V = V1 + V2
Q = C1V = 6 x 10^-6  x 2 = 12\(\mu \)C
As C₂ is in series same amount of charge will also flow through it now V₂ = Q/C₂ = (12 x 10^-6) /(12 x 10^-6) = 1 volt
Total Battery voltage, V = 2 + 1 = 3 Volt

Work done in moving a unit positive charge along distance \(\delta l\)
\(\left| { E }_{ l } \right| \delta ={ V }_{ A }-{ V }_{ B\\ }\)
\(=V-(V+\delta V)\)
  \(=-\delta V\)
\(\left| { E }_{ l } \right| =-\frac { \delta V }{ \delta l } \)

Electric field is in the direction in which the potential decreases the steepest.
(ii) The magnitude of electric field is given by the change in the magnitude of potential per unit displacement, normal to the equipotential surface at the point.

Yes. A d.c. motor generates induced e.m.f. which opposes the supplied e.m.f. It is called back e.m.f.

This impression is misleading. In fact, there is no special attractive force that keeps a person stuck with a high power line while touching that wire, whereas a current of few milliampere is enough to disorganise our nervous system. As a result of it, the affected person loses temporarily his ability to exercise his nervous control to get himself free from the high power line.

Magnetic field induction at a point P in space at a perpendicular distance a from a linear conductor carrying current i is 
\(B=\frac { { \mu }_{ o } }{ 4\pi a } \left( sin{ \phi }_{ 1 }+sin{ \phi }_{ 2 } \right) \) 
Where  \({ \phi }_{ 1 }\) and  \({ \phi }_{ 2 }\) are the angles subtended by the lines joining the ends of linear conductor with point P and perpendicular drawn on the conductor from point P.
In case of infinitely long conductor, 
\( { \phi }_{ 1 }=\frac { \pi }{ 2 } ={ \phi }_{ 2 }\)
\(\\ \therefore \quad \quad B=\frac { { \mu }_{ o } }{ 4\pi } \frac { i }{ a } \left( sin\frac { \pi }{ 2 } +sin\frac { \pi }{ 2 } \right) =\frac { { \mu }_{ o } }{ 4\pi } \frac { 2i }{ a } \)

Drift velocity is the average velocity with which the free electrons get drifted towards the positive end of the conductor under the influence of an external electric field applied.
The relation between current I and drift velocity  v_d is I = nAev_d
Where n is the number density of electrons in a conductor of area of cross-section A and e is the charge on an electron. 

\(x=0.09\)m from \(q=3\times {{10}^{-8}}C\)

(i) X-rays are produced by bombarding a metal target by high energy electrons. 
(ii) To study the atomic structures, treatment for certain forms of cancer.
(iii) Presence of mind, alertness, taking initiative, helpful and caring.

n = 300 turns/m ;
I = 5 A ; l = 0.5 m, r = 10^-2 m.
\({ B }={ \mu }_{ o }nI=\left( 4\pi \times { 10 }^{ -7 } \right) \times 300\times 5\) 
= 1.9 x 10^-3 T.

(a) To find the impedance of the circuit, we first calculate \(X_{\mathrm{L}}\) and \(X_{\mathrm{C}}\).
\( X_L=2 \pi v L \)
\(=2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} \Omega=8 \Omega \)
\(X_c=\frac{1}{2 \pi v C}  =\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}=4 \Omega\)
Therefore.
\(Z  =\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{3^2+(8-4)^2} =5 \Omega\)
\(Z  =\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{3^2+(8-4)^2} =5 \Omega\)
(b) Phase difference, \(\phi=\tan ^{-1} \frac{X_C-X_L}{R}\)
\(=\tan ^{-1}\left(\frac{4-8}{3}\right)=-53.1^{\circ}\)
Since \(\phi\) is negative, the current in the circuit lags the voltage across the source.
(c) The power dissipated in the circuit is
\(P=I^2 R\)
Now, \(I=\frac{i_m}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left(\frac{283}{5}\right)=40 \mathrm{~A}\)
Therefore, \(P=(40 \mathrm{~A})^2 \times 3 \Omega=4800 \mathrm{~W}\)
(d) Power factor \(=\cos \phi=\cos \left(-53.1^{\circ}\right)=0.6\)

Here, R = 10 cm = 0.10 cm;
C = 200 pF = 200 x 10^-12 F, 
\(\omega\)=400 rad s^-1 , V_rms = 230 V.
(i) \({ I }_{ rms }=\frac { { V }_{ rms } }{ { X }_{ C } } =\frac { { V }_{ rms } }{ 1/\omega C } ={ V }_{ rms }\times \omega C\)
= 230 x 400 x (200 x 10^-12)
= 18.4 x 10^-6 A = 18.4 \(\mu A\)
(ii) Magnetic field at a distance r from the axis of plates is
\(B=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } I\)   [See Solved Example 5]
Amplitude of \(\overrightarrow { B } \) is given by
\({ B }_{ 0 }=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } { I }_{ 0 }=\frac { { \mu }_{ 0 } }{ 2\pi } \frac { r }{ { R }^{ 2 } } { I }_{ rms }\sqrt { 2 } \)
\(\left[ \because { I }_{ rms }={ I }_{ 0 }/\sqrt { 2 } \right] \)
Here, r = 2.0 cm = 2.0 x 10^-2 m ; R = 0.10 m.
\(\therefore { B }_{ 0 }=\frac { \left( 4\pi \times { 10 }^{ -7 } \right) }{ 2\pi } \times \frac { 2.0\times { 10 }^{ -2 } }{ { \left( 0.10 \right) }^{ 2 } } \times \left( 18.4\times { 10 }^{ -6 } \right) \)
= 1.04 x 10^-11

Here, I = 3.0 x 10¹⁴ M m^-2 , E₀ = ?, B₀ = ?
Intensity of the plane electromagnetic wave is
\(I={ u }_{ av }c=\frac { 1 }{ 2 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }c\)
\(\therefore { E }_{ 0 }=\sqrt { \frac { 2I }{ { \epsilon }_{ 0 }c } = } \sqrt { \frac { 2\times 3\times { 10 }^{ 14 } }{ \left( 8.85\times { 10 }^{ -12 } \right) \times \left( 3\times { 10 }^{ 8 } \right) } } \)
= 4.75 x 10⁸ V m^-1
\({ B }_{ 0 }=\frac { { E }_{ 0 } }{ c } =\frac { 4.75\times { 10 }^{ 8 } }{ 3\times { 10 }^{ 8 } } =1.58T\)

Since drift velocity \({ v }_{ d }\) and current, I flowing in a conductor are related by the relation:
\({ v }_{ d }=\frac { I }{ neA } ......(1)\)
Also drift velocity in terms of average relaxation time T is given by
\({ v }_{ d }=\frac { eE\tau }{ m } .......(2)\)
From (1) and (2), we have
\(\frac { eE\tau }{ m } =\frac { I }{ neA } \)
\(or\quad \frac { E }{ I } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or\quad \frac { V }{ lI } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or\quad \frac { V }{ I } =\frac { ml }{ n{ e }^{ 2 }A\tau } ........(3)\)
The R.H.S. of Eq(3) is constant
\( \frac { V }{ I } = \ Constant\)
This is Ohm′s law
\((But \ \frac { V }{ I } =R,\)the resistance of the conductor)
So Equation (3) becomes
\(R=\frac { ml }{ n{ e }^{ 2 }A\tau } \)
\(But\quad R=\rho \frac { l }{ A }\)
\(\therefore \quad \rho \frac { l }{ A } =\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\(\rho =\frac { m }{ n{ e }^{ 2 }A\tau } )\)

\(P= \ 1MW={ 10 }^{ 6 }W, \ distance \ of \ town=10km,\)
\(so \ length \ of \ wires \ l=2\times 10=20km=20,000 \ m.\)
(i) Resistance of Cu wire
\(R=\rho \frac { l }{ A } =\frac { 1.7\times { 10 }^{ -8 }\times 2000 }{ \pi \times { \left( \frac { 1 }{ 2 } \right) }^{ 2 }\times { 10 }^{ -4 } } =4\Omega \)
Current drawn
\(I=\frac { P }{ V } =\frac { { 10 }^{ 6 } }{ 220 } =0.45\times { 10 }^{ 4 }A\)
\(Power \ loss={ I }^{ 2 }R={ \left( 0.45 \right) }^{ 2 }\times { 10 }^{ 8 }\times 4=0.81\times { 10 }^{ 8 }W\)
This loss is very large as compared to the power to be transmitted \(\left( { 10 }^{ 6 }W \right) \), hence this method cannot be used for transmission.
\((ii) \ Current \ drawn{ \ I }^{ \prime }=\frac { { 10 }^{ 6 } }{ 11000 } =\frac { { 10 }^{ 3 } }{ 11 } A\)
\(So \ power \ loss={ I }^{ \prime 2 }R=\frac { { 10 }^{ 6 } }{ 121 } \times 4=3.3\times { 10 }^{ 4 }W\)
\(Fraction \ of \ power \ loss=\frac { 3.3\times { 10 }^{ 4 } }{ { 10 }^{ 6 } } =3.3 \ p \ %\)%

Given, emf, E = 2.2 V
Terminal potential difference, V = 1.8 V
External resistance, R = 5Ω
∴ Internal resistance,
\(r=\left(\frac{E-V}{V}\right) R=\left(\frac{2.2-1.8}{1.8}\right) \times 5\)
\(=\frac{10}{9} \Omega\)

\({ tan\theta }_{ 1 }=\frac { V }{ { H }_{ 1 } } ,{ tan\theta }_{ 2 }=\frac { V }{ { H }_{ 2 } } ,{ tan\theta }=\frac { V }{ { H } } \) 
As H₁ & H₂ are horizontal components in two planes at 90^o to each other,
H²​​​​​​ = H₁² + H₂²
\({ \left( Vcot\theta \right) }^{ 2 } \ ={ \left( Vcot{ \theta }_{ 1 } \right) }^{ 2 }+{ \left( Vcot{ \theta }_{ 2 } \right) }^{ 2 }\) 
\(or \ { cot }^{ 2 }\theta ={ cot }^{ 2 }{ \theta }_{ 1 }+{ cot }^{ 2 }{ \theta }_{ 2 }.\)

Given, amplitude of the magnetic field part of harmonic electromagnetic wave,
B₀ = 510 nT = 510 \(\times\)10^-9 T
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀
= 3 × 10⁸ × 510 × 10⁻⁹ = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.