we know that, magnetic of the coils M = NIA. Since, the same wire is used in the three cases with same potentials, therefore, same current flows in three cases.
(i) For an equilateral triangle of side a, N = 4, as the total wire of length = 12a
Magnetic moment of the coils,

\(M=NIA=4I(\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 })\\ \Rightarrow M=I{ a }^{ 2 }\sqrt { 3 } \)
(ii) For a square of sides a, A = a²
N = 3, as the total wire of length = 12a
Magnetic moment of the coils,

\(M=NIA=3I({ a }^{ 2 })=3I{ a }^{ 2 }\)
(iii) For a regular hexagon of sides a,
n = 2, as the total wire of length = 12a
Magnetic moment of the coils,

\(M=NIA=2I(\frac { 6\sqrt { 3 } }{ 4 } { a }^{ 2 })=3\sqrt { 3{ a }^{ 2 } } I\)
M is in a geometric series.

Here, \({ B }_{ V }=0.24\sqrt { 3 } \times { 10 }^{ -4 }T\)
\( \delta ={ 30 }^{ \circ },{ B }_{ H }=?\)
Using,  
\(tan\delta =\frac { { B }_{ V } }{ { B }_{ H } }\)
\( { B }_{ H }=\frac { { B }_{ V } }{ tan\delta } =\frac { 0.24\sqrt { 3 } \times { 10 }^{ -4 }T }{ tan{ 30 }^{ \circ } } \)
\( =\frac { 0.24\sqrt { 3 } \times { 10 }^{ -4 } }{ \frac { 1 }{ \sqrt { 3 } } } T\)
\({ B }_{ H }=0.72\times { 10 }^{ -4 }T\)

Given, current, I = 4A
Number of turns per unit length, n = 600
Relative permeability, \({ \mu }_{ r }\) = 5000
Since, magnetic intensity, H = nI = 600\(\times \)4 = 2400 A\({ m }^{ -1 }\)
Since,  \({ \mu }_{ r }=1+{ x }_{ m }\Rightarrow { x }_{ m }={ \mu }_{ r }-1\)
\(=5000-1=4999\approx 5000\)
Here, \({ x }_{ m }\) = magnetic susceptibility.
Intensity of magnetisation can be given as
\(I={ x }_{ m }H=5000\times 2400\) 
\(=1.2\times { 10 }^{ 7 }A{ m }^{ -1 }\)
Therefore, magnetic field, B = \({ \mu }_{ r }{ \mu }_{ 0 }H\)
\(=5000\times \left( 4\pi \times { 10 }^{ -7 } \right) \times 2400\)
B = 15T

i)To explain qualitatively the domain picture of the irreversibility in the magetisation curve of a ferromagent, we draw the hysteresis curve for ferromagnetic substance. We can observe that the magnetisation persists even when the external fields is removed. This gives the idea of irreversibility of a ferromagnet.
ii) As we known that, in hysteresis curve, the energy dissipated per cycla is directly proportional to the area of hysteresis loop. So, as according to the question, the area of hysteresis loop is more for carbon steel, thus carbon steel piece will dissipate greater heat energy.
iii) the magnetisation of a ferromagnet depends not only on the magnetising fields, but also on the history of magnetisation (i.e how many times it was already magnetised in the past).Thus, the value of magnetisation of a specimen is a record of memory of the cycles of magnetisation, it had undergone.The system displaying such a hysterasis loop can thus act as a device for storing memory.
iv)The ferromagnetic materials which are used for coating magnetic tapes in a cassette player or for building memory stores in the modern computer are ferrites.The most commonly ferrites used are MnFe₂O₄, FeFe₂O₄, CoFe₂O₄, NiFe₂O₄, etc.(1/2)
v)To shield any space from magnetic field, surround the space with soft iron ring. As the magnetic field lines will be drawn into the ring, the enclosed region will become free of magnetic field.

Moment of inertia, I₁ = I
Time period, T₁ = T
Magnetic field = B, mass of magnet = x
  
As the magnet is cut into two pieces, mass of each piece
   \(\frac { x }{ 2 } \)
Initial moment of inertia,
 \(I=\frac { 1 }{ 12 } \times mass\times { \left( length \right) }^{ 2 }\)
\(I=\frac { 1 }{ 12 } \times \ x \ \times { \left( l \right) }^{ 2 }\)
where l = length of the magnet
Moment of inertia of half piece,
\({ I }^{ ' }=\frac { 1 }{ 12 } \times \frac { x }{ 2 } { \left( \frac { l }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 12 } \times \frac { x{ l }^{ 2 } }{ 4 } \)
The magnetic moment of half piece, \({ m }^{ ' }=\frac { m }{ 2 } \)
Using the formula of time period,
\(T=2\pi \sqrt { \frac { I }{ mB } } \)
\( \frac { T }{ { T }^{ ' } } =\sqrt { \frac { I{ m }^{ ' }B }{ { I }^{ ' }mB } } =\sqrt { \frac { I }{ { I }^{ ' } } .\frac { { m }^{ ' } }{ m } }\)
\( \frac { T }{ { T }^{ ' } } =\sqrt { \frac { x{ l }^{ 2 }\times 12\times 4\times 2 }{ 12\times x{ l }^{ 2 } } \times \frac { m }{ 2m } } =\sqrt { 2\times 2 }\)
\({ T }^{ ' } \ =\frac { T }{ 2 } \)
The new time period of each piece is \(\frac { T }{ 2 } \)