Coulomb's inverse square law states that the force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them.
\(\vec { F } =k \frac { { q }_{ m_{A} }{ q }_{ m_{B} } }{ { r }^{ 2 } } \hat { r } \)

Magnetic flux is defined as the number of magnetic field lines crossing per unit area kept normal to the direction of lines of force.

(i) When a substance is placed in a uniform magnetising field, the substance gets magnetised.
(ii) The total magnetic field inside the specimen is equal to the sum of the magnetic field produced in vacuum due to the magnetising field and the magnetic field due to the induced magnetism of the substance.

The total flux emanating from the closed surface S enclosing the dipole is zero. So,
\({ \Phi }_{ B }=\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
Here the integral is taken over closed surface. Since no isolated magnetic pole (called magnetic monopole) exists, this integral is always zero,
\(\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
This is similar to Gauss’s law in electrostatics.

Magnetic field lines can be entirely confined within the core of a toroid since the toroid has no ends. θ solenoid is open ended and the field lines inside it which are parallel to the length of the solenoid cannot form closed curves inside the solenoid.

(i) Horizontal component:
The Earth's magnetic field is parallel to the surface of the Earth (i.e., horizontal) which implies that the needle of the magnetic compass rests horizontally at an angle of dip, I = 0⁰ as shown in the figure.
B_H = B_E
B_v = 0
This implies that the horizontal component is maximum at the equator and the vertical component is zero at the equator.

(ii) Vertical component: The Earth's magnetic field is perpendicular to the surface of the Earth (i.e., vertical) which implies that the needle of magnetic compass rests vertically at an angle of dip, I = 90° as shown in Figure
Hence,
B_H = 0
B_v = B_E
This implies that the vertical component is maximum at poles and the horizontal component is zero at poles.

Density of the magnet is
Density = \(\frac { Mass }{ volume } \Rightarrow Volume=\frac { Mass }{ Density } \)
\(Volume=\frac { 200\times 1{ 0 }^{ -3 }kg }{ \left( 8\times 1{ 0 }^{ -3 }kg \right) \times 1{ 0 }^{ 6 }{ m }^{ -3 } } =25\times { 10 }^{ -6 }{ m }^{ 3 }\)
Magnitude of magnetic moment p_m = 2A m²
Intensity of magnetization,
\(I=\frac { magnetic\ moment }{ Volume } =\frac { 2 }{ 25\times { 10 }^{ -6 } } \)
M = 0.8 x 10⁵ Am^-1

The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T
Deflection is 
\(\theta =60°\Rightarrow tan60°=\sqrt { 3 } =1.732\)
\(I=\frac { 2R{ B }_{ H } }{ { \mu }_{ ° }N } tan\theta \)
\(=\frac { 2\times 0.12\times 25\times 1{ 0 }^{ -6 } }{ 4\times 1{ 0 }^{ -7 }\times 3.14\times 100 } \times 1.732=0.82\times 1{ 0 }^{ -1 }A\)
I = 0.082 A

Let  \(\vec{p}_m\)be the dipole and before switching ON the external magnetic field, there is no orientation. Therefore, the energy U = 0.
As soon as external magnetic field is switched ON, the magnetic dipole orient parallel (θ = 0^o) to the magnetic field with energy,
U_parallel = U_minimum = -p_mBcos 0
U_parallel = -p_mB
since cos 0^o = 1
Otherwise, the magnetic dipole orients anti-parallel (θ = 180^o) to the magnetic field with energy,
U _{anti-parallel} = U _maximum = -p_mBcos 180
\(\Rightarrow \) U_{anti-parallel} = P_mB
since cos 180^o = -1

Let the currents flowing through wires XY and PQ be I₁ and I₂
Magnetic field along PQ is \(\mathrm{B}_{1}=\frac{\mu_{o} I_{2}}{2 \pi r}\)
Force per unit length on PQ is \(\frac{F_{2}}{l}=\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\)
If the wire XY is slightly displaced and released, it executes simple harmonic motion with the condition that acceleration is directly proportional to the displacement y
\(\therefore a=-\omega^{2} y\)  .....(1) 
The distance between two wires = d
Time period
\(T=\frac{2 \pi}{\omega} \)
\(a=\frac{g}{d} y \) .....(2)
By comparing the equations (1) and (2) we get
\(\omega^{2} =-\frac{g}{d} \ \therefore \omega=\sqrt{\frac{g}{d}} \)
\(\text { Time period } =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{d}{g}} \)
\(\therefore T =2 \pi \sqrt{\frac{d}{g}} \)

There are three quantities required to specify the magnetic field of the Earth on its surface, which are often called as the elements of the Earth's magnetic field. They are:
(a) magnetic declination (D)
(b) magnetic dip or inclination (I)
(c) the horizontal component of the Earth's magnetic field (B_H)

Let B_E be the net Earth's magnetic field at any point P on the surface of the Earth. B_E can be resolved into two perpendicular components.
Horizontal component,  B_H = B_E cos I .... (1)
Vertical component,  B_V = B_E sin I     .....(2)
Dividing equation (1) and (2), we get,
\(=\frac{B_{V}}{B_{H}} ...(3)\)
(i) At magnetic equator:
The Earth's magnetic field is parallel to the surface of the Earth (i.e., horizontal) which implies that the needle of magnetic compass rests horizontally at an angle of dip, I = 0^o  Hence, B_H  = B_E
B_V = 0
This implies that the horizontal component is maximum and vertical component is zero at equator.
(ii) At magnetic poles:
The Earth's magnetic field is perpendicular to the surface of the Earth (i.e, vertical) which implies that the needle of magnetic compass rests vertically at an angle of dip, I = 90^o
Hence, B_H = 0
B_V = B_E
This implies that the vertical component is maximum at poles and horizontal component is zero at poles.

When the current is passed through the loop, the torque is produced. If the torque acting on the loop is increased then the loop will start to rotate. The loop will start to lift if and only if the magnitude of magnetic torque due to current applied equals to the gravitational torque as shown in Figure
T_magnetic = T_{gravitational}
IAB = mgR
\(But\quad { p }_{ m }=IA=I(\pi { R }^{ 2 })\)
\(IR^{ 2 }B=mgR\)
\(\Rightarrow \frac { mg }{ \pi { R }B } \)
The current estimate using this equation should be applied so that one edge of loop is lifted off the surface.

Lorentz force F = Bqv
Energy w = qV
Energy is equal to kinetic energy,
\(qV=\frac{1}{2}mv^2\)
\(v=\sqrt { \frac {2qV }{ m } } \)
\(\therefore Lorentz \ force \ F= Bq\times \sqrt \frac{2qV}{m}=\sqrt { \frac { 2{ B }^{ 2 }{ q }^{ 3 }V }{ m } } \)

Magnetic filed at the centre of a circular
loop, B = \(\frac{μ_oI}{2\pi R}\)
From the figure, R =\(\frac{2r}{\pi}\)
\(\therefore B'=\frac{μ_oI}{2\pi \times\frac{2r}{\pi}}=\frac{μ_oI}{4r}\)
\(B'=\frac { { \mu }_{ ° }I }{ 4r } \bigotimes \)