Power factor will be maximum, when Φ = 0 i.e., \(\tan ^{-1}\left(\frac{X_L-X_C}{R}\right)=0\)
\(\therefore X_L=X_C \Rightarrow L \omega=\frac{1}{C \omega}  \)
\(\therefore \omega=\frac{1}{2 \pi \sqrt{L C}}\)
∴ Current I be max \(I_m=\frac{V_m}{R}\)
Hence power factor of a RLC series circuit becomes maximum, when
(i) \(X_L=X_C\)
(ii) Current \(I_m=\frac{V_m}{R}\) will be maximum
(iii) Frequency \(\omega_r=\frac{1}{2 \pi \sqrt{L C}}\)

In the inductor voltage leads current by 90°

In the capacitor current leads voltage by 90°

Hence in series LC circuit, the voltage across L and C are 180° out of phase

The plate A is positively charged
The plate B is negatively charged.

The direction of induced current is, a → b, b → c, c → d and d → a

(i) Current is induced in the anticlockwise direction in the upper coil. Then only as per Lenz's law, the induced current will oppose the motion of the magnet.
(ii) In the lower coil, current is induced in the clockwise direction. As per Lenz's law the direction of induced emf will in opposite direction of motion of the magnet.

Magnitude of induced \(\mathrm{emf}|\mathrm{e}|=\left|-\frac{d \phi}{d t}\right|\)
\(\therefore c=\frac{d \phi}{d t}\)
(i) In the region ab,
\(c_1=\frac{d \phi_1}{d t}=\frac{4-0}{1-0}=\frac{4}{1}=4 \mathrm{~V}\)
(ii) In the region bc,
\(c_2=\frac{d \phi_2}{d t}=\frac{4-4}{3-1}=\frac{0}{2}=0 \mathrm{~V}\)
(iii) In the region cd,
\(e_3=\frac{d \phi_2}{d t}=\frac{4-2}{4-3}=\frac{2}{1}=2 \mathrm{~V}\)
(iv) In the region de,
Hence,
\(e_4=\frac{d \phi_4}{d t}=\frac{2-0}{7-4}=\frac{2}{3}=0.66 V  \)
\(\mathrm{e}_2<\mathrm{e}_4<\mathrm{e}_3<\mathrm{e}_1\)
In the ascending order of the magnitude of induced emf
Region bc < Region de < Region cd < Region ab.

Angle, \( \theta=60^{\circ}\)
Peak value of current, \(I_{P_m}=20 \mathrm{~A}\)
(i) Instantaneous value of current at 60^o,
\(i  =I_m \sin \theta  \)
\(=20 \times \sin 60^{\circ}=\frac{20 \times \sqrt{3}}{2}  \)
\(=10 \times 1.732  \)
\(i  =17.32 \mathrm{~A}\)
(ii) Average value of current,
\(I_{\mathrm{av}}=0.637 I_m  \)
\(I_{\mathrm{av}}=0.637 \times 20  \)
\(I_{\mathrm{av}}=12.74 \mathrm{~A}\)
(iii) RMS value of current,
\(I_{\mathrm{rms}}=0.707 I_m=0.707 \times 20  \)
\(I_{\mathrm{rms}}=14.14 \mathrm{~A}\)

Current, I = 4 A, Magnetic flux, \(\phi=6 \times 10^{-5} \mathrm{~Wb}\), Number of turns, N = 200
Self inductance, \(L=\frac{N \phi}{I}\)
\(\therefore L  =\frac{200 \times 6 \times 10^{-5}}{4}  \)
\(L  =300 \times 10^{-5} \mathrm{H}=3 \times 10^{-3} \mathrm{H}\)
∴ Magnetic energy, \(U_s=\frac{1}{2} L I^2\)
\(U_s  =\frac{1}{2} \times 3 \times 10^{-3} \times(4)^2  \)
\(=\frac{1}{2} \times 3 \times 10^{-3} \times 16  \)
\(=24 \times 10^{-3} \mathrm{~J}=0.024 \mathrm{~J}\)
Magnetic energy = 0.024 J

Relative permeability of air-core solenoid is μ_r = 1
Length of the solenoid, I = 2 m, Number of turns, N = 4000
Diameter of the solenoid, d = 4 x 10^-2m
Area of solenoid, A = \(\frac{\pi d^2}{4}\)
\(A =\frac{3.14 \times\left(4 \times 10^{-2}\right)^2}{4}  \)
\(=12.56 \times 10^{-4} \mathrm{~m}^2\)
Self inductance of a solenoid is, \(L=\frac{\mu_0 \mu_r N^2 A}{l}\)
\(\therefore L  =\frac{4 \pi \times 10^{-7} \times 1 \times(4000)^2 \times 12.56 \times 10^{-4}}{2}  \)
\(=2 \times 3.14 \times 10^{-7} \times 16 \times 10^6 \times 12.56 \times 10^{-4}  \)
\(=1262 \times 10^{-5} \mathrm{H}=12.62 \times 10^{-3} \mathrm{H}  \)
\(\therefore L  =12.62 \mathrm{mH}\)
∴ Self inductance of a solenoid L =12.62 mH

Horizontal component of Earth's magnetic field, \(B_H=4 \times 10^{-5} \mathrm{~T}\)
Length of a spoke, l = 1 m,
Induced emf, \(e = 31.4 \times 10^{-3} \mathrm{~V}\)
Change in magnetic flux, \(d \phi=B d s\)
\(\therefore d \phi=B \times \pi l^2\)
we know that, \(d t=\frac{2 \pi}{\omega}\)
\(\therefore \text {Magnitude of induced emf, } e  =\left|-\frac{d \phi}{d t}\right|  \)
\(e  =\frac{d \phi}{d t}=\frac{B \pi l^2}{2 \pi / \omega}=\frac{1}{2} B l^2 \omega  \)
\(e  =\frac{1}{2} B l^2 \times 2 \pi v=\pi B l^2 v \quad(\because \omega=2 \pi v)\)
∴ Rate of revolution, \(v=\frac{e}{\pi B l^2}\)
\(v  =\frac{31.4 \times 10^{-3}}{3.14 \times 4 \times 10^{-5} \times(1)^2} \)
\(=\frac{10 \times 10^{-3+5}}{4}=\frac{10^3}{4}=250 \mathrm{rps}\)
∴ Rate of rotation of the wheel v = 250 revolutions/second.

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

Resistance of a conductor R = 100 Ω
Induced current, i = 2.5 x 10^-3 A
Induced emf, e = iR
e = 2.5 x 10^-3 x 100
e = 250 x 10^-3 V
But, \(e=-\frac { d\Phi }{ dt } =|-\frac { d\Phi }{ dt } |=\frac { d\Phi }{ dt } \)
∴ Rate of change of flux is \(\frac { d\Phi }{ dt } =e\)
250 x 10^-3 wb/s = 250mWb/s
∴ Rate of change of flus = 250 m Wb/s

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 field, dB = (8000 - 2000) = 6000 Wb
Change in time, dt = 6s
Radius of the coil, r = 2 x 10^-2 m
Induced emf, e = 44 V,
Area of the coil A =πr² = \(\frac { 22 }{ 7 } \) x (2 x 10^-2)²
= 12.56 x 10^-4 m²
∴ Number of turns in the coil,
\(N=\frac { e }{ A\frac {dB}{dt }} \frac{44}{12.56\times10^{-4} \times(6000/6)}\)
\(N=\frac { 44 \times10 }{12.56 }=3.503\times10=35 \)
∴ Number of turns in the coil = 35

Induced emf, e = \(\frac { d\Phi _{ B } }{ dt } \)
=\(\frac {- d }{ dt } \) (2t³ + 4t² + 8t + 8) = -(6t² + 8t + 8)
When t= 3 s
e = -(6(3)² + 8(3) + 8) = -(54 + 24 + 8) = -86
Magnitude of emf, e =|-86| = 86 V
Resistance of the coil, R = 5Ω
∴ Induced curent, I =\(\frac { e }{ R } \)
I=\(\frac { 86 }{ 5 } \) = 17.2 A
∴ Induced current through the coil = 17.2 A

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

Whenever energy is given to a circuit containing a pure indicator L and a capacitor of capacitance C, the energy oscillates back and fourth between the magnetic field of the indicator and the electric field of the capacitor. Thus the electrical oscillations of definite frequency are generated. these oscillations are called LC oscillations.

Power factor is defined as the ratio of resistance to the impedance of an AC circuit
Power factor \(\cos \phi=\frac{R}{2}\)
\(=\frac{Resistance}{Impedance}\)

The current in an AC circuit is said to be wattless current if the power consumed by it is zero.

Q factor is defined as the ratio of voltage across L or C to resonance to the applied voltage
Q - factor = \(\frac{Voltage \ across \ L \ or \ C \ resonance }{Applied \ voltage}\)
\(Q-factor=\frac{X_{L}}{R}=\frac{1}{R}\sqrt\frac{{L}}{C}\)

Resonant frequency is the frequency of the applied alternating source at which the current in the circuit reaches its maximum value.

When the frequency of the applied alternating source (ω_r) is equal to the natural frequency \(\left[\frac{1}{\sqrt{L C}}\right]\) of the RLC circuit, the current in the circuit reaches its maximum value. Then, the circuit is said to be in electrical resonance.

A sinusoidal alternating voltage (or current) can be represented by a vector which rotates about the origin in anti-clockwise; direction at a constant angular velocity ω. Such a rotating vector is called a phasor.

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

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.

AC generator work on the principle of electromagnetic induction. The relative motion between a conductor and a magnetic field changes the magnetic flux linked with the conductor which in turn, induces an emf

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.

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

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.

Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

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.

First law:
Whenever magnetic flux linked with a closed circuit changes, an emf is induced in the circuit. Which lasts in the circuit as long as the magnetic flux is changing.
Second law:
The magnitude of induced emf in a closed circuit is equal to the time rate of change of magnetic flux linked with the circuit.

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.

\(C=\frac { { 10 }^{ 2 } }{ \pi } \times { 10 }^{ -6 }F,{ V }_{ RMS }=220V;f=50Hz\)
(i) Capacitive reactance,
\({ X }_{ c }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi fC } \)
\(=\frac { 1 }{ 2\times \pi \times 50\times \frac { { 10 }^{ -4 } }{ \pi } } =100\Omega \)
(ii) RMS value of current,
\(\\ { I }_{ RMS }=\frac { { V }_{ RMS } }{ { X }_{ c } } =\frac { 220 }{ 100 } =2.2\ A\)
(iii) V_m = 220 x \(\sqrt{2}\) = 311V
I_m = 2.2 x \(\sqrt{2}\) = 3.1A
Therefore,
v = 311sin314t
i = 3.1sin\((314t+\frac { \pi }{ 2 } )\)

L = 400 x 10^-3 H; I_eff = 6 x 10^-3A
f = 1000 Hz
Inductive reactance, X_L= L\(\omega\) = L x 2\(\pi\)f
= 2 x 3.14 x 1000 x 0.4
= 2512 Ω
Voltage across L,
V = I X _L = 6 x 10^-3 x 2512
V = 15.072 V (RMS)

Case (i):
di₁ = 10 – 2 = 8 A; dt = 0.4 s;
ε₂ = 60 x 10^-3V
Case(ii):
di₁ = 16 – 4 = 12 A; dt = 0.03 s
(i) Mutual inductance between the coils.
\({ M }=\frac { { \epsilon }_{ 2 } }{ \frac { { di }_{ 1 } }{ dt } } \)
\(=\frac { 60\times { 10 }^{ -3 }\times 0.4 }{ 8 } \)
\({ M }=3\times { 10 }^{ -3 }H\)
(ii) Induced emf in the second coil due to the rate of change of current in the first coil is
\({ \epsilon }_{ 2 }={ M }=\frac { { di }_{ 1 } }{ dt } \)
\(=\frac { 3\times { 10 }^{ -3 }\times 12 }{ 0.03 } \)
ε₂ = 1.2V

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
\(\epsilon =\int { d\epsilon } =\int _{ 0 }^{ l }{ B\omega xdx } =B\omega { { \left[ \frac { { x }^{ 2 } }{ 2 } \right] } }_{ 0 }^{ l }\)
\(\epsilon =\frac { 1 }{ 2 } B\omega { l }^{ 2 }\)

\(\Phi_B\) = (2t³ + 3t² + 8t + 5)mWb; N = 1;t = 3 s
i) \(ε=\frac { d(N{ \Phi }_{ B }) }{ dt } \)
\(=\frac { d }{ t } \left( { 2t }^{ 3 }+{ 3t }^{ 2 }+8t+5 \right) \times { 10 }^{ -3 }\)
= (6t² + 6t + 8) x 10^-3 V
At t = 3 s,
ε = [( 6 x 9) + (6 x 3) + 8] x 10^-3
= 80 x 10^-3V = 80mV
(ii) As time passes, the magnetic flux linked with the loop increases. According to Lenz’s law, the direction of the induced current should be in a way so as to oppose the flux increase. So, the induced current flows in such a way to produce a magnetic field opposite to the given field. This magnetic field is perpendicularly outwards. Therefore, the induced current flows in anticlockwise direction.

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.

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
\(ε =N\frac { d{ \Phi }_{ B } }{ dt } \)
\(=\frac { 40\times (4\times { 10 }^{ -2 }-2\times { 10 }^{ -2 }) }{ 0.2 } =4V\)

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.

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 \(\mu \mathrm{Wb}\).