Step down transformer:

Principle: When the current flowing through the primary coil changes, an emf is induced in the secondary coil due to the change in magnetic flux linked with it i.e., it works on the principle of mutual induction. For step down transformer,
N_s < N_p, hence \({\epsilon}_{s}<{\epsilon}_{p}\).
\({ { {\epsilon}_{s} }\over{ {\epsilon}_{p} } }={{{N}_{s}}\over{{N}_{p}}}\)
For an ideal transformer, P_in = P_out
\({\epsilon }_{p }{ I}_{p }={ \epsilon}_{s }{I }_{s }\Rightarrow{{{I}_{p}}\over{{I}_{s}}}={{{\epsilon}_{s}}\over{{\epsilon}_{p}}}={{{N}_{s}}\over{{N}_{p}}}\)
P_in = P_out = 550 W = \({\epsilon}_{p}{I}_{p}\) = 550
\(220\times{I}_{p}=550\Rightarrow{I}_{p}={{550}\over{220}}={{5}\over{2}}=2.5 \ A\)

(a) Student B had concentrated in the class room teaching and also had studied again to remember what was taught.
(b) H = I²Rt

(a) Electric flux is defined as the number of electric field lines passing through an area normal to the surface.
Alternatively
Surface integral of the electric field is defined as the electric flux through a closed surface
\(\phi= \oint \overrightarrow{E}. \overrightarrow{ds}\)
SI unit :  \(\frac{N.m^2}{C}\) or volt. metre
(a)


Outward flux through the Gaussian surface, is
\(2EA =\sigma A/\epsilon_0\)
\(\therefore\)  \(E =\sigma A/2\epsilon_0\)
Vectorically \(\overrightarrow{E} = \frac{\sigma}{2\epsilon_0}\hat{n}\),
where \(\hat{n}\) is a unit vector normal to the plane, away from it.  
Hence, electric field is independent of the distance from the sheet.
(c) For positively charged sheet  \(\longrightarrow\) away from the sheet  
For negatively charged sheet \(\longrightarrow\) towards the plane sheet  

Let R be the resistance of voltmeter. Let n be the number of divisions in the voltmeter. The voltage (V) recorded by each division of voltmeter when current i_g flows through it is
i_g R/n = V     .......(i)
When resistance is connected in series with voltmeter, then
 i_g(R + 1980)/n = 100 V  ...(ii) 
Dividing (ii) by (i), we get
R + 1980 = 100 R or R = 1980/99 = 20 \(\Omega \)

Capacitor of 10\(\mu\)F can withstand only 50 V, therefore to be connected across a 200 V battery, four capacitors must be connected in series in a row. Capacitor C₁ of each row of four capacitors is
\({1\over C_1}={1\over 10}+{1\over 10}+{1\over10}+{1\over 10}={4\over 10}\)
\(C_1={10\over 4}=2.5\mu F\)
For a total capacity of 10\(\mu F\), four such rows of capacitors must be connected in parallel so that
\(C_p=4C_1=4\times 2.5=10\mu F\)
Hence, we need 16 capacitors with 4 capacitors in series in each row and 4 such rows in parallel.

\(Here, \ R=200 \ ohm, \ C=15.0\mu F=15\times { 10 }^{ -6 }F\)
\({ E }_{ v }=220V, \ v=50Hz, \ { I }_{ v }=? \ { V }_{ R }=?, \ { V }_{ C }=?\)
\( Now \ { X }_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } =\frac { 1 }{ 2\times 3.14\times 50\times 15\times { 10 }^{ -6 } } =212.2\Omega \)
\((a) \ Impedance \ of \ the \ circuit,\)
\(Z=\sqrt { { R }^{ 2 }+{ X }_{ C }^{ 2 } } =\sqrt { { 200 }^{ 2 }+\left( 212.3 \right) ^{ 2 } } =291.7 \ ohm\)
\( \therefore \ \ Current \ in \ the \ circuit, \ { I }_{ v }=\frac { { E }_{ v } }{ Z } =\frac { 220 }{ 291.7 } =0.75A\)
\( (b) \ { V }_{ R }={ I }_{ v }\times R=0.75\times 200=150.8 \ V\)
\({ V }_{ C }={ I }_{ v }{ X }_{ C }=0.75\times 212.3=159.2 \ V\)
\( { V }_{ R }+{ V }_{ C }=150.8+159.2=310V, \ which \ is \ more \ than \ the \ source \ voltage \ of \ 220 \ V.\)
This paradox is resolved by the fact that the two voltage are not in same phase. Therefore, they cannot be added like ordinary numbers. As V_R and V_C are out of phase by \({ 90 }^{ \circ }\), therefore
\({ V }_{ RC }=\sqrt { { V }_{ R }^{ 2 }+{ V }_{ C }^{ 2 } } =\sqrt { \left( 150.8 \right) ^{ 2 }+\left( 159.2 \right) ^{ 2 } }\)
\(=220V,\ the\ source\ voltage\)

\(Here, \ { E }_{ v }=60V, \ P=10W\)
\(R=\frac { { E }_{ v }^{ 2 } }{ P } =\frac { 60\times 60 }{ 10 } =360\Omega \)
\(Current \ through \ the \ lamp,\)
\({ I }_{ v }=\frac { { E }_{ v } }{ R } =\frac { 60 }{ 360 } =\frac { 1 }{ 6 } A\)
\(Let \ L \ be \ the \ inductance \ of \ the \ choke \ required.\)
\( \therefore \ \ Z=\frac { { E }_{ v } }{ { I }_{ v } } =\frac { 100 }{ 1/6 } =600\Omega \)
\( From \ \ { R }^{ 2 }+{ X }_{ L }^{ 2 }={ Z }^{ 2 }\)
\({ X }_{ L }=\sqrt { { Z }^{ 2 }-{ R }^{ 2 } } =\sqrt { { \left( 600 \right) }^{ 2 }-{ \left( 360 \right) }^{ 2 } } =480\Omega \)
\({ X }_{ L }=\omega L=2\pi vL=480\)
\(L=\frac { 480 }{ 2\pi v } =\frac { 480 }{ 2\times 3.14\times 60 } \)
\(L=1.274 \ H\)
\(To \ achive \ the \ same \ result \ resistance \ required\)
\({ R }^{ ' }=Z-R=600-360=240\Omega \)

Here, \({ I }_{ D }=2.0A, \ C=1\mu F={ 10 }^{ -6 }F.\)
We know, \({ I }_{ D }={ \epsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }\frac { d }{ dt } \left( EA \right) \)
\(={ \epsilon }_{ 0 }A\frac { dE }{ dt } ={ \epsilon }_{ 0 }A\frac { d }{ dt } \left( \frac { V }{ d } \right) ={ \epsilon }_{ 0 }\frac { A }{ d } \frac { dV }{ dt } =C\frac { dV }{ dt } \)
\(\left( \because E=\frac { V }{ d } \right) and \ \left( C=\frac { { \epsilon }_{ 0 }A }{ d } \right) \)
\(or \ \frac { dV }{ dt } =\frac { { I }_{ D } }{ D } =\frac { 2.0 }{ { 10 }^{ -6 } } =2\times { 10 }^{ -6 }V{ s }^{ -1 }\)
Thus a displacement current of 2.0 A can be set up by changing the potential difference across the parallel plates of capacitor at the rate of 2 x 10⁶ Vs^-1 .

Given 
\(\lambda _{ m }=1\mu m=10^{ -6 }\) 
\(b \ =0.2a \ cm \ k\)
\(=0.29\times 10^{ -2 }mK\)
Since \(\lambda _{ m }T=b\) 
\(\therefore \)  \(T=\frac { b }{ \lambda _{ m } } \) 
 \(=\frac { 0.29\times 10^{ -2 } }{ 10^{ -6 } } =2900 \ K\)

When key is open, then
\( R_{e q}=(12+6) \|(6+12)=9 \Omega \)
\(\therefore I_{1}=\frac{V}{9} \)
When key is closed, then
\( R_{e q \boldsymbol{x}} =(12 \| 6)+(6 \| 12)=8 \Omega \)
\( I_{2} =\frac{V}{8} \)
\(\therefore \ \frac{I_{2}}{I_{1}} =\frac{V}{8} \times \frac{9}{V} \)
\(I_{2}=\frac{9}{8} I_{1}, \text { i.e. current becomes } \frac{9}{8} \text { times. }\)

Given, number of electrons, n = 1 x 10¹²
\( \therefore \) Charge transferred, Q = ne = 1 x 10¹² x 1.6 x 10^-19
= 1.6 x 10^-7 \(C\left[ \because e=1.6\times { 10 }^{ -19 }C \right] \)
Capacitance between two conductors,
\(C=\frac { Q }{ v } =\frac { 1.6\times { 10 }^{ -7 } }{ 10 } =1.6\times { 10 }^{ -8 }F\) 

Using Eq, the magnitude of B is
\(B=\frac { E }{ c } \) 
\(=\frac { 6.3V/m }{ 3\times { 10 }^{ 8 }m/s } =2.1\times { 10 }^{ -8 }T\)
To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.
Since, \((+\overrightarrow{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\overrightarrow{\mathbf{i}}, \mathbf{B}\) is along the z-direction.
Thus, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)

The resistance of thick metallic strips is negligible and hence do not affect the resistance of the potentiometer wire. So we can increase the length of potentiometer wire hence sensitivity of potentiometer can be increase d which leads to accurate measurement.

Ampere circuit law,
\(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }I\)
\(b/\omega \) the plates of capacitor,
I = 0
\(\oint { \overrightarrow { B } .\overrightarrow { dl } } =0\)
Which is impossible.
Modified ampere circuital law,
\(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\left( I+{ I }_{ D } \right) \)
Where I_D = Displacement current  = \({ \varepsilon }_{ 0 }\frac { d\Phi E }{ dt } \)

No. there is no such device that can control DC without any energy loss. Even a choke coil cannot control DC.

Here \(\frac{X}{R}=\frac{1}{3}\)
Let the balance point be obtained at length l of bridge wire from the end having resistance X. Then
\(\frac{X}{R}=\frac{l}{(100-l)} \text { or } \frac{1}{3}=\frac{l}{(100-l)} \text { or } l=25 \mathrm{~cm}\)

Zero. As \(W = q\Delta V= 500\mu C \times 0 = 0 \)

When a point P lies on the axis of wire conductor in air, then  \(I\overrightarrow { dl } \ and \ \overrightarrow { r } \)  for each element of the straight wire conductor are parallel. Therefore, \(I\overrightarrow { dl } \times \overrightarrow { r } =0,\) so the magnetic field induction at given point is zero.