Coulomb's law of magnetic force is inversely proportional to the squared distance between the magnetic poles and directly proportional to the product of magnetic poles.

On bending the bar magnet, the length of the bar magnet,
\(A C=A O+O C=2 L \sin \left(\frac{60^{\circ}}{2}\right)+2 L \sin \left(\frac{60^{\circ}}{2}\right)\)
\(=4 L \sin 30^{\circ}=4 L \times \frac{1}{2}=2 L\)

Geometric length of a magnet is \(\frac{6}{5}\) times its magnetic length.
\(\therefore\) Geometric length = \(\frac{6}{5}\) x 20
= 24 cm

Induced e.m.f., \(e=V=\frac { work }{ charge } =\frac { M{ L }^{ 2 }{ T }^{ -2 } }{ AT } \)
\(=[{ M }^{ 1 }{ L }^{ 2 }{ T }^{ -3 }{ A }^{ -1 }]\)
Rate of change of magnetic flux\(=\frac { d\Phi }{ dt } =\frac { BA }{ t } \)
 \(=\frac { FA }{ q\upsilon t } =\frac { { (MLT }^{ -2 })({ L }^{ 2 }) }{ (AT)({ L }T^{ -1 })(T) } \ (\because F=Bq\upsilon )\\ \ \ =[{ M }^{ 1 }{ L }^{ 2 }T^{ -3 }A^{ -1 }]\)
Both have the same units/dimensions.

They are both, useful and harmful.

\(\text {As } L \propto N^2 \text {, }\)
when N is double, L becomes four times.

Induced emf is called back emf as it opposes the growth as well as decay of current in the circuit.

(i) The self inductance is defined on the magnetic flux linked with the coil when unit current flows through it. The self inductance is defined as the emf induced in the coil, when the rate of change of current in the coil is 1 ampere/second. The SI unit of self inductance is henry (H).
(ii) The self-inductance of a solenoid is defined as the magnetic flux to current ratio via the solenoid. It is given by,
\(\mathrm{L}=\frac{\phi}{I}\)
Total number of turns in the solenoid,
\(\mathrm{N}=\mathrm{nl}\)
The magnetic field inside the long solenoid, \(B=\mu_0 n i\)
Flux through one turn, \(\phi_1=B A=\mu_0 n i A\)
Thus, total flux through N turns,
\(\phi_t=N \phi_1=n l \times \mu_0 n i A=\mu_0 n^2 l A i\)
Using \(\phi_t=L i\)
where L is the self-inductance of the coil.
\(\therefore \mu_0 n^2 l A i=L i \)
\(\mathrm{~L}=\mu_0 n^2 A l\)

(a) From Eq, the magnetic energy is
\(U_{B}=\frac{1}{2} L I^{2}\)
\(=\frac{1}{2} L\left(\frac{B}{\mu_{0} n}\right)^{2} \ \left(\text { since } B=\mu_{0} n I, \text { for a solenoid }\right)\)
\(=\frac{1}{2}\left(\mu_{0} n^{2} A l\right)\left(\frac{B}{\mu_{0} n}\right)^{2}\)   [from Eq.]
\(=\frac{1}{2{\mu }_{0}}{B}^{2}Al\)
(b) The magnetic energy per unit volume is,
\({U}_{B}=\frac{{U}_{B}}{V}\)  (where V is volume that contains flux)
\(=\frac{{U}_{B}}{Al}\)
\(=\frac{{B}^{2}}{2{\mu }_{0}}\)
We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor.
\({U}_{E}=\frac{1}{2}{\epsilon }_{0}{E}^{2}\)
In both the cases energy is proportional to the square of the field strength. Equation have been derived for special cases: a solenoid and a parallel plate capacitor, respectively. But they are general and valid for any region of space in which a magnetic field or/and an electric field exist.

(a) Here, H = I = 5000 x 2 = 10⁴ A/m
and  I = XH
\(\therefore\) \(\chi=\frac{I}{H}\) 
\(=\frac{2 \times 10^{-2}}{10^{4}}=2 \times 10^{-6}\)
(b) According to Curie law.
\(x=\frac{c}{T}\)
\(\Rightarrow \quad \frac{\chi_{2}}{\chi_{1}}=\frac{T_{2}}{T_{1}}\)
\(\chi_{2}=\frac{T_{2}}{T_{1}} \chi_{1}=\frac{320}{300} \times 2 \times 10^{-6}\)
= 2.13 x 10^-6
\(\therefore\) Magnetisation at 320 K,
I = X₂H = 2.13 x 10^-6 x 10⁴
= 2.13 x 10^-2A/m

Number of turns in the circular coil, N = 16
Radius of the coil, r = 10 cm = 0.1 m
Cross-section of the coil, A = \(\pi r\)² = n x (0.1)² m²
Current in the coil, I = 0.75 A
Magnetic field strength, B = 5.0 x 10^-2 T
Frequency of oscillations of the coil, v = 2.0 s^-1
∴ Magnetic moment, M = NIA = N I \(\pi r^{2}\) 16 x 0.75 x n x (0.1)²
= 0.377 J T^{- 1}
Frequency is given by the relation:
\(v=\frac{1}{2 \pi} \sqrt{\frac{M B}{I}}\)
Where,
I = Moment of inertia of the coil
Rearranging the above formula, we get:
\(\therefore I=\frac{M B}{4 \pi^{2} v^{2}}\)
\(=\frac{0.377 \times 5 \times 10^{-2}}{4 \pi^{2} \times(2)^{2}}\)
= 1.2 x 10^-4 kg m²
Hence, the moment of inertia of the coil about its axis of rotation is 1.19\times 10^-4 kg m².

(a) The field H is dependent of the material of the core, and is
H = nI = 1000 x 2.0 = 2 x 10³ A/m.
(b) The magnetic field B is given by
B = μ_r μ₀ H
= 400 x 4π x 10^-7 (N/A²) x 2 x 10³ (A/m)
= 1.0 T
(c) Magnetisation is given by
M = (B - μ₀ H)/ μ₀
= (μ_r μ₀ H - μ₀ H) / μ₀ = (μ_r – 1)H = 399 x H
\(\cong\)8 \(\times\)10⁵ A/m
d) The magnetising current I_M is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = μ_r n (I + I_M). Using I = 2A, B = 1 T, we get I_M = 794 A.

(a) 
From the phesor diagram
\(\overrightarrow { V } =\ \overrightarrow { { V }_{ L } } +\overrightarrow { { V }_{ R } } +\overrightarrow { { V }_{ C } } \)
\({ V }_{ m }=\sqrt { [{ R }^{ 2 }+({ X }_{ C }-{ X }_{ L }{ ) }^{ 2 }] } \)
\(\Rightarrow { I }_{ m }=\frac { { V }_{ m } }{ \sqrt { [{ R }^{ 2 }+({ X }_{ C }-{ X }_{ L }{ ) }^{ 2 }] } } \)
From the figure tan \(\theta\) =\(\frac{V_Cm-V_Lm}{V_Rm}\)
\(=\frac{I_m(X_C-X_L)}{I_{m}R}\)
\(\theta=tan^{-1}(\frac{X_C-X_L}{R})\)
(b) At resonance, Im is maximum.
\(\Rightarrow\) X_L = X_C,

Quality factor of LCR circuit is defined as= \(\frac{{\omega}_{o}}{2\triangle\omega}\)
=\(\frac{{\omega}_{o}L}{R}\)
A larger value of quality factor corresponds to a sharper resonance.