Here, total number of turns,
\(N=100, A=3 \mathrm{~m}^{2}, \omega=60 \mathrm{rads}^{-1}, B=0.04 \mathrm{~T}\)
(i) Maximum emf produced in the coil,
\(e_{0}=N B A \omega=100 \times 0.04 \times 3 \times 60\)
\(e_{0}=720 \mathrm{~V}\)
Since, resistance of the coil is 500\(\Omega\),  the maximum current drawn from the generator is
\(I_{0}=\frac{e_{0}}{R}=\frac{720}{500}=1.44 \mathrm{~A}\)
(ii) Maximum power dissipation in the coil
\(P=e_{0} I_{0}=720 \times 1.44=1036.8 \mathrm{~W}\)

If the field is non-uniform, the net force will be non-zero.

(i) The figure is shown as below :1

equipotential surfaces of two identical positive charges
(ii) By definition, electric potential energy of any charge q placed in the region of electric field is equal to the work done in bringing charge q from infinity to that point and given by
U = qV
where, V is the electric potential (as potential at infinity is assumed to be zero) where, the charge q is placed. Now, considering the electric potentials at positions r₁ and r₂ as V₁ and V₂
respectively. Therefore, total potential energy of the system of two charges q₁ and q₂ placed at points with position vectors r₁ and r₂ in the region of E is given by U = work done in bringing charge q from infinity to that position in E is equal to work done for charge q₂ from infinity to that position in E + work done to that of charge q₂ at these positions in presence of q₁.
\({ U }={ q }_{ 1 }{ V }_{ 1 }+{ q }_{ 2 }{ V }_{ 2 }\Rightarrow U=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { { q }_{ 1 }{ q }_{ 2 } }{ \left| { r }_{ 2 }-{ r }_{ 1 } \right| } \) 

\(2.4 \times { 10 }^{ -7 }Wb;6.37 \times { 10 }^{ -5 }{ m }^{ 2 }\)

\((i) \ B={ \mu }_{ o }\frac { NI }{ l } ;\)
\( here,l=0.20 \ m;N=120;I=2.5A\)
\( (ii) \ B=\frac { { \mu }_{ o }NI }{ 2l } \)

\(Given \ P=100W, \ V=220V\)
\((a) \ R=\frac { V^{ 2 } }{ P } =\frac { { \left( 220 \right) }^{ 2 } }{ 100 } =484\Omega \)
\((b) \ { E }_{ 0 }=\sqrt { 2 } { E }_{ v }=\sqrt { 2 } \times 220=311V\)
\((c) \ I=\frac { P }{ V } =\frac { 100 }{ 220 } =0.45A\)
The average power dissipated,
P = (i²R) = (\({ i }_{ m }^{ 2 } \)R sin² \(\omega t\) ) =  \({ i }_{ m }^{ 2 }R({sin}^{2}\omega t)\)
\(\because\)  \({sin}^{2}\omega t = {{1}\over{2}}(1-cos\ 2\omega t)\)
\(\therefore\) \(({sin}^{2}\ \omega t)={{1}\over{2}}[1-(cos\ 2\omega t)]={{1}\over{2}}\)  \((\because \ cos\ 2\omega t=0)\)
\(\therefore\)   \(\bar{P}={{1}\over{2}}{ i }_{ m }^{ 2 }R\)

Relaxation time is inversely proportional to the velocities of electrons and ions. The applied electric field produces the insignificant change in velocities of electrons of the order of 1 mm/s, whereas the change in temperature T affects velocities at the order of 10² m/s
This decreases the relaxation time cinsiderably in metals and consequently resistivity of metal or conductor increases as, \(\rho=\frac{1}{\sigma}=\frac{m}{n e^2 \tau}\)

Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T₂ = 100°C
Resistance of the silver wire at T₂, R₂ = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as 
\(\alpha=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}\)
\(=\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1}\)
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.

(i) The emf of a cell is greater than its terminal voltage because there is some potential drop across the cell due to its small internal resistance.
(ii) ∵ \(V=\left(\frac{E}{R+r}\right) R=\frac{E}{1+r / R}\)
i.e. with the increase of R, V increases

One can determine the emf of cell by finding terminal potential difference when current I becomes zero.
(iii) In these, type of questions, we have to look out the connections of different cells, if the opposite terminals of all the cells are connected, then they support each other, i.e. these individual emfs are added up. If the same terminals of the cells are connected, then the equivalent emf is obtained by taking the difference of emfs. Net emf of combination
= E - 2E + 5E = 4 E
Net resistance of current
= r + 2r + 3r + R = 6r + R
∴ Current, \(I=\frac{V}{R}\) (from Ohm's law)
⇒ \(I=\frac{4 E}{6 r+R}\)

Given, applied voltage, V = 140sin100\(\pi t\) V
\(C=\frac{50}{\pi} \mu \mathrm{F}=\frac{50}{\pi} \times 10^{-6} \mathrm{~F}\)
\(L=\frac{5}{\pi} \mathrm{H}, R=400 \Omega\)
Comparing with V = V₀ sin \(\omega\)t, we get
V₀ = 140 V and \(\omega\) = 100\(\pi\)
Inductive reactance, \(X_{L}=\omega L=100 \pi \times \frac{5}{\pi}=500 \Omega\)
Capacitive reactance, \(X_{C}=\frac{1}{\omega C}=\frac{1}{100 \pi \times \frac{50}{\pi} \times 10^{-6}}\)
= 200 \(\Omega\)
Impedance of the circuit, \(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
\(=\sqrt{(400)^{2}+(500-200)^{2}}\)
\(=\sqrt{160000+90000}=500 \Omega\)
Maximum current in the circuit,
\(I_{0}=\frac{V_{0}}{Z}=\frac{140}{500}\)
\(I_{\mathrm{rms}}=\frac{I_{0}}{\sqrt{2}}=\frac{140}{500 \times \sqrt{2}}=0.2 \mathrm{~A}\)
V_rms across resistor, V_R = I_rmsR
= 0.2 x 400 = 80V 
V_rms across inductor, V_L = I_rms X_L
= 0.2 x 500 = 100V
V_ems across capacitor, V_c = I_rms X_C
= 0.2 x 200 = 40 V
Now, V \(\ne \) V_R + V_L + V_C
Because V_C V_L and V_R are not in same phase, instead
\(V=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)
\(=\sqrt{80^{2}+(100-40)^{2}}=100 \mathrm{~V}\)
which is same as that of applied rms voltage.

(a) Given, V = V₀ sin \(\omega\)t
\(I=I_{0} \sin \left(\omega t+\frac{\pi}{2}\right)\)
As it is clear that, the current leads the voltage by a phase angle \(\frac{\pi}{2}\)
\(\therefore\)The device X is a capacitor.
(b)
  
(c) The reactance of the capacitance is given as
\(X_{C}=\frac{1}{\omega C}\)
where, \(\omega\) = angular frequency and C = capacitance of capacitor.
\(\therefore \quad X_{C}=\frac{1}{2 \pi v C}\)
where, V = frequency of AC or X_C \(\propto \frac{1}{V}\)
\(\therefore\) The graphical representation between reactance of capacitance and frequency is given as

(d) Phasor diagram

(a) Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop,
L = 2 π r
I_e = Current enclosed by the loop = I
The result is the familiar expression for a long straight wire
B (2π r) = μ₀I
\(B=\frac{\mu_{0} I}{2 \pi r}\)
\(B \propto \frac{1}{r} \quad(r>a)\)
Now the current enclosed I_e is not I, but is less than this value. Since the current distribution is uniform, the current enclosed is
\(I_{e}=I\left(\frac{\pi r^{2}}{\pi a^{2}}\right)=\frac{I r^{2}}{a^{2}}\)
Using Ampere’s law, \(B(2 \pi r)=\mu_{0} \frac{I r^{2}}{a^{2}}\)
\(B=\left(\frac{\mu_{0} I}{2 \pi a^{2}}\right) r\)
B ∝ r (r < a)

Figure shows a plot of the magnitude of B with distance r from the centre of the wire. The direction of the field is tangential to the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section.
This example possesses the required symmetry so that Ampere’s law can be applied readily.

Statement for Biot-Savart Law: The magnitude of magnetic field \(d\overrightarrow { B } \) due to current element is directly proportional to the current I, the elements length \(\left| dl \right| \) and inversely proportional to the square of the distance r of the field point. Its direction is perpendicular to the plane containing \(\overrightarrow { dl } \)and \(\overrightarrow{r}\).
\(d\overrightarrow B\alpha\frac{I\overrightarrow dl \times \overrightarrow r}{r^{ 3}}\)
Or \(d\overrightarrow B=\frac {\mu_0}{4 \pi}\frac{I\overrightarrow dl \times \overrightarrow r}{r^{ 3}}\)

The magnetic field due to \(\overrightarrow {dl}\) is given by Biot Savart law as
\(dB=\frac {\mu_0}{4\pi}.\frac { I\left| \overrightarrow { dl } \times \overrightarrow { r } \right| }{ { r }^{ 3 } } \)
Now dB_x= Db Cos \(\theta\) = \(\frac { \mu _{ 0 } }{ 4\pi } .\frac { Idl }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) } \cos { \theta } \)
\(=\frac { \mu _{ 0 } }{ 4\pi } .\frac { Idl }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) } \frac { R }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 1/2 } } \)
So, \({ B }_{ x }=\int { dB_{ s }=\frac { { \mu }_{ 0 } }{ 4\pi } } \frac { IR }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \int { dl } \)
\(=\frac { { \mu }_{ 0 }IR^{ 2 } }{ 2\left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \)
(The y-components, of the field, add up to zero,due to symmetry)
\(\therefore\)Magnetic field at P due to a circular loop
\(=B={ B }_{ x }\overrightarrow { i } =\frac { { \mu }_{ 0 }IR^{ 2 } }{ 2\left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \overrightarrow { i } \)
Explanation: A circular current loop produces magnetic field and its magnetic moment is the product of current and its area \(\overrightarrow M=\overrightarrow {LA}\)

Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V¹
R is connected to the storage battery in series. Hence, it can be written as
V¹ = V - E
V¹ = 120 - 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
\(I=\frac{V^{1}}{R+r}\)
\(=\frac{112}{15.5+5}=\frac{112}{16}=7 A\)
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R
Terminal voltage of battery = 120 - 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.

Since, the electric field has only x-component, for faces normal to x- direction, the angle between E and \(\triangle\)S is \(\pm \frac{\pi}{2} .\) Therefore, the flux is separately zero for each of the cube except the shaded ones.
The magnitude of the electric field at the left face is 
E _L = 0  [as x = Oat the left face]
The magnitude of the electric field at the right face is 
E _R = 3a  [as, x = a at the right face]
The corresponding fluxes are
\(\phi_{L}=\mathbf{E}_{L} \cdot \Delta \mathbf{S}=0\)
and \(\phi_{R}=\mathbf{E}_{L} \cdot \Delta \mathbf{S}=E_{R} \Delta S \cos \theta=E_{R} \Delta S\)     [\(\because\)  \(\theta\)= 0°]
\(\Rightarrow\)  \(\phi\)R = E_Ra²
Net flux (\(\phi\)) through the cube
= \(\phi_L\)L+ \(\phi_R\)
= 0 + E _Ra² = E_Ra²
 \(\Rightarrow\) q = 2a(a)² = 2a³
We can use Gauss' law to find the total charge q inside the cube

\(\phi=\frac{q}{\varepsilon_{0}} \Rightarrow \phi=\frac{2 a^{3}}{\varepsilon_{0}}\)

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.
The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

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

P.E. of dipole = -MB cos 0^o = -MB
Work done = MB (cos 0^o - cos 180^o)
= MB (1+1) = 2 MB.

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. 

Given,
Capacitance of the capacitor, C = 12pF = 12 x 10^-12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
\(\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 12 \times 10^{-12} \mathrm{\times}(50)^{2} \mathrm{~J}=1.5 \times 10^{-8} \mathrm{~J}\)
Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10^-8 J. was disconnected.

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

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

(b): Microwaves are the electromagnetic waves of wavelength of the order of a few millimetres, which is less than those of T.V. signals. On account of smaller wavelength, the microwaves can be transmitted as beam signals in a particular direction and are much better than radiowaves because microwaves do not spread or bend around the corners of any obstacle coming in their way. Therefore microwaves are better carriers of signals than radiowaves.

1.B) The amount of work done per unit charge in bringing a positive test charge from infinity to that point.

2.B) As the work done in assembling the system of charges from infinity.

3.C) A device that stores electrical energy in an electric field.

4.B) The ratio of the amount of charge stored on one conductor to the potential difference between the conductors.
5.A) Two conductors separated by an insulator.