Angle of dip is 45°.

Bismuth is a diamagnetic material. So, when it is kept in an external magnetic field the field lines are repelled and the field inside the material is reduced.

I is the total magnetic field.
Now, \(I \cos 60^{\circ}=B \Rightarrow I=\frac{B}{\cos 60^{\circ}}=\frac{B}{1 / 2}=2 B\)
At equator, dip angle is 0^o
\(\therefore B H=1 \cos 0^{\circ}=1=2 B .\)

Magnetic field strength, B = 0.25 T 
Magnetic moment, M = 0.6 T^-1
The angle θ, between the axis of the solenoid and the direction of the applied field, is 30°.
Therefore, the torque acting on the solenoid is given as:
\(\tau\) = MB sinθ
= 0.6 x 0.25 sin30°
= 7.5 x 10^-2 J

Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T =  4.5 x 10^-2 J
The angle between the bar magnet and the external magnetic field, θ = 30°
Torque is related to magnetic moment (M) as:
\(T=M B \sin \theta \therefore M=\frac{T}{B \sin \theta}\)
\(=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}=0.36 J T^{-1}\)
Hence, the magnetic moment of the magnet is 0.36  J T^-1.

Consider an electron revolving around the nucleus of an atom. The electron is in a uniform circular motion around the nucleus of charge +Ze. This constitutes a current.
\(\therefore \ I=\frac{e}{T}\)     .......(i)

If r is the orbital radius of the electron and 'v' is the orbital speed, then the time period is given as
\(T=\frac{2 \pi r}{v}\)      ............(ii)
From equations (i) and (ii), we get
\(I=\frac{e v}{2 \pi r}\)     ............(iii)
As the magnetic moment is given by
\(\mu_{l}=I \pi r^{2}\)
We have \(\mu_{l}=\left(\frac{e v}{2 \pi r}\right) \pi r^{2}=\frac{e v r}{2}\)
Multiplying and dividing by me in above equation, we get
\(\mu_{l}=\frac{e v m_{e} r}{2 m_{e}}=\frac{e}{2 m_{e}} l\)
where I is the angular momentum of the electron.
Since magnetic dipole moment and angular momentum are oppositely directed,

\(\therefore \ \overrightarrow{\mu_{e}}=-\frac{e}{2 m_{e}} \vec{L}\)
According to the Bohr hypothesis, the angular momentum can have discrete values only.
\(\text { i.e. } l=\frac{n h}{2 \pi}\)
So, \(\mu_{l}=\frac{e n h}{2 \pi(2 m)}=\frac{n e h}{4 \pi m}\)
In this case it is directed into the plane of the paper.

(i) Material X is paramagnetic substance. When a specimen of a paramagnetic substance is placed in a magnetising field, the lines of force prefer to pass through the specimen rather than through air. Thus, magnetic induction inside the sample is more than the magnetic intensity. 
(ii) Material Y is ferromagnetic substance. These are the substances in which a strong magnetism is produced in the same direction as the applied magnetic field, these are strongly attracted by a magnet, exhibits highly concentrated lines of force.

We know that magnetic moment of the coil M = NIA. Since, the same wire is used in three cases with same potentials, therefore, same current flows in three cases.
The different shapes form figures of different area and hence, their magnetic moments vary.
(i) For an equilateral triangle of side a,
N = 4, as the total wire of length = 12a Magnetic moment of the coil,

\(M=N I A=4 I\left(\frac{\sqrt{3}}{4} a^{2}\right)\)
\(\Rightarrow\) \(M=I a^{2} \sqrt{3}\)
(ii) For a square of side a, A = a²
N = 3, as the total wire oflength = 12a Magnetic moment of the coil,

M = NLA = 3I (a² ) = 3I a²
(iii) For a regular hexagon of sides a,
N = 2, as the total wire' of length = 12a Magnetic moment of the coils,
\(M=N I A=2 I\left(\frac{6 \sqrt{3}}{4} a^{2}\right)=3 \sqrt{3} a^{2} I\)
\(\therefore\) M is in a geometric series.

Given, pole strength, m = 6.64 A-m
r = 3mm = 3 x 10^-3m
\(\text { Since, force, } F=\frac{\mu_{0}}{4 \pi} \times \frac{m_{1} m_{2}}{r^{2}}\)
\(\therefore \quad F=\frac{\mu_{0}}{4 \pi} \times \frac{m^{2}}{r^{2}}\) \(\left[\because m_{1}=m_{2}=m\right]\)
\(=\frac{\mu_{0}}{4 \pi} \times \frac{(6.64)^{2}}{\left(3 \times 10^{-3}\right)^{2}}\)
\(=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{44.0896}{9 \times 10^{-6}}=10^{-1} \times 4.8988\)
= 0.49N

Horizontal component of earth’s magnetic field, H ​= 0.35G
Angle made by the needle with the horizontal plane = Angle of dip = δ = 22°
Earth’s magnetic field strength = B
We can relate B and B_H​ as:
\(B_{H}=B \cos \delta \therefore B=\frac{B_{H}}{\cos \delta}\)
\(=\frac{0.35}{\cos 22^{\circ}}=0.38 \mathrm{G}\)
Hence, the strength of the earth’s magnetic field at the given location is 0.38 G.