(i) R is not affected by frequency so current does not change on an increasing frequency.
(ii) Inductive reactance X_L = 2π γL γ when frequency increases ~ increases and hence current in the circuit decreases.
(iii) Capacitive reactance X_c = \({ X }_{ c }=\frac { 1 }{ 2\pi \gamma c } \)
As the frequency ⋎ is increased, X_c decreases, and hence current increases.

If the loop moves within the region of a uniform magnetic field no current will be induced in the loop. Hence no direction.

The current in the coil will flow clockwise the direction of current will be from P to Q in coil PQ and from C to D in coil CD.

The current in the series RLC circuit becomes maximum at resonance. Due to the increase in current, the voltage across L and C are also increased. This magnification of voltages at series resonance is termed as Q-factor. It gives a measure of sharpness of current peak in the resanance condition of LCT circuit

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.

Magnetic field B = 4 x 10^-3 T
Induced emf, e = 0.02V = 2 x 10^-2 V
Length of a blade, I= 0,4; m = 4 x 10^-1 m
Change in magnetic flux, dΦ = Bds
∴ dΦ = B x πl²
we know that, dt = \(​​\frac{2\pi}{\omega}\)
∴ Magnitude of induced emf, e \(=​​|-\frac{dΦ}{dt}|\)
\(e=\frac{dΦ}{dt}=\frac{B\pi l^2}{ 2\pi l \omega} =\frac{1}{ 2}Bl^2\omega\)
\(=\frac{1}{ 2}Bl^2 \times 2\pi v= \pi Bl^2v \quad (\because \omega=2\pi v)\)
Rate of rotation of the blade, \(v=\frac{e}{\pi Bl^2}\)
\(v=\frac { 2.10^{-2} }{ 3.14\times 4\times 10^{ -3 }\times(4 \times10^{-1})^2 } \)
\(=\frac { 2 \times10^{-2} }{ 3.14 \times64\times10^{-5} } =\frac{2\times10^{-2+5}}{200.96}\) 
v = 0.00995 x 10³
v = 9.95 rev/s
Rate of rotation of the blade = 9.95 revolutions /second

Area of a rectangular coil, \(A=6 \times 10^{-4} \mathrm{~m}^2, Resistance \ \mathrm{R}=35 \Omega\)
Number of turns of the coil, N = 3500 turns
Magnetic field, B = 0.4 T
Angle of orientation, \(\theta=\pi-\frac{\pi}{2}=\frac{\pi}{2} \mathrm{rad}=90^{\circ}\)
Induced emf, e=N A B sin θ
\(e=3500 \times 6 \times 10^{-4} \times 0.4 \times \sin 90^{\circ}\)
\(=3500 \times 2.4 \times 10^{-4} \times 1  \)
\(=840 \times 10^{-3} \mathrm{V} \)
\(I=\frac{e}{R}=\frac{1440 \times 10^{-1}}{35}  \)
\(I=24 \times 10^{-3} A\)
Charge, Q = It
\(=24 \times 10^{-3} \times 2  \)
\(=48 \times 10^{-3} \mathrm{C} \)
\(\therefore Charge Q=48 \times 10^{-3} \mathrm{C}\)

Change in magnetic flux, dф = 4 x 10^-3 Wb
Change in time, dt = 0.4 s
Magnitude of Induced emf \(= |\frac { -d\Phi }{ dt }|=\frac { d\Phi }{ dt }\)
\(=\frac { 4\times 10^{ -3 } }{ 0.4 } \) = 10 x 10^-3 V = 10 mv
∴ Magnitude of induced emf =10 mV

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
\(I_{RMS}=\sqrt\frac{\text {Area of one cycle of squared wave}}{\text {Base length of one cycle}}\)

(i) The current is drawn directly from fixed terminals on the stator without the use of brush contacts.
(ii) The insulation of stationary armature winding is easier.
(iii) The number of sliding contacts (slip rings) is reduced. Moreover, the sliding contacts are used for low-voltage DC Source.
(iv) Armature windings can be constructed more rigidly to prevent deformation due to any mechanical stress.

An electric current flowing through a coil will set up a magnetic field around it. Therefore the magnetic flux of the magnetic field is linked with that coil it self. If this flux is changed by changing the current, an emf is induced in that same coil. This phenomenon is known as self-induction.

Emf can be produced by changing magnetic flux in any of the following ways:
(i) By changing the magnetic field B
(ii) By changing the area A of the coil and
(iii) By changing the relative orientation θ of the coil with magnetic field.

The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions. If the index finger points the direction of the magnetic field and the thumb indicates the direction of motion of the conductor, then the middle finger will indicate the direction of the induced current.

L_air  = 4.8 x 10^-3H
L_iron = 1.8H
L_air = \(\mu_{o}\)n²Al = 4.8 x 10^-3H
L_iron = \(\mu_{o}\)n²Al = \(\mu_{o}\mu_r\)n²Al = 1.8H
\(\therefore { \mu }_{ r }=\frac { { L }_{ iron } }{ { L }_{ air } } =\frac { 1.8 }{ 4.8\times { 10 }^{ -3 } } =375\)

Consider a small element of length dx at a distance x from the centre of the circle described by the rod. As this element moves perpendicular to the field with a linear velocity v = xω, the emf developed in the element dx is dε = Bvdx = B(xω)dx
This rod is made up of many such elements, moving perpendicular to the field. The emf developed across two ends is
\(\epsilon =\int { d\epsilon } =\int _{ 0 }^{ l }{ B\omega xdx } =B\omega { { \left[ \frac { { x }^{ 2 } }{ 2 } \right] } }_{ 0 }^{ l }\)
\(\epsilon =\frac { 1 }{ 2 } B\omega { l }^{ 2 }\)

N = 40 turns; B = 2 Wb m^-2
A = 200 cm² = 200 x 10^-4 m²;
Initial flux, \(\Phi_i\) = BA cos\(\theta\)
= 2 x 200 x 10^-4 x cos60^o
since θ = 90°− 30°= 60°
\(\Phi_i\)= 2 x 10^-2 Wb
Final flux, \(\Phi_f\) = BA cos\(\theta\)
= 2 x 200 x 10^-4 x cos0^o since \(\theta\) = 0^o
\(\Phi_f\) = 4 x 10^-2Wb
Magnitude of the induced emf is
\(ε =N\frac { d{ \Phi }_{ B } }{ dt } \)
\(=\frac { 40\times (4\times { 10 }^{ -2 }-2\times { 10 }^{ -2 }) }{ 0.2 } =4V\)

B = 4.1 x 10^–5 T; θ = 90^o – 47^o = 43° ;
A = 3m²
We know that   \(\Phi_{B}=B A \cos \theta\) 
\(\Phi_{\mathrm{B}}\) = 4.1 x 10^–5 x 3 x cos 43^o  
= 4.1 x 10^–5 x 3 x 0.7314
= 89.96 \(\mu \mathrm{Wb}\).

Self inductance of a coil is the inductance that enables to poduce an opposing induced emf in it, when the current in the coil changes 
Physical significances :
The inductance plays the same role in a circuit as mass and moment of inertia play at a mechanical motion. When a circuit is switched on, the increasing current induces an emf which opposes the growth of current in a circuit (Figure (a)). Likewise, when circuit is broken, the decreasing current induces an emf in the reverse direction. This emf now opposes the decay of current (Figure (b)).
Thus, the inductance of the coil opposes any change in current and tries to maintain the original state

A = 70 x 10^-4m²; N = 600 turns
B = 0.4 Wbm^-2; f = 500 rpm
The instantaneous emf is
ε = ε_msinωt
since \({ \epsilon }_{ m }=N{ \Phi }_{ m }\omega =N(BA)(2\pi f)\)
ε = NBA x 2\(\pi\)f x sinωt
(i) When ωt = 0^o,
ε = ε_m sin0 = 0
(ii) When ωt = 90^o,
ε = ε_msin90^o = NBA x 2\(\pi\)f x 1
= 600 x 0.4 x 70 x 10^-4 x 2 x \(\frac{22}{7}\times(\frac{500}{60})\)
= 88V
(iii) When ωt = 90° – 60° = 30°,
ε = ε_m sin30^o = 88 x \(\frac{1}{2}\) = 44V

\(Q_{midpoint}=\frac{Q}{\sqrt{1^2+1^2}}=\frac{Q}{\sqrt2}\)

\(\omega=250 \mathrm{rad} / \mathrm{s}, C=4 \times 10^{-} \mathrm{F}  \)
\(R=100 \Omega, \quad \mathrm{V}_{\mathrm{R}}=40 \mathrm{~V}  \)
\(\therefore I_{\mathrm{R}}=\frac{V_R}{100}=\frac{40}{100}=0.4 \mathrm{~A}  \)
\(\omega=\frac{1}{\sqrt{L C}}  \)
\(\omega^2=\frac{1}{L C}  \)
\((250)^2=\frac{1}{L \times 4 \times 10^{-6}}  \)
\(L=\frac{1}{4 \times(250)^2 \times 10^{-6}}  \)
\(=\frac{1}{4 \times 250 \times 250 \times 10^{-6}}  \)
\(=\frac{1}{1000 \times 10^{-6} \times 250}  \)
\(=\frac{10^3}{250}=\frac{1000}{250}=4 \mathrm{H}\)
Voltage acnoss L, V_t = IX_L
V_L = l x L x ω
= 0.4 x 4 x 250
0.4 x 1000 = 400 V

\(A_1  =4 \times 10^{-4} \mathrm{~m}^2  \)
\(N_1  =10 \text { turns }  \)
\(A_2  =4 \times 10^{-4} \mathrm{~m}^2  \)
\(N_2  =15 \times 10^{-}=1500 \mathrm{turn} / \mathrm{m}  \)
\(\phi  =B_2 A_2=\left(\mu_0 \mathrm{n}_2 I_2\right) \mathrm{A}_1 \)
\(Where, \mathrm{n}_2=\frac{\mathrm{N}_2}{l}=1500 \mathrm{turn} / \mathrm{m}\)
The mutual Inductance is,
\(M  =\frac{N_1 o_{12}}{I_2}=\mu_0 n_2 N_1 A_1  \)
\(=4 \pi \times 10^{-7} \times 1500 \times 10 \times 4 \times 10^{-4}  \)
\(=7.54 \times 10^{-6} \mathrm{H}=7.54 \mu \mathrm{H}\)

\(\text {emf } e=8 \mathrm{~V}  \)
\(d I=I_1-I_0=2-(-2)=4 \mathrm{~A}  \)
\(\text {dt }=0.05 \mathrm{~s}  \)
\(L=\frac{-e}{d I / d t}=\frac{-8}{4 / 0.05}  \)
\(=\frac{-8 \times 0.05}{4}=\frac{-0.40}{4}  \)
=-0.1 H
-ve sign indicates that self-induced emf always opposes the current w.r.t. time.

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

I = 40cm = 0.4m
A = 20 cm² = 20 x 10^-4 m²
N = 800, L =?
\(L=\frac { { \mu }_{ o }{ N }^{ 2 }{ A } }{ l } \) = \(\frac { 4\pi \times 3\times { 10 }^{ -7 }\times { \left( 800 \right) }^{ 2 }\times 20\times { 10 }^{ -4 } }{ 0.4 } \)
L = 4.02 x 10^-2H

Given:
We know that e = -\(\frac { d\Phi }{ dt } \)
e = \(\frac{d}{dt}\) (5t³ + 4t² + 2t)
e = 15t² + 8t + 2
for t = 2s, e = 15 x (2)² + 8 (2) + 2
e = 78V

Given:
B = 10T, N = 50 turns
A = 100 cm² = 10^-2m²
dt = 0.15S
Magnetic flux linked with the coil initially,
Φ₁ = NBA = 50 x 10 x 10^-2= 5wb
But magnetic flux linked with the coil finally, i.e (when removed from 'B')
Φ₂ = 0
Formula:
∴ emf induced, e = - \(\frac { d\phi }{ dt } \)
= - \(\frac { { (\phi }_{ 2 }-{ \phi }_{ 1 }) }{ dt } =-\frac { (0-5) }{ 0.15 } \)
e = 33.33V

(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. Let us first calculate the power at an instant and then it is averaged over a complete cycle.
(ii) The alternating voltage and alternating current in the series inductive RLC circuit at an instant are given by
v=V_m sinωt and i=I_m=(ωωt+\(\phi \))t+\(\phi \))
(iii) where \(\phi \) is the phase angle between v and i. The instantaneous power is then written as
P=vi =V_mI_m sinωt sin(ωt + \(\phi \))
=V_mI_m sinωt [sin ωt cos\(\phi \) - cosωt sin\(\phi \)]
P=V_mI_m [cos\(\phi \) sin²ωt - sinωt cosωt sin\(\phi \)]   ....(1)
(iv) Here the average of sin²ωt over a cycle is\(\frac{1}{2}\)and that of sin ωt cos ωt is zero. Substituting these values, we obtain average power over a cycle.
P_av =V_mI_m cos\(\phi \) x \(\frac { 1 }{ 2 } \)
=\(\frac { { V }_{ m } }{ \sqrt { 2 } } \frac { { I }_{ m } }{ \sqrt { 2 } } cos​​\phi​​\)
P_av = V_RMS I_RMS cos\(\phi \)  ....(2)
(v) where V_RMS I_RMS is called apparent power and cos\(\phi \) is power factor. The average power of an AC circuit is also known as the true power of the circuit.
Special Cases:
(i) For a purely resistive circuit, the phase angle between voltage and current is zero and cos\(\phi \)=1
∴ P_av =V_RMS I_RMS
(ii) For a purely inductive or capacitive circuit, the phase angle is ± \(\frac { \pi }{ 2 } \) and cos\(\left( \pm \frac { \pi }{ 2 } \right) \)=0
∴ P_av =0
(iii) For series RLC circuit, the phase angle
\(\phi \) =tan^-1\(\left( \frac { { X }_{ L }-{ X }_{ C } }{ R } \right) \)
∴ P_av =V_RMS I_RMS cos\(\phi \)
(iv) For series RLC circuit at resonance, the phase angle is zero and cos\(\phi \)=1
∴ P_av =V_RMS I_RMS

(i) Consider a rectangular coil of turns kept in a uniform magnetic field \(\vec{B}\) as shown in Figure (a). The coil rotates in anti-clockwise direction with an angular velocity  ω about an axis, perpendicular to the field, and to the plane of the paper.
(ii) At time = 0, the plane of the coil is perpendicular to the field and the flux linked with the coil has its maximum value  \({ \Phi }_{ m }=BA\) (where A is the area of the coil).
(iii) In a time t seconds, the coil is rotated through an angle θ (= ωt) in anti-clockwise direction. In this position, the flux linked is NBA cos ωt, is due to the component of B normal to the plane of the coil (Figure(b)). The component parallel to the plane  (B sin ωt) has no role in electromagnetic induction. Therefore, the flux linkage with the coil  at this deflected position is
NΦ_B = NBA cosθ = NBA cos ωt
According to Faraday's law, the emf induced at that instant is

\(ε=- \frac { d }{ dt } (N{ \Phi }_{ B })=-\frac { d }{ dt } (NBAcos\omega t)\)
= -NBA (-sin ωt)ω
= NBAω sin ωt
(iv) When the coil is rotated through 90⁰ from initial position, sin ωt = 1. Then the maximum value of induced emf is
ε_m = NBAω
Therefore, the value of induced emf at that instant is then given by,
ε = ε_m sinωt   ....(1)

(vii) It is seen that the induced emf varies as sine function of the time angle ωt. The graph between induced emf and time angle for one rotation of coil will be a sine curve (Figure) and the emf varying in this manner is called sinusoidal emf or alternating emf.
If this alternating voltage is given to a closed circuit, a sinusoidally varying current flows in it. This current is called alternating current and is given by,
i = I_m sinωt ....(2)
Where, I_m is the maximum value of induced current.

(a) Induction stove
(i) Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire.
(ii) The cooking pan made of suitable material, is placed over the cooking zone. When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan.
(iii) The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.
(b) Eddy current brake
(i) This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails.
(ii) To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.
(c) Eddy current testing
(i) It is one of the simple non-destructive testing methods to find defects like surface cracks, air bubbles present in a specimen.
(ii) A coil of insulated wire is given an alternating electric current, so that it produces an alternating magnetic field.
(iii) When this coil is brought near the test surface, eddy current is induced in the test surface.
(iv) The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.
(d) Electro magnetic damping:
(i) The armature of the galvanometer coil is wound on a soft iron cylinder.
(ii) Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder.
(iii) The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.
 

(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.