The magnetic flux through an area A in a magnetic field is defined as the number of magnetic field lines passing through that area normally and is given by the equation.
\({ \Phi }_{ B }=\int _{ A }^{ }{ \vec { B } .d\vec { A } =BAcos\theta } \)
where the integral is taken over the area A and θ is the angle between the direction of the magnetic field and the outward normal to the area.

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

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 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.

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 \(\mu \mathrm{Wb}\).

Current lags behind the applied voltage by \(\frac{\pi}{2}\) in an inductive circuit.

Current leads the applied voltage by \(\frac{\pi}{2}\) in a capacitive circuit.
This is the resistance offered by the capacitor, called capacitive reactance (X_c). It measured in ohm.
\({ X }_{ c }=\frac { 1 }{ \omega C } \)
The capacitive reactance (X_c) varies inversely as the frequency. For a steady current, f = 0
∴\({ X }_{ c }=\frac { 1 }{ \omega C } -\frac { 1 }{ 2\pi fC } =\frac { 1 }{ 0 } =\infty \)
Thus a capacitive circuit offers infinite resistance to the steady current.

Current lags behind the applied voltage  \(\frac{\pi}{2}\) in an inductive circuit. This fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by 90°.

Phasor: A sinusoidal alternating voltage (or current) can be represented by a vector that rotates about the origin in an anti-clockwise direction at a constant angular velocity ω. Such a rotating vector is called a phasor. A phasor is drawn in such a way that
(i) the length of the line segment equals the peak value Vm (or I_m) of the alternating voltage (or current)
(ii) its angular velocity w is equal to the angular frequency of the alternating voltage (or current)
(iii) the projection of phasor on any vertical axis gives the instantaneous value of the alternating voltage (or current)
(iv) the angle between the phasor and the axis of reference (positive x-axis) indicates the phase of the alternating voltage (or current).
The notion of phasors is introduced to analyze the phase relationship between voltage and current in different AC circuits.
Phasor diagram
The diagram which shows various phasors and their phase relations is called the phasor diagram.

A = 5 x 10^-2 m²; B = 0.2 T
(i) θ = 0°;
\({ \Phi }_{ B }=BAcos\theta =0.2\times 5\times { 10 }^{ -2 }\times { cos }0^{ o }\)
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }Wb\)
(ii) θ = 90° – 60° = 30°;
\(\Phi_B\) = BAcosθ  = 0.2 x 5 x 10^-2 x cos 30^o
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }\times \frac { \sqrt { 3 } }{ 2 } =8.66\times { 10 }^{ -3 }Wb\)
(iii) θ = 90°;
\({ \Phi }_{ B }\) = BA cos90^o = 0

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

Solution
e = -L\(\frac{dl}{dt}\)

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

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

The direction of conventional current is always opposite to the direction of flow of electrons.

(b) Self Inductance

(d) Power loss