Let v = speed of electron revolving around a neucleus (proton)
r = orbital radius.
According to the Bohr's second postulate,

\(m v r=n \frac{h}{2 \pi}\)
\(\therefore \ m v r=\frac{h}{2 \pi}\) (∵ in ground state n = 1)
\(\Rightarrow \ v=\frac{h}{2 \pi m r}\)     ........(i)
Therefore, the centripetal force for revolution of an electron is provided by the electrostatic force of attraction between nucleus (proton) and an electron.
\(\therefore \frac{m v^{2}}{r}=\frac{e^{2}}{4 \pi \varepsilon_{0} r^{2}} \Rightarrow v^{2}=\frac{e^{2}}{4 \pi \varepsilon_{0} m r}\)      ..........(ii)
Using (i) and (ii), we get
\(\frac{h^{2}}{4 \pi^{2} m^{2} r^{2}}=\frac{e^{2}}{4 \pi \varepsilon_{0} m r}\)
\(\Rightarrow \ r=\frac{\varepsilon_{0} h^{2}}{\pi m e^{2}}\)     ............(iii)
and \(v=\frac{h}{2 \pi m \frac{\varepsilon_{0} h^{2}}{\pi m e^{2}}}=\frac{e^{2}}{2 \varepsilon_{0} h}\)      ...........(iv)
∴ Magnetic field at the site of a point nucleus in a hydrogen atom due to the circular motion of the electron is given by
\(B=\frac{\mu_{0} i}{2 r}=\frac{\mu_{0}}{2 r} \cdot \frac{e}{T}\)     .........(v)
∵ T = Time period of revolution of the electron
\(T=\frac{\text { circumference }}{\text { speed }} \Rightarrow T=\frac{2 \pi r}{v}\)   ........(vi)
Using (iii) to (vi); we get
\(B=\frac{\mu_{0}}{2 r} \cdot \frac{e v}{2 \pi r}=\frac{\mu_{0} e\left(\frac{e^{2}}{2 \varepsilon_{0} h}\right)}{4 \pi\left(\frac{\varepsilon_{0} h^{2}}{\pi m e^{2}}\right)^{2}}=\frac{\mu_{0} \pi m^{2} e^{7}}{8 \varepsilon_{0}^{3} h^{5}}\)

(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 on the solenoid, n = 2000
Area of cross-section of the solenoid, A = 1.6 x 10^-4 m²
Current in the solenoid, I = 4.0 A
(a) The magnetic moment along the axis of the solenoid is calculated as:
M = nAI
= 2000 x 4 x 1.6 x 10^-4
= 1.28 Am²
(b) Magnetic field, B = 7.5 x 10^-2
The angle between the magnetic field and the axis of the solenoid, θ = 30°
Torque, \(\tau\) = MB sinθ
=1.28 x 7.5 x 10^-2 sin30°
= 0.048J
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 0.048 J

(a) Magnetic moment, M = 1.5 J T^-1
Magnetic field strength, B = 0.22 T
(i) Initial angle between the axis and the magnetic field, θ₁​ = 0°
Final angle between the axis and the magnetic field, θ₂ ​= 90°
The work required to make the magnetic moment normal to the direction of the magnetic field is given as:
W = -MB (cosθ₂​ - cosθ_1​)
= -1.5 x 0.22(cos90°- cos0°)
= - 0.33 (0 - 1)
= 0.33 J
(ii) Initial angle between the axis and the magnetic field, θ₁ ​= 0°
Final angle between the axis and the magnetic field, θ₂ ​= 180°
The work required to make the magnetic moment opposite to the direction of the magnetic field is given as:
W = - MB (cosθ₂ ​- cosθ₁​)
= -1.5 x 0.22 (cos180° - cos0°)
= - 0.33 (- 1 - 1)
= 0.66 J
(b) For case (i):
θ = θ₂ ​= 90°
∴ Torque, \(\tau\) =MB sinθ
= MB sin 90°
= 0.33 J
The torque tends to align the magnitude moment vector along B.
For case (ii):
θ = θ₂ ​= 180° ∴ Torque, \(\tau\) = MB sinθ
= MB sin180°
= 0

(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.

(i) We know that \(\tau=M B \sin \theta\)
Magnitude of the magnetic field is calculated as
\(B=\frac{\tau}{M \sin \theta}=\frac{0.063}{0.9 \times \sin 30^{\circ}}=0.14 \mathrm{~T}\)
(ii) When the magnetic moment vector and the magnetic field vectors are in the same direction,
i.e. θ = 0°
It's so because this configuration corresponds to a minimum energy.
U = -MB

B_H = Bcos \(\delta \), B_v = Bsin \(\delta \)
\({ Also } \tan \delta=\frac{B_{V}}{B_{H}}\)

Given, magnetic length of bar magnet, 2l = 6 cm
\(\Rightarrow\) l = 3 cm = 3 x 10^-2 m
Distance, d = 60cm = 0.6m
Magnetic moment, M = 0.60 A - m²
\(\therefore \text { Magnetic field, } B=\frac{\mu_{0}}{4 \pi} \times \frac{M}{\left(d^{2}+l^{2}\right)^{3 / 2}}\)
\(=\frac{\mu_{0} M}{4 \pi d^{3}} \) [∵ l < < d]
\(=\frac{4 \pi \times 10^{-7} \times 0.60}{4 \pi \times(0.6)^{3}}=2.7 \times 10^{-7} \mathrm{~T}\)

Let m and 5m be the Pole strength of the two poles.
Here, F = 0.8 x 10^-3kg - wt = 0.8 x 10^-3 x 9.8N,
r = 10 cm = 0.1m
\(\therefore \quad F=\frac{\mu_{0}}{4 \pi} \cdot \frac{m_{1} m_{2}}{r^{2}}\)
\(\Rightarrow 0.8 \times 10^{-3} \times 9.8=\frac{10^{-7} \times m \times 5 m}{(0.1)^{2}} \Rightarrow m=12.52 \mathrm{~A}-\mathrm{m}\)
and 5m = 5 x 12.52 A-m = 62.6A-m

Number of turns in the solenoid, n = 800
Area of cross-section, A =  2.5 × 10^-4 m² 
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis, i.e., along with its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M = n I A
= 800 x 3 x 2.5 x 10^-4
= 0.6 J T^-1

Here, v = 2.18 x 10⁶ mis, r = 5.3 x 10^-11 m
e = 1.6 x 10^-19 C
(i) Time period of revolution of electron is given by
\(T=\frac{2 \pi r}{v}\)
\(=\frac{2 \pi \times 5.3 \times 10^{-11}}{2.18 \times 10^{6}}\)
= 1.528 x 10^-16 s
\(\text { Equivalent current, } I=\frac{\text { Charge }}{\text { Time }}=\frac{e}{T}\)
\(I=\frac{1.6 \times 10^{-19}}{1.528 \times 10^{-16}}\)
\(\Rightarrow\) 1 = 1.05 x 10^-3 A
(ii) Field at proton due to orbiting electron is
\(B=\frac{\mu_{0} I}{2 r} \text { or } B=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi I}{r}\)
\(B=\frac{10^{-7} \times 2 \pi \times 1.05 \times 10^{-3}}{5.3 \times 10^{-11}}=12.4 \mathrm{~T}\)

(i) \(\begin{aligned} & \therefore B=\frac{\mu_0}{4 \pi} \frac{2 M}{d^3} \end{aligned}\)
\(\begin{aligned} \Rightarrow B=\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^3}=0.96 \times 10^{-4} \mathrm{~T} \end{aligned}\)
\(\Rightarrow\) B = 0.96 G
The magnetic field is along the S-N direction.
(ii) \(\therefore B=\frac{\mu_0}{4 \pi} \times \frac{M}{d^3}=\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi \times(0.1)^3}=0.48 \mathrm{G}\)
The magnetic field is along the N-S direction.

The statement is correct. The number of magnetic field lines leaving a surface is balanced by the number of lines entering it. The net magnetic flux is zero.

In both the cases
(i) and
(ii) we get two magnets, each with a North and South pole.

Magnetic lines of force come out from North pole and enter into the South pole outside the magnet and travels from South pole to North pole inside the magnet. So, magnetic lines of force form closed loop, hence magnetic monopoles do not exist.

The magnetic moment of a magnet is a quantity that determines the torque, it will experience in an external magnetic field. Its SI unit is A - m ².

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.

The pole strength of a magnet may depend on its cross-section, nature of material.

Given, \(\theta\) = 90°, \(\tau\) = 3 x 10^-4 N-m
and B = 0.4 x 10^-4 T
Since, torque \(\tau\) = MB sin \(\theta\)
\(\therefore \quad M=\frac{\tau}{B \sin \theta}=\frac{3 \times 10^{-4}}{0.4 \times 10^{-4} \sin 90^{\circ}}=7.5 \mathrm{JT}^{-1}\)

If length of wire is 2l , then magnetic moment
M = m x 2l = 4\(\pi\) A-m²  [given]
As wire is bent in the form of semicircle, effective distance between the ends is 2r.
So, new dipole moment
\(M^{\prime}=m \times 2 r=m \times 2 \times \frac{2 l}{\pi}=\frac{2}{\pi}(m \times 2 l) \quad[\because \pi r=2 l]\)
\(=\frac{2}{\pi} M=\frac{2}{\pi} 4 \pi\)
= 8  A - m²

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

(i) Since, diamagnetic subtances gets magnetised in opposíte direction, therefore it will try to reduce the magnctíc field in the direction of magnetising field. Hence, diamagnetic material is oriented towards the region in which magnetising field is weaker.
(ii) Given, B = 6.28 \(\times\)10^-2 T
H = 100 Am^-1
\(\therefore\) Magnetic permeability.
\(\mu=\frac{B}{H}=\frac{6.28 \times 10^{-2}}{100}\)
= 6.28 \(\times\)10^-4 TmA^-1
Relative magnetic permeability,
\(\mu_r=\frac{\mu}{\mu_0}=\frac{6.28 \times 10^{-4}}{4 \pi \times 10^{-7}}=5 \times 10^2=500\)
\(\therefore\) Magnetic susceptibility
X_m = \(\mu\)_r - 1 = 500 - 1 = 499
(iii) 

(iv) There are some ferromagnetic materials in which the magnetisation disappears on the removal of external magnetic field, e.g. soft iron. Such materials are called soft magnetic materials or soft ferromagnets. e.g. Iron, cobalt, nickel, gadolinium, etc.

(i) (b)
(ii) (d): Magnetic permeability - Henry m^-1
(iii) (d): Given, L= 3 cm, A = 2 cm², M = 3 A m²
.Intensity of magnetisation \(=\frac{M}{l A}=\frac{3}{3 \times 10^{-2} \times 2 \times 10^{-4}}\)
\(=\frac{1}{2 \times 10^{-6}}=0.5 \times 10^{6}=5 \times 10^{5} \mathrm{~A} / \mathrm{m}\)
(iv) (b): Here, n = 500 turns/m
\(I=1 \mathrm{~A}, \mu_{-}=500\)
Magnetic intensity \(H=n I=500 \mathrm{~m}^{-1} \times 1 \mathrm{~A}=500 \mathrm{~A} \mathrm{~m}^{-1}\)
As \(\mu_{r}=1+\chi \quad \text { or } \chi=\left(\mu_{r}-1\right)\)
Magnetisation, M = XH
\(=\left(\mu_{r}-1\right) H=(500-1) \times 500 \mathrm{~A} \mathrm{~m}^{-1}\)
\(=2.495 \times 10^{5} \mathrm{~A} \mathrm{~m}^{-1} \approx 2.5 \times 10^{5} \mathrm{~A} \mathrm{~m}^{-1}\)
(v) (a): Relative permeability of iron \(\mu_{r}=6000\)
Magnetic susceptibility \(\chi_{m}=\mu_{r}-1=5999\)