Number of turns, N = 500
Area of cross section, A = 30 x 30 x 10^-4 
= 900 x 10^-4 m²
Magnetic field, B = 0.4 T
Angle of inclination θ = 90⁰ - 30⁰= 60⁰
∴ Magnetic flux Φ = NAB cos θ
∴ Φ = 500 x 900 x 10^-4 x 0.4 x cos60⁰
= 45 x 10⁴ x 10^-4 x 4 x 10^-1 x \(\frac{1}{2}\)
Φ= 9 x 10^-1 = 9.0 Wb
∴ Magnetic flux Φ = 9.0 Wb

The average value of alternating current is defined as the average of all values of current over a positive half-cycle or negative half-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.

When an electric current passing through a coil changes with time, an emf is induced in the neighboring coil. This phenomenon is known as mutual 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.

Whenever the magnetic flux linked with a closed coil changes, an emf is induced and hence an electric current flows in the circuit. This current is called an induced current and the emf giving rise to such current is called an induced emf. This phenomenon is known as electromagnetic induction.

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 \(\mu \mathrm{Wb}\).

i = 77 sin 314t ; t = 2 ms = 2 x 10^-3 s
The general equation of an alternating current is i = I_m sinωt. On comparison,
(i) Peak current, I_m = 77A
(ii) Frequency, \(f=\frac { \omega }{ 2\pi } =\frac { 314 }{ 2\times 3.14 } =50Hz\)
(iii) Time period, \(T-\frac { 1 }{ f } =\frac { 1 }{ 50 } =0.02s\)
(iv) At t = 2 m s, Instantaneous current, i = 77sin(314 x 2 x 10⁻³)
\( =77 \sin \left(314 \times 2 \times 10^{-3} \times \frac{180^{\circ}}{3.14}\right) \)
\(=77 \sin 36^{\circ}=77 \times 0.5878 \)
= 45.26 A

\(L=\frac{20}{\pi^2} \mathrm{H}, \mathrm{f}=50 \mathrm{~Hz}  \)
\(f=\frac{1}{2 \pi \sqrt{L C}}  \)
\(50=\frac{1}{2 \pi \sqrt{\frac{20}{\pi^2} \times C}}  \)
\(50=\frac{1}{2 \times \sqrt{20 C}}  \)
\(\therefore(50)^2=\frac{1}{4 \times 20 C}  \)
\(\therefore C=\frac{1}{2500 \times 4 \times 20}=5 \times 10^{-6}=5 \mu \mathrm{F}\)

L = 20 x 10^-3H. C = 50 x 10^-6 F, R= 40Ω
enf V = 10 sin 340 t
\(\therefore V_0=10 \mathrm{~V}, \omega=340  \)
\(X_1=1 \omega^{\prime}=20 \times 10^3 \times 340  \)
\(=6800 \times 10^{-1}=6.8 \Omega  \)
\(X_C=\frac{1}{C .}  \)
\(=\frac{1}{50 \times 10^{-\alpha} \times 340}=\frac{10^{\circ}}{17000}=\frac{10^{\prime}}{17}=58.823 \Omega   \)
\(Z=\sqrt{R^2+\left(X_6-X_1\right)^2}  \)
\(=\sqrt{(40)^2+(58.82-6.8)^2}  \)
\(=\sqrt{(40)^2+(52.02)^2}  \)
\(=65.62 \Omega\)
The peak current in the circuit is,
\(I_0=\frac{V_0}{Z}=\frac{10}{65.62}  \)
\(\cos 0=\frac{R}{Z}=\frac{40}{65.62} \)
\(\text{Power loss in A.C. circuit }=V_{r m} 1_{r \rightarrow \infty} \cos \phi \)
\(=\frac{1}{2} V_{\mathrm{o}} I_{\mathrm{c}} \cos \phi  \)
\(=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}\)
\(\frac{2000}{4305.98}\)
= 0.46 W

\(\Phi = \frac{\pi}{3}-\frac{\pi}{3}=0 \)
Power factor = cosФ = cos 0 = 1

\(\phi_B =10 t^2-50 t+250  \)
\(e  =\frac{-d \phi_B}{d t}  \)
\(=\frac{-d}{d t}\left(10 t^2-50 t+250\right) \)
=-(20 t - 50) 
=-20 t + 50
When, t = 3 s, e =-20(3) + 50 = -60 + 50
e = -10V

(i) There is a difficulty during power transmission. A sizable fraction of electric power is lost due to Joule heating (FR) in the transmission lines which are hundreds of kilometer long. This power loss can be tackled either by reducing current I or by reducing resistance R of the transmission lines. The resistance R can be reduced with thick wires of copper or aluminium. But this increases the cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable.

(ii) Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing power losses to a greater extent.
(iii) At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer.
(iv) Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss.
Illustration:
An electric power of 2 MW is transmitted to a place through transmission lines of total resistance, say R = 40 Ω, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases.
Case (i):
\(P=2 \mathrm{MW} ; R=40 \Omega ; V=10 \mathrm{kV}\)
Power, P = V I
∴ Current, \(I=\frac{P}{V}=\frac{2 \times 10^6}{10 \times 10^3}=200 \mathrm{~A}\)
Power loss = Heat produced \(=\mathrm{I}^2 \mathrm{R}=(200)^2 \times 40=1.6 \times 10^6 \mathrm{~W}\)
\(% of power loss =\frac{1.6 \times 10^6}{2 \times 10^6} \times 100 \%\)\(\% of \ power \ loss =\frac{1.6 \times 10^6}{2 \times 10^6} \times 100 \%\)
\(=0.8 \times 100 \%=80 \%\)
Case (ii):
\(P=2 \mathrm{MW} ; R=40 \Omega ; V=100 \mathrm{kV}  \)
\(\therefore  \text {Current, } I=\frac{P}{V}=\frac{2 \times 10^5}{100 \times 10^3}=20 \mathrm{~A}\)
Power loss = PR= (20)² x 40 = 0.016 x 10⁶ W
\(\% of \ power \ loss =\frac{0.01.6 \times 10^6}{2 \times 10^6} \times 100 \%\times 0.008 \%\times 100 \%=0.8\%\)
Thus, it is clear that when an electric power is transmitted at higher voltage, the power loss is reduced to a large extent.

Principle:
The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction:
(i) In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core.
(ii) The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.
(iii) The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.
Working:
(i) If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core.
(ii) If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coil.
(iii) This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.
(iv) As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil v_p or back of ε_p is given by,
v_p = ε_p = -N_p \(\frac{dФ_B}{dt}\)             ........(1)
(vi) The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil ε_s is given by ,
ε_s = -N_s\(\frac{dФ_B}{dt}\)  
Where N_p and N_s are the number of turns in the primary and secondary coil respectively. If the secondary circuit is open, then ε_s = ሀ_s  where u is the voltage  ሀ_s across secondary coil
ሀ_s = ε_s = -N_s \(\frac{dФ_B}{dt}\)               ........(2)
From equations (1) and (2),
\(\frac{v_s}{v_p}=\frac{N_s}{N_p}\) = K                    ........(3)
(vii) This constant K is known as voltage transformation ratio. For an ideal transformer, input power v_pi_p= Output power v_si_s 
where i_p and i_s are the currents in the primary and secondary coil respectively.
(ix) Therefore,
\(\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}\)             ........(4)
Equation (4) is written in terms of amplitude of corresponding quantities
\(\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}\) = K
i) If N_s> N_p (or K > 1), ∴ V_s > V_p and I_s <  I_p This is the case of step-up transformer in which voltage is increased and the corresponding current is decreased.
ii) If N_s< N_p (or K < 1), ∴ V_s < V_p and I_s >  I_p This is step-down transformer where voltage is decreased and the current is increased.
Efficiency of a transformer:
The efficiency η of a transformer is defined as the ratio of the useful output power to the input power. Thus,
\(η=\frac{Outpur \ power}{Input \ power}\times100% \) %          ....(5)

(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.
 

(b) Self Inductance

(d) Power loss

(d) None

(a) I only