When the frequency of the applied alternating source (ω_r) is equal to the natural frequency \(\left[ \frac { 1 }{ \sqrt { LC } } \right] \) of the RLC circuit, the current in the circuit reaches its maximum value. Then the circuit is said to be in electrical resonance. The frequency at which resonance takes place is called resonant frequency.
Resonant angular frequency, ω_r = \(\frac { 1 }{ \sqrt { LC } } \)

The resistance offered by the inductor, called inductive reactance (X_L). It is measured in ohm.
X_L = ωL
X_L = 2πfL
where f is the frequency of the alternating current. For a steady current, f = 0. Therefore, X_L = 0. Thus an ideal inductor offers no resistance to steady DC current.

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.

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.

From right hand rule, the magnetic field by the straight wire is directed into the plane of the square loop perpendicularly and its magnetic flux is decreasing. The decrease in flux is opposed by the current induced in the loop by producing a magnetic field in the same direction as the magnetic field of the wire. Again from right hand rule, for this inward magnetic field, the direction of the induced current in the loop is clockwise.

Yes! An emf will be induced in the wire because it moves perpendicular to the horizontal component of Earth’s magnetic field and hence it cuts the magnetic lines of Earth's magnetic field.

The magnetic field of a cylindrical magnet is symmetrical about its axis. As the magnet is rotated along the axis of the solenoid, there is no induced current in the solenoid because the flux linked with the solenoid does not change due to the rotation of the magnet.

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.

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

\(\mathrm{V}_{\mathrm{P}}=220 \mathrm{~V}, \mathrm{~V}_{\mathrm{s}}=11 \mathrm{~V}  \)
\(\mathrm{I}_{\mathrm{P}}=6 \mathrm{~A}, \mathrm{I}_{\mathrm{s}}=100 \mathrm{~A} .  \)
\(\text {Efficiency }=\frac{\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}  \)
\(=\frac{11 \times 100}{220 \times 6}=\frac{1100}{220 \times 6}=\frac{5}{6}=0.83\)

\(A_1  =4 \times 10^{-4} \mathrm{~m}^2  \)
\(N_1  =10 \text { turns }  \)
\(A_2  =4 \times 10^{-4} \mathrm{~m}^2  \)
\(N_2  =15 \times 10^{-}=1500 \mathrm{turn} / \mathrm{m}  \)
\(\phi  =B_2 A_2=\left(\mu_0 \mathrm{n}_2 I_2\right) \mathrm{A}_1 \)
\(Where, \mathrm{n}_2=\frac{\mathrm{N}_2}{l}=1500 \mathrm{turn} / \mathrm{m}\)
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.

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

(i) S₁ and S₂ are two long solenoids each length l. The solenoid S₂ is wound closely over the solenoid S_1.
(ii) N₁ and N₂ are the number of turns in the solenoids S₁ and S₂ respectively. Both the solenoids are considered to have the same area of cross-section A as they are closely wound together.

(iii) I₁ is the current flowing through the solenoid S₁ The magnetic field B₁ produced at any point inside the solenoid S₁ due to the current I₁ is
\({ B }_{ 1 }={ \mu }_{ o }\frac { { N }_{ i } }{ l } { I }_{ 1 }\) ...(1)
(iv) The magnetic flux linked with each turn of S₂ is equal to B₁A.
Total magnetic flux linked with solenoid S₂ having N₂ turns is
Φ₂ = \(\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }{ AI }_{ 1 } }{ l } \) ...(2)
But Φ₂ = MI₁ ...(3) 
where M is the coefficient of mutual induction between S₁ and S₂
From equations (2) and (3),
MI₁ = \(\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }{ AI }_{ 1 } }{ l } \); M = \(​​\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }A }{ l } \)
(v) If the core is filled with a magnetic material of permeability μ,
M = \(​​\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }A }{ l } \)

By differentiating equation twice, we get
q(t) Q_m cos(ωt + Φ) ...(1)
\(\frac { { d }^{ 2 }q }{ { dt }^{ 2 } } \) = - Q_m ω² cos (ωt + Φ) ...(2)
Substituting equations (1) and (2) in equation
\(\frac { dU }{ dt } =\frac { 1 }{ 2 } L\left( 2i\frac { di }{ dt } \right) +\frac { 1 }{ 2C } \left( 2q\frac { dq }{ dt } \right) =0\)
(or) \(L\frac { { d }^{ 2 }q }{ { dt }^{ 2 } } +\frac { 1 }{ C } q=0\)
we obtain
L[- Q_m ω² cos (ωt + Φ)] + \(\frac { 1 }{ C } \) Qm cos (ωt + Φ) = 0
Rearranging the terms, the angular frequency of LC oscillations is given by
ω = \(\frac { 1 }{ \sqrt { LC } } \) 
This equation is the same as that obtained from qualitative analogy.