \times \sin 45^{\circ}$

$\tau=0.27\, \mathrm{Nm}$

$=\mathrm{Nl}\left(\pi \mathrm{r}^{2}\right) \hat{\mathrm{j}} $

Torque on loop (coil) $=\overline{\mathrm{M}} \times \overline{\mathrm{B}}$

$=\mathrm{Nl}\left(\pi \mathrm{r}^{2}\right) \mathrm{B} \sin 90^{\circ}(-\hat{\mathrm{k}})$

$=\mathrm{NI} \pi \mathrm{r}^{2} \mathrm{B}(-\hat{\mathrm{k}})$

$=\left(\frac{\mathrm{d} \mathrm{q} \omega}{2 \pi}\right) \pi \mathrm{x}^{2} $

$=(\rho \mathrm{d} \mathrm{x}) \frac{\omega}{2 \pi} \pi \mathrm{x}^{2} $

$\mathrm{M} =\int_{0}^{\mathrm{L}} \mathrm{dM}$

Current through shunt $\left(I-I_{g}\right)=2\, A$

Galvanometer resistance $\mathrm{R}_{\mathrm{g}}=25\, \Omega$

Resistance of shunt, $\mathrm{S}=?$

$\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$

$ \Rightarrow S = \frac{{{{10}^{ - 3}} \times 25}}{2}$

$S \simeq 1.25 \times {10^{ - 2}}\,\Omega $

$\mathrm{I}=\frac{\varepsilon}{\mathrm{R}+\mathrm{G}} \quad \mathrm{G}=\frac{1}{9}\, \mathrm{K}\, \Omega$

$\frac{1}{2}=\frac{\varepsilon}{R+\frac{G S}{G+\bar{S}}} \times \frac{S}{S+G} \Rightarrow \frac{1}{2}=\frac{\varepsilon S}{R(S+G)+G S}$

$S=\frac{R G \times \frac{1}{2}}{\varepsilon-\frac{(R+G) I}{2}}$

$S=\frac{11 \times 10^{3} \times \frac{1}{2} \times 10^{2} \times 270 \times 10^{-6}}{6-\left(\frac{6}{2}\right)}=110\, \Omega$

or,  $\mu \,\, = \,\,\frac{{qr\omega }}{{2\pi r}}(\pi {r^2})\, = \,\frac{1}{2}\,q{r^2}\omega $

Dipole moment of circular loop is $m=I A$

$\mathrm{m}_{1}=\mathrm{I} \cdot \mathrm{A}=\mathrm{I} \pi \mathrm{R}^{2} \quad\{\mathrm{R}=\text { Radius of the loop }\}$

If moment is doubled (keeping current constant) $R$ becomes $\sqrt{2 \mathrm{R}}$

${{\text{m}}_2} = {\text{L}}\pi {(\sqrt 2 {\text{R}})^2} = 2.{\text{I}}\pi {{\text{R}}^2} = 2{{\text{m}}_1}$

$B_{2}=\frac{\mu_{0} I}{2(\sqrt{2} R)}$

Galvanometer resistance, $G=15\, \Omega$

Let resistance $R$ to be put in series with the galvanometer to convert it into a voltmeter.

$V=i_{g}(R+G)$

$10= 5 \times 10^{-3}(R+15) $

$ \therefore \quad  R=2000-15=1985 $

$=1.985 \times 10^{3} \,\Omega $

magnetic moment of current carrying coil is a vector and its direction is given by right hand thumbrule, for rectangular loop, $\overrightarrow{B}$ at centre due to current in loop and $\overrightarrow{\mathrm{M}}$ are always parallel.

Hence, $(c)$ corresponds to stable equilibrium. 

In case $1,$ when $\mathrm{R}=2400 \Omega$ and deflection of 40 divisions present.

$\therefore \quad \frac{40}{50} \mathrm{I}=\frac{V}{G+R}$

$\Rightarrow \quad \frac{4}{5} \mathrm{I}=\frac{2}{G+2400}\dots (1)$

In case $2,$ when $\mathrm{R}=4900 \Omega$ and deflection of 20 divisions present

$\therefore \quad \frac{20}{50} \mathrm{I}=\frac{V}{G+R}$

$\Rightarrow \quad \frac{2}{5} \mathrm{I}=\frac{2}{G+4900}\dots (2)$

From $(1)$ and $(2)$ we get,

$\frac{4}{2}=\frac{G+4900}{G+2400}$

$\Rightarrow 2 G+4800=G+4900$

$\Rightarrow \mathrm{G}=100 \Omega$

Putting value of G in equation (1), we get.

$\frac{4}{5} \mathrm{I}=\frac{2}{100+2400}$

$\Rightarrow \mathrm{I}=1 \mathrm{mA}$

Current sensitivity $=\frac{\mathrm{I}}{\text { number of divisions }}$

$=\frac{1}{50}$

$=0.02 \mathrm{mA} /$ division

$=20 \mu \mathrm{A} /$ division

Resistance required for deflection of 10 divisions

$\frac{10}{50} \mathrm{I}=\frac{V}{G+R}$

$\Rightarrow \frac{1}{5} \times 1 \times 10^{-3}=\frac{2}{100+R}$

$\Rightarrow \mathrm{R}=9900 \Omega$

$I' = 0.099$

When Galvanometer is connected

$R_{e q}=50+\frac{100 S}{100+S}=\frac{V}{I}$

$\Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=\frac{5}{0.099}-50$

$\Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=50.50-50 \Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=0.5$

$\Rightarrow 100 \mathrm{S}=50+0.55 \Rightarrow 99.5 \mathrm{S}=50$

$S=\frac{50}{99.05}=0.5 \,\Omega$

So, shunt of resistance $=0.5\, \Omega$ is onnected in parallel with the galvanometer.

$\mathrm{I}_{\mathrm{g}}=\frac{\mathrm{V}}{\mathrm{R}+\mathrm{G}}$

where, $I $$_{g}$ - Galvanometer current, $G-$Galvonometer resistance

When shunt of resistance $S$ is connected parallel to the Galvanometer then

$\mathrm{G}=\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}$

$\therefore I=\frac{V}{R+\frac{G S}{G+S}}$

Equal potential difference is given by

$\mathrm{I}_{\mathrm{g}}^{\prime} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{B}}^{\prime}\right) \mathrm{S}$

$\mathrm{I}_{\mathrm{g}}^{\prime}(\mathrm{G}+\mathrm{S})=\mathrm{IS}$

$\Rightarrow \frac{\mathrm{I}_{\mathrm{g}}}{2}=\frac{\mathrm{IS}}{\mathrm{G}+\mathrm{S}}$

$\Rightarrow \frac{\mathrm{v}}{2(\mathrm{R}+\mathrm{G})}=\frac{\mathrm{v}}{\mathrm{R}+\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}} \times \frac{\mathrm{S}}{\mathrm{G}+\mathrm{S}}$

$ \Rightarrow \frac{1}{{2(R + G)}} = \frac{S}{{R(G + S) + GS}}$

$\Rightarrow R(G+S)+G S=2 S(R+G)$

$\Rightarrow \mathrm{RG}+\mathrm{RS}+\mathrm{GS}=2 \mathrm{S}(\mathrm{R}+\mathrm{G})$

$\Rightarrow \mathrm{RG}=2 \mathrm{S}(\mathrm{R}+\mathrm{G})-\mathrm{S}(\mathrm{R}+\mathrm{G})$

$\therefore \mathrm{RG}=\mathrm{S}(\mathrm{R}+\mathrm{G})$

$\therefore 10^{-3} \times 100=\left(10-10^{-3}\right) \times \mathrm{S}$

$\therefore \mathrm{S} \approx 0.01 \,\Omega$

$W=\int_{0}^{2} F d x$

$=\int_{0}^{2} 3.0 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x$

$=9 \times 10^{-3} \int_{0}^{2} e^{-0.2 x} d x$

$=\frac{9 \times 10^{-3}}{0.2}\left[-e^{-0.2 \times 2}+1\right] B=3.0 \times 10^{-4} e^{-0.2 x}$

(By exponential function)

$=\frac{9 \times 10^{-3}}{0.2} \times\left[1-e^{-0.4}\right]$

$=9 \times 10^{-3} \times(0.33)=2.97 \times 10^{-3}\, \mathrm{J}$

Power required to move the conductor is, $P=\frac{W}{t}$

$P=\frac{2.97 \times 10^{-3}}{(0.2) \times 5 \times 10^{-3}}=2.97 \,\mathrm{W}$

$ \text { As } I_{8} =\frac{1 S}{G+S} $

$=\frac{5.5 \times 1}{120+1}=0.045 \text { ampere } $

For ammeter, shunt resistance, $\mathrm{S}=\frac{\mathrm{IgG}}{\mathrm{I}-\mathrm{Ig}}$

Therefore for $I$ to increase, $S$ should decrease, So additional $S$ can be connected across it.

$\text { in equilibrium } \tau_{\text {net }}=0$

$\tau_0= MB \sin (90-\theta)- mg \cdot r \sin \theta=0$

$\text { I. } \pi r ^2 \cdot B _0 \cos \theta= mg r \cdot \sin \theta$

$\tan \theta=\frac{\pi rIB }{ mg }$

$\frac{ R _1}{ R _2}=\frac{\int_0^{0.5} \frac{\rho d x}{\pi r _x^2}}{\int_{0.5}^1 \frac{\rho dx }{\pi r _x^2}}$

$\frac{ R _1}{ R _2}=\frac{+\left[\frac{1}{ r _x}\right]_0^{0.5}}{+\left[\frac{1}{ r _{ x }}\right]_{0.5}^1}$

$\therefore R _1=5 R _2=5 \Omega$

$\Rightarrow i =\frac{\mu_0 m \pi a ^2}{2 \pi r ^3 L }$

$\Rightarrow i \propto \frac{ m }{ r ^3}$

$m ^{\prime}=\pi a ^2 i =\frac{\mu_0 m ^2 a ^4}{2 \pi r ^3 L }$

$F =\frac{ km ^2 \pi^2 a ^4}{2 \pi r ^7 L }$

$W =\int Fdr \propto \int \frac{ m ^2 dr }{ r ^7}$

$W \propto \frac{ m ^2}{ r ^6}$

$\text { where } \quad \vec{M} \text { is magnetic moment }$

$\vec{B} \text { is magnetic field }$

$\therefore \tau=i \pi R ^2 N B B _0 \text { [at the instant shown } \theta=\pi / 2 \text { ] }$

$\therefore \vec{\tau} dt = d \overrightarrow{ L }= i \pi R ^2 NB _0 dt = Q \pi R ^2 N B _0[ idt = Q ]$

$0.1=2 \times 10^{-6}\left(10+R_V\right)$

$\therefore R_V=49990 \Omega$

(image)

$2 \times 10^{-6} \times 10=10^{-3} R _{ A } \therefore R _{ A }=0.02 \Omega$

(image)

$y 50000=(x-y) 1000$

$\therefore 51 y=x$

Reading $=\frac{y 50000}{x} \approx 980$

Area $=2 \times 10^{-4} m ^2$

$B=0.02 T$

Torsional constant $C =10^{-4}$

At full scale deflection $\theta=0.2 rad$

Resistance $R =50 \Omega$

Torque $\tau= C \theta=$ niAB

Full scale current $i =\frac{ C \theta}{ nAB }=\frac{0.2 \times 10^{-4}}{50 \times 2 \times 10^{-4} \times 0.02}$

$i =0.1 amp$

$0.9 \times S =0.1 \times R$

$S =\frac{ R }{9}=\frac{50}{9} \Omega$

$S =5.56 \Omega$

$\Rightarrow G+4990=\frac{30,000}{6}=5000 $

$\Rightarrow G=10 $

$\frac{6}{1000} \times 10=\left(1.5-\frac{6}{1000}\right) S $

$\Rightarrow S =\frac{60}{1494}=\frac{2 n}{249} $

$\Rightarrow=\frac{249 \times 30}{1494}=\frac{2490}{498}=5$

$\vec{B}_1=\vec{B}$ due to wire $-1$

$\vec{B}_2=\vec{B}$ due to wire $-2$

In magnitudes $B _1= B _2=\frac{\mu_0 I }{2 \pi r }$

Resultant of $B _1$ and $B _2=2 B _1 \cos \theta=\frac{\mu_0 Ia }{\pi \pi^2}$

$B _{ R }=\frac{2 \mu_0 I a ^2}{4 \pi r ^3}$

For zero magnetic field at $P$

$\frac{\mu_0 Ia }{\pi r ^2}=\frac{2 \mu I \pi a ^2}{4 \pi r ^3} $

$\Rightarrow h \approx 1.2 a$

$Image$

$2.$ Magnetic field at mid point of two wires $=\frac{\mu_0 I }{\pi d } \otimes $

Magnetic moment of loop $= I a ^2 $

Torque on loop $= MB \sin 150^{\circ} $

$\qquad=\frac{\mu_0 I ^2 a ^2}{2 d }$

$=a^2+\frac{\pi a^2}{2} $

$A=\left(1+\frac{\pi}{2}\right) a^2 \hat{k}$

$R =0.005 \Omega, r _2=0.1 \ m$

$I = I _0 \cos (300 t )$

So we get $\omega=300$

The magnetic field of the tube, $B=\frac{\mu_0 NI }{ L }$

In the problem $N =1, \quad L=10 m, \quad I = I _0 \cos (300 t$ )

So, $B=\frac{\mu_0 I_0 \cos (300 t)}{10}$

$Q=\frac{\mu_0 I_0 \cos (300 t)}{10} \times \pi(0.1)^2$

$\text { Induced emf, } e=-\frac{d Q}{d t}=300 \mu_0 \pi I_0 \sin (300 t) \times 10^{-3}$

$e=0.3 \mu_0 \pi I_0 \sin (300 t)$

The current in the ring $i =\frac{ e }{ R }$

$i =\frac{0.3 \pi \mu_0 I _0 \sin (300 t)}{0.005}$

$\text { or } i =60 \mu_0 \pi I _0 \sin (300 t)$

$\text { Magnetic moment, } M = iA = i \pi r ^2$

$M =60 \mu_0 \pi I _0 \sin (300 t) \times \pi(0.1)^2$

$=0.6 \mu_0 \pi^2 I _0 \sin (300 t) A$

$=\left(0.6 \times \pi^2\right) \mu_0 I _0 \sin (300 t)$

$=6 \mu_0 I _0 \sin (300 t)$

So, the value of $N =6$

$ \Rightarrow $ $I =$ $\left( {1 + \frac{G}{S}} \right){I_G}$

$ \Rightarrow $ $ I = 100.1\, mA$

Magnetic moment $M = iA = \frac{1}{2}q\omega {r^2}$

Angular moment $L = I\omega = m{r^2}\omega $ $==>$ $\frac{M}{L} = \frac{q}{{2m}}$

Resistance of the ammeter is

$R_{A}=\frac{(480\, \Omega)(20\, \Omega)}{(480\, \Omega+20\, \Omega)}=19.2\, \Omega$

(As $480 \,\Omega$ and $20\, \Omega$ are in parallel)

As ammeter is in series with $40.8\, \Omega$,

$\therefore \quad$ Total resistance of the circuit is

$R=40.8\, \Omega+R_{A}=40.8\, \Omega+19.2\, \Omega=60\, \Omega$

By Ohm's law,

Current in the circuit is

$I=\frac{V}{R}=\frac{30\, \mathrm{V}}{60\, \Omega}=\frac{1}{2} \mathrm{A}=0.5\, \mathrm{A}$

Thus the reading in the ammeter will be $0.5\, \mathrm{A}$.

where $N$ is the number of turns in the coil, $I$ is the current through the coil, $B$ is the uniform magnetic field, $A$ is the area of the coil and $\theta$ is the angle between the direction of the magnetic field and nomal to the plane of the coil. Here, $N=50, I=2\, \mathrm{A},$

$A=0.12\, \mathrm{m} \times 0.1\, \mathrm{m}=0.012\, \mathrm{m}^{2}$

$B=0.2\, \mathrm{Wb} / \mathrm{m}^{2} \text { and } \theta=90^{\circ}-30^{\circ}=60^{\circ}$

$\therefore \quad \tau =(50)(2\, \mathrm{A})\left(0.012\, \mathrm{m}^{2}\right)\left(0.2\, \mathrm{Wb} / \mathrm{m}^{2}\right) \sin 60^{\circ} $

$=0.20\, \mathrm{N} \mathrm{m}$