Electric dipole moment is a measurement of the strength of electric dipole. It is given by \(\vec{p}=q(\overrightarrow{2 a})\) cm, where \(\vec{p}\) is the electric dipole moment and 2a is the separation between the charges. It is a vector quantity directed from negative to positive charge on the line joining them.

Let the dipole be made of two equal and opposite charges +q and -q, separated by 2a. Consider a point P at a distance r from the mid-point. Field at P due to each charge will be of equal magnitude \(\left|\vec{E}_{\pm q}\right|=\frac{k q}{\left(r^{2}+a^{2}\right)}\) pointing as shown.
Resolving electric fields due to two charges. We can see that Y-axis components get cancelled out.

∴ Net field at P,E = 2E_q cos θ
\( E= \frac{2 k q}{\left(r^{2}+a^{2}\right)} \cdot \frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}} \)
\(= \frac{2 a q k}{\left(r^{2}+a^{2}\right)^{3 / 2}}=\frac{k p}{\left(r^{2}+a^{2}\right)^{3 / 2}} \)
\(\left(\because \cos \theta=\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\right)\)
(pointing anti-parallel to dipole moment) 
If r >> a, i.e. a² can be neglected in comparison to r².
\(\therefore \ E=\frac{k p}{r^{3}}\) (anti-parallel to \(\vec{p}\))

Let us consider a charge q is placed at the centre of a cubic Gaussian surface. As per the question,
q = 2 \(\mu\)C = 2 \(\times\)10^-6 C
Length of edge = 9 cm

According to Gauss' theorem, the net electric flux (\(\phi\)) through the surface is given by
\(\phi=\frac{q}{\varepsilon_{0}}=\frac{2\times 10^{-6}}{8.854\times 10^{-12}} \)
= 2.26 \(\times\)10⁵ N-m² /C
Thus, the net electric flux through the surface is 2.26 \(\times\)10⁵ N-m²/C.

\(v={ v }_{ m }sin \ \omega t\)
\(i={ i }_{ m }sin \ (\omega t+\Phi )\)
Power at any instant
\(P=vi={ v }_{ m }{ i }_{ m }sin \ \omega t \ sin{ (\omega }_{ t }+\Phi )\)
\(P=\frac { { v }_{ m }{ i }_{ m } }{ 2 } [cos \ \Phi -cos(2\omega t+\Phi )\)
The average of second term in the above expression is zero over a full cycle
\(\therefore \ Average \ Power=\overset { - }{ P } =\ \frac { { v }_{ m }{ i }_{ m } }{ 2 } cos\quad \Phi\)
\(\overset { - }{ P } =\frac { { v }_{ m } }{ \sqrt { 2 } } \times \frac { { i }_{ m } }{ \sqrt { 2 } } cos\Phi\)
\(\overset { - }{ P } ={ V }_{ rms } \ { I }_{ rms }\)
Wattless current is that which flows in the circuit but no power dissipation occurs. It is realized only when circuit is purely inductive or capacity.i.ve, when \(cos\Phi =0\ or\ \Phi =\frac { \overset { + }{ - } \pi }{ 2 } \)

Current in each bulb,
I = \(\frac{P}{V}=\frac{60}{220}=\frac{3}{11}A\)
No. of bulbs used,
n = \(\frac{9}{3/11}\)= 33

f₁=7.5×10⁶ Hz
f₂=12×10⁶ Hz
λ₁=c/f₁=40 m
λ₂=c/f₂=25 m
So the range is 40m to 25m

\(Here,\upsilon =500m/s \ \ e=? \ l=25m\)
\(R=5 \times { 10 }^{ -4 }T, \ \delta ={ 30 }^{ \circ }\)
\(e=Bl\upsilon =Vl\upsilon =(R \ sin \ \delta )l\upsilon \)
\(=5 \times { 10 }^{ -4 }sin{ 30 }^{ \circ } \times 25 \times 500=3.125 \ V\)

\((a) \ { I }_{ 0 }=\frac { \sqrt { 2 } \times 100 }{ \sqrt { 1600+\frac { 1 }{ 4{ \pi }^{ 2 }\times 144\times { 10 }^{ 6 }\times { 10 }^{ -6 } } } } =3.9A\)
Team for C is negligible at high frequency.
(b) \(tan\phi =-\frac { 1 }{ 2\pi \times 12\times { 10 }^{ 3 }\times { 10 }^{ -4 }\times 40 } =-\frac { 1 }{ 96\pi } \)
\(\delta \) is nearly zero at high frequency.
We see that high-frequency  C acts like a conductor. For a d.c. circuit after steady-state, \(\omega =0\) and C amount to an open circuit.

Given, resistance of heating element at temperature 27°C,   
R₂₇ = 100 \(\Omega\)
Resistance of heating element at temperature t°C,
R_t = 117 \(\Omega\)
\(\alpha\)=1.70 \(\times\)10^-4 °C^-1, t = ?
By using the formula of temperature coefficient of resistance,
\(\alpha=\frac{R_2-R_1}{R_1\left(t_2-t_1\right)}\)   ...(i)
Here, R₁ = R_t, R₁ = R₂₇, t₂ = t and t₁ = 27°C
Such that, \(\alpha=\frac{R_t-R_{27}}{R_{27}(t-27)}\)
 Substituting given values in Eq. (i), we get
\(\begin{aligned} 1.70 \times 10^{-4} & =\frac{117-100}{100(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{17}{100 \times 1.70 \times 10^{-4}} \end{aligned}\)
or t = 1000 + 27 = 1027°C 

Here , q = 25 x10^-9C, 2a =6 m, r=4 m,
p=q(2a)= 25x10^-9x 6  = 1.5 x10^-7 C-m
i) \( E_{\text {axil }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 p r}{\left(r^2-a^2\right)^2} \)i
Now, \(=\frac{9 \times 10^9 \times 2 \times 1.5 \times 10^{-7} \times 4}{\left(4^2-3^2\right)^2}=\frac{2700 \times 4}{49} \)
\(=220.4 \mathrm{NC}^{-1} \)
ii)\( \therefore E_{\text {equitorial }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{\left(r^2+a^2\right)^{3 / 2}} =\frac{9 \times 10^9 \times 1.5 \times 10^{-7}}{\left(4^2+3^2\right)^{3 / 2}}=\frac{1350}{125}=10.8 \mathrm{~N} \mathrm{C}^{-1} \)

R₀ = 5 Ω, R₁₀₀ = 5.23 Ω and R_t = 5.795 Ω
Now, \(t=\frac{R_{t}-R_{0}}{R_{100}-R_{0}} \times 100, \quad R_{t}=R_{0}(1+\alpha t)\)
\(=\frac{5.795-5}{5.23-5} \times 100\)
\(=\frac{0.795}{0.23} \times 100=345.65^{\circ} \mathrm{C}\)

Here v = 0.5 Hz; N = 100, A = 0.1 m² and B = 0.01 T. Employing Equation.
\(\varepsilon_0=N B A(2 \pi v)\)
= 100 \(\times\)0.01 \(\times\)0.1 \(\times\) 2 \(\times\) 3.14 \(\times\) 0.5
= 0.314 V
The maximum voltage is 0.314 V.
We urge you to explore such alternative possibilities for power generation.

The One Which is moved Rapidly

(i) The earth's magnetic field at a place can be completely described by three parameters which are called elements of earth's magnetic field. They are as follows:
(a) Angle of declination \(\left( \theta \right) \) 
(b) Angle of dip \((\delta )\) or magnetic inclination )
(c) Horizontal component of earth's magnetic field (H_e) 

\( L_0=0.01 \mathrm{mH}=10^{-5} H \\ L=10 \mathrm{mH}=10^{-2} H \\ \mu_r=?, \mu_r=\frac{L}{L_0}=\frac{10^{-2}}{10^{-5}}=10^3 \\ \mu_r=1000 \)

Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
E = Ir
\(I=\frac{E}{r}\)
\(=\frac{12}{0.4}=30 A\)
The maximum current drawn from the given battery is 30 A.

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

(i) Magnetic flux, \(\phi_1 \hat{\mathrm{i}}=N \mathrm{~B} \mathrm{~A} \cos \theta\)
\(=N B A \cos 0^{\circ}=N B A \)
\(=140 \times 0.09 \times 5 \times 10^{-4}=63 \times 10^{-4} \quad \mathrm{~Wb} .\)
\(\phi_2 \hat{\mathbf{i}} =N B A=0 \quad\{\because B=0\}\)
therefore Change in magnetic flux
\(=\phi_2-\phi_1=-63 \times 10^{-2} \mathrm{~Wb} \text {. }\)
(ii) Emf induced \(=\frac{-\Delta \phi}{\Delta t}=\frac{-63 \times 10^{-4}}{\Delta t}\)

i) Consider an electric dipole consisting of two point charges + q and -q separated by a small distance AB =2l  with centre at O and dipole moment, p= (2) as shown in the figure

Resultant electric field intensity at the point Q.
\(\mathrm{E}_Q=\mathrm{E}_A+\mathrm{E}_B\)
The vectors E_A and E_B are acting at an angle 2\(\theta\)
Here, \( E_A=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{\left(x^2+l^2\right)} \text { and } E_E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{\left(x^2+l^2\right)}\)
On resolving  E_A and E_B into two rectangular components, the vectors \(E_A \sin \theta \text { and } E_B \sin \theta \) are equal
in magnitude and opposite to cach other and hence cancel out.
The vectors \(E_A \cos \theta \text { and } E_B cos \theta \) are acting along the same direction and hence add up.
\(\begin{aligned} \therefore \quad E_Q & =E_A \cos \theta+E_g \cos \theta=2 E_A \cos \theta \\ & =\frac{2}{4 \pi \varepsilon_0} \cdot \frac{q}{\left(x^2+l^2\right)} \cdot \frac{l}{\left(x^2+l^2\right)^{1 / 2}} \\ & {\left[\because E_A=E_b\right] } \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 q l}{\left(x^2+l^2\right)^{3 / 2}} \end{aligned}\)
But q x 2/=|p|, the dipole moment,
\(E_Q=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{|\mathrm{P}|}{\left(x^2+l^2\right)^{3 / 2}}\)
The direcion of E is along QE || BA, i.e. opposite to AB. Invector form, we can rewrite as
\(\mathrm{E}_Q=\frac{-\mathrm{p}}{4 \pi E_0\left(x^2+l^2\right)^{3/ 2}}\) 
Obviously, E, is in a direction opposite to the direction of p. If the dipole is short, i.e. 2/ < \(\therefore \quad E_Q=\frac{1}{4 \pi e_0} \cdot \frac{|\mathrm{p}|}{x^3}\)
\(Clearly, \quad E_Q \propto \frac{1}{x^3}\)
\(From Eqs. (i) and (ii), we get \)
\(\frac{E_{\text {arial }}}{E_{\text {equaturial }}}=2\)
Both the magnitude and the direction of dipole field depend not only on the distance r, but also on the angle between the position vector r and dipole moment p.
The electric field due to a dipole falls off at large distances, at a much faster rate \(\left(\propto \frac{1}{r^3}\right)\) than the electric field due to a single charge \(\left(\propto \frac{1}{r^2}\right)\)
ii) Let P be thepoint at which the system of charges isin equilibrium, then 
F(x)= F(2-x)

\(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q Q}{x^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q Q}{(2-x)^2}\)
\(\begin{array}{ll} \Rightarrow & \frac{1}{x^2}=\frac{1}{(2-x)^2} \\ \Rightarrow & x=(2-x) \Rightarrow x=1 \end{array}\)
Thus,the charpe ) should be placed at the centre of line joining two given charges, Also, the two given charges are identical, i.e, having same nature, so the third charge could be of any nature (positive or negative), As the forces on it at the centre are equal and opposite,

Given: \(C_{1}=C_{2}=C_{3}=C_{4}=C_{5}=C_{6}=C_{7}=C_{8}=3 \mu \mathrm{F}\)

Capacitors C₁, C₂ and C₃ are in series.
\(\therefore \ C_{123}=\frac{3}{3}=1 \mu \mathrm{F} \ \left[C_{5}=\frac{C}{n}\right]\)
Now, capacitors C₁₂₃ and C₄ are in parallel.
\(\therefore \ C_{1234}=C_{123}+C_{4}=1+3=4 \mu \mathrm{F}\) ...(i)
C₅ and C₆ are in parallel.
\(C_{56}=C_{5}+C_{6}=(3+3) \mu \mathrm{F}=6 \mu \mathrm{F}\)
C₇ and C₅₆ are in series
\(\therefore \ C_{756}=\frac{3 \times 6}{3+6}=2 \mu \mathrm{F}...(ii)\)
Now, C₁₂₃₄ and C₇₅₆ are in parallel.
\(C_{p}=(2+4)=6 \mu \mathrm{F}\)
Finally, C, C_p and C₈ are in series.
\( \therefore \frac{1}{C}+\frac{1}{C_{p}}+\frac{1}{C_{8}} =\frac{1}{1} \Rightarrow \frac{1}{C}+\frac{1}{6}+\frac{1}{3}=\frac{1}{1} \frac{1}{C} =1-\frac{1}{2}=\frac{1}{2} \Rightarrow C=2 \mu \mathrm{F} \)

Let the net electric field be zero at point P at a distance x from charge +4\(\mu\)C, then
\(\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \times 10^{-6}}{x^{2}}-\frac{1 \times 10^{-6} \times 1}{4 \pi \varepsilon_{0} \times(2-x)^{2}}=0\)
\(\Rightarrow \ \frac{4}{x^{2}}=\frac{1}{(2-x)^{2}}\)
\(\Rightarrow \ \frac{2}{x}=\frac{1}{2-x}\)
\(\Rightarrow\)  x = 4-2x
\(\Rightarrow \ x=\frac{4}{3} \mathrm{~m}\)

Let VL, VR, Vc and V represent the voltage across the inductor, resistor, capacitor and the source respectively. VR is parallel to I. Vc is pi/2 behind I and VL is pi/2 ahead of I.
Clearly,
\(={ i }_{ 0 }^{ 2 }[{ R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 }]\)
\({ i }_{ o }=\frac { { V }_{ 0 } }{ \sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } } } \)
\(Impedence=\frac { { V }_{ 0 } }{ { i }_{ 0 } } =\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } } \)
\(\\ =\sqrt { { R }^{ 2 }+\left( \omega L-\frac { 1 }{ \omega C } \right) ^{ 2 } } \)

The capacitance of a capacitor in the tuning circuit is varied such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal to be received. When this happens, the amplitude of the current becomes maximum in the receiving circuit.

Here,
I = 1 A,
t = 1 s,
e = 1.6 x 10^-19 C
I = \(\frac{ne}{t}\)
or  n = \(\frac{It}{e}\)
= \(\frac{1 \times 1}{1.6 \times 10^{-19}}\) 
= 6.25 x 10¹⁸

\(Here, \ L=4.0H, \ C=100\mu F={ 10 }^{ -4 }F, \ R=60\Omega \)
\({ E }_{ v }=240V, \ \omega =?, \ { I }_{ v }=?, \ { V }_{ L }=?\)
\((i) \ \omega =\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 4.0\times { 10 }^{ -4 } } } =50rad/s.\)
\((ii) \ At \ resonance, \ Z=R;\)
\({ I }_{ v }=\frac { { E }_{ v } }{ Z } =\frac { { E }_{ v } }{ R } =\frac { 240 }{ 60 } =4A\)
\((iii) \ { V }_{ L }={ I }_{ v }{ X }_{ L }=4\left( \omega L \right) =4\times 50\times 4.0 \ =800V\)

(1) Given,
Radius of spherical conductor, r = 12cm = 0.12m
Charge is distributed uniformly over the surface, q = 1.6 x 10^-7 C.
The electric field inside a spherical conductor is zero.
(2) Electric field E, just outside the conductor is given by the relation
\(\mathrm{E}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{r^{2}}\)
Here, permittivity of free space and \(\frac{1}{4 \pi \epsilon_{o}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\)
Therefore,
\(\mathrm{E}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}}=10^{5} \mathrm{NC}^{-1}\)
Therefore, just outside the sphere the electric field is 4.4 x 10⁴ NC^-1.
(3) From the centre of the sphere the electric field at a point 18m = E₁.
From the centre of the sphere, the distance of point d = 18 cm = 0.18m
\(\mathrm{E}_{1}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{d^{2}}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{\left(1.8 \times 10^{-2}\right)^{2}}=4.4 \times 10^{4} \mathrm{NC}^{-1}\)
So, from the centre of sphere the electric field at a point 18 cm away is 4.4 x 10⁴ NC^-1.

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(d): Velocity of light has different values in different media. It depends on the refractive index of the medium. Related by formula
\(v_{\text {medium }}=\frac{\text { velocity in vacuum }}{\text { refractive index of medium }}\)

(d): The whole charge of a conductor can be transferred to another isolated conductor, if it is placed inside the hollow insulated conductor and connected with it.

c): The charge q on a body is given as q = ne where n is any integer positive or negative. The charge on the electron is q = 1.6 X 10^-19 C whichis less than 1 C.

(i) (c) :Inductive reactance 
\(X_{L}=\omega L=2 \pi v L=2 \pi \times 100 \times 14 \times 10^{-3}\)
\(X_{L}=8.8 \Omega\)
(ii) (b)
:(iii) (b) In an inductor voltage leads the current by \(\frac{\pi}{2}\) or current lags the voltage by \(\frac{\pi}{2}\).
(iv) (b): The current in the inductor coil is given by \(I_{0}=\frac{E_{0}}{X_{L}}=\frac{\sqrt{2} E_{v}}{2 \pi v L}\)
\(L=\frac{\sqrt{2} E_{v}}{2 \pi v I_{0}}=\frac{1.414 \times 200}{2 \times 3.14 \times 50 \times 0.9}=1 \mathrm{H}\)
(v) (a): Inductive reactance
\(X_{L}=\omega L=2 \pi v L=2 \times 3.14 \times 50 \times 2=628 \Omega\)
\(I_{0}=\frac{E_{0}}{X_{L}} \Rightarrow I_{0}=\frac{\sqrt{2} \times E_{v}}{X_{L}}=\frac{\sqrt{2} \times 150}{628}=0.337 \mathrm{~A}\)

(i) (b): Temperature coefficient of resistivity is defined as the fractional increase in resistivity per unit increase in temperature.
(ii) (d): Nichrome (which is an alloy of nickel, iron and chromium) exhibits a very weak dependence of resistivity with temperature.
(iii) (d): Using, \(R_{T}=R_{0}(1+\alpha T)\)
\(\therefore \quad \frac{R_{T_{2}}}{R_{T_{1}}}=\frac{R_{0}\left(1+\alpha T_{2}\right)}{R_{0}\left(1+\alpha T_{1}\right)}=\frac{2}{1}=\frac{\left(1+\alpha T_{2}\right)}{(1+\alpha \times 300)}\)
\(\Rightarrow \quad 2+\alpha \times 600=1+\alpha T_{2}\)
\(\Rightarrow \quad 1=\alpha\left(T_{2}-600\right) \Rightarrow \frac{1}{0.00125}=\left(T_{2}-600\right)\)
\(\Rightarrow \quad 800^{\circ} \mathrm{C}=T_{2}-600\)
T₂ = 800 - 273 + 600
T₂ = 1127 K
(iv) (a): The temperature coefficient of resistance of an alloy used for making resistors is small and positive.
(v) (b): The resistance of a metallic wire at temperature t^oC is given by
\(R_{t}=R_{0}(1+\alpha t)\) where \(\alpha\) is the temperature coefficient of resistance and R_o is the resistance of a wire at O°c.
For metals,\(\alpha\) is positive. Hence, resistance of a wire increases with increase in temperature.
Also, from Ohm's law
\(\frac{V}{I}=R\)
Hence on increasing the temperature, the ratio \(\frac{V}{I}\) increases.