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

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

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.

N_A = 200 turns; N_B = 1000 turns;
r_A = 20 x 10^-2 m; r_B = 2 x 10^-2 m;
dt = 0.04 s; di_A = 6−2 = 4A
Let i_A be the current flowing in coil A, then the magnetic field B_A at the centre of the circular coil A is
\({ B }_{ A }=\frac { { \mu }_{ o }{ N }_{ A }{ i }_{ A } }{ { 2r }_{ A } } =\frac { 4\pi \times { 10 }^{ -7 }{ N }_{ A }{ i }_{ A } }{ { 2r }_{ A } } \)
\(=\frac { { 10 }^{ -7 }\times 2\times 3.14\times 200 }{ 20\times { 10 }^{ -2 } } \times { i }_{ A }\)
= 6.28 x 10^-4i_A Wbm^-2
The magnetic flux linkage of coil B is
\({ N }_{ B }{ \Phi }_{ B }={ N }_{ B }{ B }_{ A }{ A }_{ B }\)
= 1000 x 6.28 x 10^-4 x i_A x 3.14 x (2 x 10^-2)²
= 7.89 x 10^-4i_AWb turns
The mutual inductance between the coils
\(\\ { M }=\frac { { N }_{ B }{ \Phi }_{ B } }{ { i }_{ A } } =7.89\times { 10 }^{ -4 }H\)
Induced emf in coil B is
ε_B = - M\(\frac { { di }_{ A } }{ dt } \)
ε_B = \(\frac { 7.89\times { 10 }^{ -4 }\times (6-2) }{ 0.04 } \)(magnitude only)
ε_B = 78.9mV
The rate of change of magnetic flux of coil B is
\(\frac { d\left( { N }_{ B }{ \Phi }_{ B } \right) }{ dt } ={ \epsilon }_{ B }=78.9m{ Wbs }^{ -1 }\)

l = 0.5 m; h = 7.2 m; u = 0 m s^–1;
g = 10 m s^–2; B_H = 4.04 x 10^–5 T
The final velocity of the rod is
V² = u² = + 2g h = 0 + (2 x 10 x 7.2) =144
v = 12 ms^-1
The magnitude of the induced emf when the rod is about to touch the ground is
ε = B_H lv = 4.04 × 10^–5 × 0.5 × 12
= 242.4 µV

Construction:
Alternator consists of two major parts, namely stator and rotor. As their names suggest, stator is stationary while rotor rotates inside the stator. In any standard construction of commercial alternators, the armature winding is mounted on stator and the field magnet on rotor.
The construction details of stator, rotor and various other components involved in them are given below.

(a) Stator:
The stationary part which has armature windings mounted in it is called stator. It has two  components, namely stator frame, stator core and armature winding.
Stator core:
Stator core or armature core is made up of iron or steel alloy. It is a hollow cylinder and is laminated to minimize eddy current loss. The slots are cut on inner surface of the core to accommodate armature windings.
Armature winding:
Armature winding is the coil, wound on slots provided in the armature core. 
(b) Rotor
Rotor contains magnetic field windings. The magnetic poles are magnetized by DC source. The ends of field windings are connected to a pair of slip rings, attached to a common shaft about which rotor rotates. Slip rings rotate along with rotor. To maintain connection between the DC source and field windings, two brushes are used which continuously slide over the slip rings.

(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 rectangular conducting loop of width 1 in a uniform magnetic field \(\vec { B } \)  which is perpendicular to the plane of the loop and is directed inwards.
(ii) A part of the loop is in the magnetic field while the remaining part is outside the loop as shown in Figure.
(iii) When the loop is pulled with a constant velocity \(\vec { v } \) to the right, the area of the portion of the loop within the magnetic field will decrease.
(iv) Thus, the flux linked with the loop will also decrease. According to Faraday's law, an electric current is induced in the loop which flows in a direction so as to oppose the pull of the loop.
(v) Let x be the length of the loop which is still within the magnetic field, then its area is lx. The magnetic flux linked with the loop is
\({ \phi }_{ B }=\int _{ A }^{ }{ B.d } \vec { A } =BAcos\theta \)
Here θ = 0^o and cos  0^o = 1
= BA
\({ \phi }_{ B }=Blx\)
(vi) As this magnetic flux decreases due to the movement of the loop, the magnitude of the induced emf is given by
\(\varepsilon =\frac { d{ \Phi }_{ B } }{ dt } =\frac { d }{ dt } (Blx)\)
(vii) Here, both B and 1 are constants. Therefore,
\(\varepsilon =Bl\frac { dx }{ dt } \)
ε = Blv
where \(v=\frac { dx }{ dt } \) is the velocity of the loop. This emf is known as motional emf since it is produced due to the movement of the loop in the magnetic field.