1) The frequency of incident radiation was kept constant.
2) de Broglie wavelength,
\(\lambda =\frac { h }{ \sqrt { 2mqV } } \alpha \frac { 1 }{ V } \)
If potential difference V is doubled, the de-Broglie wavelength is decreased to \(\frac { 1 }{ \sqrt { 2 } } \) time.

\(\text { (i) } \lambda_{\mathrm{d}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV_1}}} ; \lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{\mathrm{me}}}\)
\(\frac{\lambda_{\mathrm{d}}}{\lambda_{\alpha}}=\frac{\frac{1}{\sqrt{2 m e}}}{\frac{1}{\sqrt{4 m 2 e}}} \Rightarrow \lambda_{\mathrm{d}}=2 \lambda_{\alpha}\)
\(\text { (ii) } =\frac{K.E_d}{K.E_a}=\frac{eV}{2eV} =\frac{1}{2}\Rightarrow K.E_d=\frac{1}{2}K.E_a\)
∴ K.E of deuteron is half of K.E of α-particle.

(i) The phenomenon of spontaneous emission of highly penetrating radiations such as α, β, γ rays by heavy elements having atomic number greater than 82 is called radioactivity and the substances which emit these radiations are called radioactive elements.
(ii) The radioactive phenomenon spontaneous and is unaffected by any external agent like temperature, pressure, and magnetic fields.

Wavelength of incident radiation = 97.5 nm = 97.5 x 10^-9 m
Energy of hydrogen atom in its ground state = -13.6 eV
(a) Principal quantum number n = ?
(b) (i) Number of possible transitions = ?
(ii) Total number possible lines = ?
(a) Energy absorbed by Hydrogen atom
\(E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{97.5 \times 10^{-9}} \mathrm{~J} \)
\(E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{97.5 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
E = 12.74 eV
Energy of the electron in first orbit of Hydrogen is -13.6 ev
E_n = -13.6 + 12.74 = -0.86 eV
We know that
\(\mathrm{E}_{\mathrm{n}} =-\frac{13.6}{\mathrm{n}^{2}} \)
\(-0.86 =-\frac{13.6}{\mathrm{n}^{2}} \)
\(\mathrm{n}^{2} =15.88 \)
\(\mathrm{n} \cong 4 \)
(b) (i) By using arithmetic progression, For the principle quantum number "n",
Total number of possible transition form level n is \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)
(ii) Total number of possible transitions form level 4 is 3
Total number of possible transitions form level 3 is 2
Total number of possible transition form level 2 is 1
Hence total number of possible transitions is 3 + 2 + 1 = 6

(i) u - v rays
(ii) X - rays
(iii) ૪ - rays.

The like currents i.e current in both the wire are in the same direction attracts each other. The force is repulsive when the current flows in opposite direction through the wires.
F = \(\frac { { \mu }_{ 0 }{ I }_{ 1 }{ I }_{ 2 }dl }{ 2\pi r } \)
i.e \(F\alpha \frac { 1 }{ r } \)

As the wire of the loop carrying the opposite current is near so the net force acting on the loop is repulsive.

(i) Acceleration to the right-hand thumb rule, when the current flows through the circular loop in the direction.

(ii) The direction of magnetic field (B) is perpendicular to the plane of the loop and in the outward direction.

If we rotate aright-handed screw by ascrew driver, then the direction of current is same as the direction in which screw advances and the direction of rotation of the screw gives the direction of the magnetic field.

(i) The total resistance offered to the flow of current due to resistance R, inductive resistance X_L, and capacitive reactance X_c in a circuit is called impedance.
It is given by \(z=\sqrt { { R }^{ 2 }+({ { X }_{ L }-{ X }_{ C }) }^{ 2 } } \)
(ii) At resonance, when X_L= X_C

i) Maxwell's prediction was experimentally confirmed by Heinrich Rudolf Hertz in 1888. The experimental set up used is shown in Figure.
ii) It consists of two metal electrodes which are made of small spherical metals. These are connected to larger spheres and the ends of them are connected to induction coil with very large number of turns. This is to produce very high electromotive force (emf).
iii) Since the coil is maintained at very high potential, air between the electrodes gets ionized and spark (spark means discharge of electricity) is produced.
iv) The gap between electrode (ring type - not completely closed and has a small gap in between) kept at a distance also gets spark. This implies that the energy is transmitted from electrode to the receiver (ring electrode) as a wave, known as electromagnetic waves.

v) If the receiver is rotated by 90° - then no spark is observed by the receiver. This confirms that electromagnetic waves are transverse waves as predicted by Maxwell.
vi) Hertz detected radio waves and also computed the speed of radio waves which is equal to the speed of light (3 x 10⁸m S^-1).

L = 40 x 10^-3 H; i = 0.3 sin (200t – 40^o)
X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90^o Therefore,
v = V_msin (ωt + 90^o)
v = 2.4sin (200t - 40^o + 90^o)
v = 2.4sin (200t + 50^o)volt

V_p = 12 V; V_s = 240 V
I_s = 50 mA; N_p = 100 turns
\(\frac { { V }_{ s } }{ { V }_{ P } } =\frac { { N }_{ s } }{ { N }_{ p } } =\frac { { I }_{ P } }{ { I }_{ S } } =K\)
Transformation ratio, K = \(\frac{240}{12}=20\)
The number of turns in the secondary
N_S = N_P x K = 100 x 20 = 2000
Primary current,
I_P = K x I_s = 20 x 50 mA = 1 A

We can denote the current that flows from 9V battery as I₁ and it splits up nto I₂ and (I₁ – I₂) at the junction E according Kirchoff’s current rule (KCR).
Now consider the loop EFCBE and apply KVR, we get
1I₂ + 3I₁ + 2I₁ = 9
5I₁ + I₂ = 9          (1)
Applying KVR to the loop EADFE, we get
3 (I₁ – I₂ ) – 1I₂ = 6
3I₁ – 4I₂ = 6           (2)
Solving equation (1) and (2), we get
I₁ = 1.83 A and I₂ = -0.13 A
It implies that the current in the 1 ohm resistor flows from F to E.

P_av = \(\frac{1}{2}\)V₀I₀cosΦ
\(= \frac{1}{2}\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}cos\times\frac{\pi}{3}\times \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\)

R_s = 3 + 2.5 + P = 5.5 + P
V = 9 V, I = 1.0 A
R_s = \(\frac{V}{I}=\frac{9}{1}= 9 \Omega\)
∴ 9 = 5.5 + P
∴ P = 9 - 5.5 = 3.5 Ω

Dispersion is splitting of white light into its constituent colours. This band of colours of lightis called its spectrum.

The P.D. between A and B is given by
\(V =\left[V_{\mathrm{A}}-V_{\mathrm{B}}\right]-V_{\mathrm{b}}(\mathrm{Si}) \)
\(=[3.3-(-7.4)]-0.7 \)
\(=10.7-0.7=10 \mathrm{~V} \)
The value of current flowing through AB can be obtained by using Ohm’s law
\(I=\frac { V }{ R } =\frac { 10}{ 1\times { 10 }^{ 3 } } ={ 10 }^{ -2 }A=10mA\)

Energy produced per second in reactor = 1 W = 1 J/s
Energy produced per fission = 200 MeV = 200 x 10⁶ x 1.6 x 10^-19 J
= 3.2 x 10¹¹ J
Number of fissions per second required \(=\frac{\text { Energy produced per second in reactor }}{\text { Energy produced per fission }} \)
\(=\frac{1}{3.2 \times 10^{11}}=\frac{10 \times 10^{10}}{3.2}=3.125 \times 10^{10} \)
Number of fissions per second = 3.125 x 10¹⁰ 

i) Noise : Interference of undesirable electrical signal with the transmitted signal.
ii) Attenuation: Loss of strength of a signal.
iii) Medium : It makes some distortion tothe signal.

The charges will flow between the two spherical conditioners till their potential become equal.
\(i.e.\frac { { Kq }_{ 1 } }{ { R }_{ 1 } } =\frac { { Kq }_{ 2 } }{ { R }_{ 2 } } (or)\frac { { q }_{ 1 } }{ { R }_{ 1 } } =\frac { { q }_{ 2 } }{ { R }_{ 2 } } \)
The ratio of the surface charge densities on the two conditioners will be
\(\frac { { \sigma }_{ 1 } }{ { \sigma }_{ 2 } } =\frac { \frac { { q }_{ 1 } }{ 4\pi { R }_{ 1 }^{ 2 } } }{ \frac { { q }_{ 2 } }{ 4\pi { R }_{ 2 }^{ 2 } } } =\frac { { q }_{ 1 } }{ { q }_{ 1 } } .\frac { { R }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 2 } } =\frac { { R }_{ 1 } }{ { R }_{ 2 } } \times \frac { { R }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 2 } } \)
\(\frac { { \sigma }_{ 1 } }{ { \sigma }_{ 2 } } =\frac { { R }_{ 1 } }{ { R }_{ 2 } } \)

(i) Angle of dip (90°) is maximum at the magnetic piles.
(ii) Vertical component of earth's magnetic field is zero at magnetic equator.

(i) Capacitive reactance \({ X }_{ c }=\frac { 1 }{ 2\pi \gamma c } \)
\({ X }_{ c }\alpha \frac { 1 }{ \gamma } \)
The graph between γ and X_L

(ii) Inductive reactance XL = 2πγL
XL α γ

The phenomenon of electrical resonance is possible when the circuit contains both L and C. Only then does the voltage across L and C cancels one another when VL and VC are 1800 out of phase and the circuit becomes purely resistive. This implies that resonance will not occur in RL and RC circuits.

Thick copper strips offer minimum resistance and hence avoid the error due to end resistance. Which have not been taken into account in the bridge formula.

At the transmitting point, the voltage is increased and the corresponding current is decreased by using a step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss. At the receiving point, the voltage is decreased and the current is increased to appropriate values by using a step-down transformer, and then it is given to consumers.

(i) To increase the resistance of the circuit the resistor are connected in series.
(ii) When they are connected in parallel resistance of the circuit is decreased

RMS value is also defined as that value of the steady current which when flowing through a given circuit for a given time produces the same amount of heat as produced by the alternating current when flowing through the same Circuit for the same time. (or)
The root mean square value of an alternating current is defined as the square root of the mean of the squares of all currents over one cycle
\(I_{RMS}=\sqrt\frac{\text {Area of one cycle of squared wave}}{\text {Base length of one cycle}}\)

Inductor is a device used to store energy in a magnetic field, when an electric current flows through it. Examples: coils, solenoids and toroids

Using right hand rule, current flows upwards.

(i) Electrical power P = VI
(ii) Erectrical power \(P=V\left(\frac{V}{R}\right)=\frac{V^{2}}{R} \quad \therefore P=\frac{V^{2}}{R}\)
(iii) P = Iv = I(IR) = I²R
(iv) P = I²R

Electric power is the rate at which the electrical potential energy is delivered
\(P =\frac{d U}{d t} \)
\(P =\frac{VdQ}{d t}=\mathrm{V} \frac{d Q}{d t} \)
Since \(\frac{d Q}{d t}=I\), where I - electric current
∴ P = VI

de Broglie wavelength, \(\lambda =\frac { h }{ \sqrt { 2mE } } \)
Given data 
h = 6.62 x 10 - 34 Js, m = 9.1 x 10 - 31kg
E = 50 KeV = 50 x 1.6 x 10^-19J
\(\lambda =\frac { { 6.62\times 10 }^{ -34 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }\times 8\times { 10 }^{ -15 } } } m\)
\(=\frac { 6.62 }{ \sqrt { 145.6 } } \times { 10 }^{ -11 }m\)
\(=\frac { 6.62 }{ 12.07 } \times { 10 }^{ -11 }m=5.48\times { 10 }^{ -12 }m\)
Wavelength of yellow light,
λ' = 5990 x 10^-10 m = 5.99 x 10^-7 m
Now \(\frac { \lambda }{ \lambda ' } ={ 10 }^{ -5 }\)
Since resolving power is inversely proportional to wavelength, therefore, the resolving power of an electron microscope is 105 times larger than the resolving power of the optical microscope.

HaIf wave rectifier:
Only one half of the input wave reaches the output. Therefore it is called half wave rectifier.
Construction:

(i) The circuit consists of a transformer, a p-n junction diode and a resistor
(ii) In a half wave rectifier circuit, either a positive half or the negative half of the AC input is passed through by the diode while the other half is blocked
(iii) It acts as a rectifier diode.

(iv) Efficiency (η) is the ratio of the output DC power to the AC input power circuit. supplied to the circuit.
(v) The efficiency (η) of a half wave rectifier is found to be 40.6 %.

Since displacement current = conduction current
\({ I }_{ d }=\frac { V }{ { X }_{ C } } =\frac { 220 }{ 100 } =2.2A\)

The cyclotron's frequency v = 18 μHz
= 8 x 10⁶ Hz
The mass of the proton m = 1.67 x 10^-27 kg
The charge of the proton q = 1.6 x 10^-19C
Radius of the dees r = 50 cm = 50 x 10^-2 m
Magnetic field B = ?
k.E of the proton k.E = ?
Magnetic field B = \(\frac { 2\pi mv }{ q } \)
B = \(\frac { 2\times 3.14\times 1.67\times 10^{ -27 }\times 8\times 10^{ 6 } }{ 1.6\times 10^{ -19 } } \)
= \(\frac { 83.90\times 10^{ -21 } }{ 1.6\times 10^{ -19 } } \)
= 52.438 x 10^-2
B = 0.524T
k.E = \(\frac { 1 }{ 2 } \) mv²
v = rω = r x 2πv
= 0.5 x 2 x 3.14 x 8 x 10⁶
= 25.12 x 10⁶ m/s
k.E = \(\frac { 1 }{ 2 } \) x 1.67 x 10^-27 x (25.12 x 10⁶)²
=\(\frac { 41.95\times 10^{ -21 }\times 25.12\times 10^{ 6 } }{ 2 } \)
= 526.89 x 10^-15 J (or) 5.269 x 10^-13J
To convert J into MeV
= \(\frac { 526.89\times 10^{ -15 } }{ 1.6\times 10^{ -19 } } \)  = 3.29 x 10⁶ eV
k.E = 3.29 x 10⁶ MeV

Given: N = 1000
A = 100 cm² = 10-2 m²
v = 100 rpm = \(\frac{100}{60}\) rps
B = 3.6 x 10^-27
em = ?
em = NBAω = NBA (2πγ)
= 1000 x 3.6 x 10^-2 x 10^-2 x 2 x \(\frac{22}{7}\times\frac{100}{60}\)
e = 5.77 V

Phase difference between V and i = \(\frac{\pi}{2}\)red
∴ P_average = \(\frac { { V }_{ m }{ I }_{ m } }{ 2 } \). cos Φ = \(\frac { 50\times 10 }{ 2 } \) cos 0^o

For the bridge to be balanced \(\cfrac { \\ P }{ Q } =\cfrac { R }{ S } \)
Since additionally a resistance 'x' is added in series with S, equation (1) can be written as
\(\cfrac { \\ P }{ Q } =\cfrac { R }{ (S+x) } \)
\(\left( S+x \right) =\cfrac { QR }{ P } x=\cfrac { \theta R }{ P } -S\)
\(x=\cfrac { 8\times 12 }{ 7 } -7\)
\(x=\frac{96}{7}-7=6.714 \Omega\)

(i) The equivalent capacitance is,
C_p = C₁+ C₂ + C₃
= (1 + 2 + 3) = 6μF
(ii) Total charge, q = CμV
= 6 x 10^-6 x 100 = 600μC
q₁ = C₁V = 1 x 100 = 100μC
q₂ = C₂V = 2 x 100 = 200μC
q₃ = C₃V = 3 x 100 = 300μC

(i) Power of a circuit is defined as the rate of consumption of electric energy in that circuit. It is given by the product of the voltage and current. In an AC circuit, the voltage and current vary continuously with time.
(ii) The alternating voltage and alternating current in the series RLC circuit at an instant are given by
v = Vm sin ωt and i = I_m sin(ωt + Φ)
(iii) Where Φ is is the phase angle between v and i. The instantaneous power is then written as
P = v_i
= V_mI_m sin ω sin(ωt + Φ)
= V_mI_msin ωt[sin ωt cos Φ - cos ωt sin Φ)
P = V_mI_m[cos Φ sin² ωt - sin ωt cos ωt sin Φ]
(iv) Here the average of sin²ωt over a cycle is \(\frac { 1 }{ 2 } \) and that of sin wt cos wt is zero. Substituting these values, we obtain average power over a cycle.
= \(\frac { { V }_{ m } }{ \sqrt { 2 } } \frac { { I }_{ m } }{ \sqrt { 2 } } \) cos Φ
Pav = V_RMSI_RMS cosΦ
(v) where V_RMS I_RMS is called apparent power and cos Φ is power factor. The average power of an AC circuit is also known as the true power of the circuit.

By the concept of Drift velocity \(I=nAeu_{ d }\)
\({ \mu }_{ d }=\cfrac { eE }{ m } \tau \)
\(\rho =\cfrac { m }{ { ne }^{ 2 }\tau } \)
\(\therefore\) Current density
\(J=\cfrac { I }{ A } ={ \cfrac { nA{ ev }_{ d } }{ A } =ne.\cfrac { em\tau }{ m } }=\left( \cfrac { { ne }^{ 2 }\tau }{ A } \right) E\)
\(J=\cfrac { 1 }{ \rho } .E\)

(i) Drift velocity \({ u }_{ d }=\cfrac { ev }{ ml } .\tau \)
When I am doubled, drift velocity, become \(\cfrac { 1 }{ 2 } \) times the original v_d
(ii) Resistor \(R=\rho .\cfrac { l }{ A } \)
Resistor becomes doubled i.e. 2 times the original resistors.
(iii) Resistivity is not affected.

Let YY' be an infinitely long straight conductor carry current I. In order to calculate magnetic field at a point P which is at a distance a from the wire, let us consider a small line element dl (segment AB).
According to Biot Savart law, the magnetic field at a point P due to current element Idl is,
\({ d \vec B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { Idl sin \theta} }{ { r }^{ 2 } }\hat n \).
To apply trigonometry, draw a perpendicular AC to the line BP as shown in Figure.
In triangle ΔABC, \(\sin \theta=\frac{\mathrm{AC}}{\mathrm{AB}}\)
∴ AC = AB sinθ
\(\text { But, } A B  =d l \Rightarrow A C=d l \sin \theta\)
Let dΦ be the angle subtended between AP and BP
ie., \(\angle \mathrm{APB}=\angle \mathrm{APC}=d \phi\)
In a triangle \(\triangle \mathrm{APC}, \sin (d \phi) \simeq A C / A P\)
Since, dΦ is very small, \(\sin (d \phi) \simeq d \phi\)
But,  \(\mathrm{AP}  =r \Rightarrow A C=r d \phi \)
\(\therefore \mathrm{AC} =d l \sin \theta=r d \phi \)
\(\therefore d \vec{B}  =\frac{\mu_0}{4 \pi} \frac{I}{r^2}(r d \phi) \hat{n}=\frac{\mu_0}{4 \pi} \frac{I d \phi}{r} \hat{n}\)
Let Φ be the angle between AP and OP
\(\text {In a } \triangle \mathrm{OPA}, \cos \phi =\frac{\mathrm{OP}}{\mathrm{AP}}=\frac{\mathrm{a}}{\mathrm{r}}  \)
\(r  =\frac{a}{\cos \phi}  \)
\(\text {Now, } d \vec{B}  =\frac{\mu_0}{4 \pi} \frac{I}{a / \cos \phi} d \phi . \hat{n}  \)
\(d \vec{B}  =\frac{\mu_0 I}{4 \pi a} \cos \phi d \phi \hat{n}\)
The total magnetic field at P due to the conductor YY' is
\(\vec { B } = \int _{- \Phi _{ 1 } }^{ { \Phi }_{ 2} }d\vec B =\int _{ -\Phi _{ 1 } }^{ { \Phi }_{ 2 } }\frac { { \mu }_{ 0 }I }{ 4\pi a }{ cos\phi d\phi } \hat { n }\)
\(=\frac { { \mu }_{ 0 }I }{ 4\pi a }[{ sin\phi ]^{\phi_2} _{\phi_-1}} \hat { n }\)
\( \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } (sin{ \Phi }_{ 1 }+sin{ \Phi }_{ 2 })\hat { n } \)
For infinitely long conductor, Φ₁ =  Φ₂ = 90^o
\(\therefore \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } \times 2\hat{n}\Rightarrow\vec { B } =\frac { { \mu }_{ 0 }I }{ 2\pi a } \hat { n } \)
                                                                                                                 

(i) Consider a circuit containing a resistor of resistance R, a inductor of inductance L and a capacitor of capacitance C connected across an alternating voltage source (Figure ). The instantaneous value of the alternating voltage is given by the equation
ሀ = V_m sin ωt   ......(1)

(ii) Let i be the resulting circuit current in the circuit at that instant. As a result, the voltage is developed across R, Land C.
(iii) We know that voltage across R (V_R) is in phase with i, voltage across L (V_L) leads i by \(\frac { \pi }{ 2 } \) and voltage across C (V_c) lags behind i  by \(\frac { \pi }{ 2 } \)
(iv) The phasor diagram is drawn with current as the reference phasor. The current is represented by the phasor \(\vec { OI } \), V_R by \(\vec { OA } \); V_L by \(\vec { OB } \); V_c by \(\vec { OC } \) as shown in Figure.
(v) The length of these phasors are OI = I_m, OA = I_mR, OB = I_mX_L; OC = I_mX_C 
The circuit is either effectively inductive or capacitive or resistive that depends on the  value of V_L or V_C. Let us assume that V_L>V_C so that nef voltage drop across L-C combination is V_L - V_C which is represented by a phasor \(\vec { OD } \)
(vi) By parallelogram law, the diagonal \(\vec { OE } \) gives the resultant voltage ሀ of V_R and (V_L - V_C) and its length OE is equal to V_m Therefore,
V²_m = V²_R + (V_L - V_C)² = \(\sqrt { { ({ { I }_{ m }R) } }^{ 2 }+{ ({ I }_{ m }{ X }_{ L }-{ I }_{ m }{ X }_{ C }) }^{ 2 } }=I_m \sqrt {R^2+({X_L-X_C)}^2}\) 
or  \({ I }_{ m }=\frac { { V }_{ m } }{ \sqrt { R^{ 2 }+({ { X }_{ L }-{ X }_{ C }) }^{ 2 } } } \)   ......(2)
\((or) { I }_{ m }=\frac { { V }_{ m } }{ Z } \) where z = \(\sqrt { { R }^{ 2 }+({ { X }_{ L }-{ X }_{ C }) }^{ 2 } } \)   ......(3)
(vii) Z is called impedance of the circuit which refers to the effective opposition to the circuit current by the series RLC circuit. The voltage triangle and impedance triangle are given in the Figure.


(viii) From phasor diagram, the phase angle between v and i is found out from the following relation
\(tan\phi =\frac { V_{ L }-{ V }_{ C } }{ { V }_{ R } } =\frac { X_{ L }-{ V }_{ C } }{ R } \)
Special cases:
(i) If X_L > X_C (X_L - X_C) is positive and phase angle \(\phi \) is also positive. It means that the applied voltage leads the current by \(\phi \) (or current lags behind voltage by \(\phi\)). The circuit is inductive.
∴v = V_m sin ωt; i = I_m sin(ωt - \(\phi \))
(ii) If X_L < X_C (X_L - X_C) is negative and\(\phi \) is also negative. Therefore current leads voltage by \(\phi \) (or voltage lags behind current by\(\phi \)) and the circuit is capacitive.
∴ = V_m sin ωt; i = I_m sin(ωt + \(\phi \))
(ii) If X_L = X_C \(\phi \) is zero. Therefore current and voltage are in the same phase and the circuit is resistive
∴v = V_m sin ωt, i = I_m sinωt 

Conservation of energy:
(i) The truth of Lenz's law can be established on the basis of the law of conservation of energy. The explanation is as follows:
(ii) According to Lenz's law, when a magnet is moved either towards or away from a coil, the induced current produced opposes its motion
(iii) As a result, there will always be a resisting force on the moving magnet.
(iv) Work has to be done by some external agency to move the magnet against this resisting force
(v) Here the mechanical energy of the moving magnet is converted into the electrical energy which in turn, gets converted into Joule heat in the coil i.e., energy is converted from one form to another.
(vi) On the contrary to Lenz's law, let us assume that the induced current helps the cause responsible for its production.
(vii) Now When we push the magnet litle bit towards the coil, the induced current helps the movement of the magnet towards the coil.
(viii) Then the magnet starts moving towards the coil without any expense of energy. This, becomes a perpetual motion machine.
(ix) In practice, no such machine is possible. Therefore, the assumption that the induced current helps the cause is wrong.

(i) Consider a closed circuit consisting of a coil C of insulated wire and a galvanometer G. The galvanometer does not indicate deflection as there is no electric current in the circuit.
(ii) When a bar magnet is inserted into the stationary coil, with its north pole facing the coil, there is a momentary deflection in the galvanometer. This indicates that an electric current is set up in the coil. If the magnet is kept stationary inside the coil, the galvanometer does not indicate deflection.
(iii) The bar magnet is now withdrawn from the coil, the galvanometer again gives a momentary deflection but in the opposite direction. So, the electric current flows in opposite direction. Now if the magnet is moved faster, it gives a larger deflection due to a greater current in the circuit.
(iv) The ar magnet is reversed, i.e., the south pole now faces the coil. When the above experiment is repeated, the deflections are opposite to that obtained in the case of north pole.
(v) If the magnet is kept stationary and the coil is moved towards or away from the coil, similar results are obtained. It is concluded that whenever there is a relative motion between the coil and the magnet, there is deflection in the galvanometer, indicating the electric current setup in the coil.

MaxWell's equations in integral form
i) Gauss law in electricity, \(\oint _s\vec{E} \vec{d} A=\frac{Q_{\text {enclosed }}}{\varepsilon_{o}}\)
ii) Gauss law in magnetism \(\oint _s \vec{B} \cdot \vec{d} A=0\)
iii) Faraday's law \(\oint_l \vec E. \vec {d l}=-\frac{d \phi _B}{d t}\)
iv) Ampere-Maxwell's law \(\oint_l \vec {B}. \vec {d l}=\mu_{o} i_c+\mu_{o} \varepsilon_{o} \frac{d}{d t} \oint_s \vec{E} \cdot {d} \vec A\)