Hysteresis is the phenomenon of lagging of magnetic induction behind the magnetising field.

Ampere's circuital law states that the line integral of magnetic field over a closed loop is μ₀, times net current enclosed by the loop.
\(\oint _{ c }^{ }{ \vec { B } \vec { dl } } \) = μ₀I _enclosed.

Biot-Savart's law states that, the magnitude of magnetic field \(d\vec { B } \) at a point P at a distance of r from the small elemental length taken on a conductor carrying current varies
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length of element \(\vec { dl } \)
(iii) directly as the sine of the angle θ between \(\vec { dl } \) and \(\hat { r } \).
(iv) inversely as the square of the distance r between the point P and length of element \(\vec { dl } \).
\(d\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { I\vec { dl } \times \hat { r } }{ { r }^{ 2 } } \)

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 magnitude of the force between two poles is given by
\( F =k\frac { { q }_{ m_A }{ q }_{ { m }_{ B } } }{ { r }^{ 2 } } \)
(Given : F = 9 × 10^–3 N, r = 10 cm = 10 × 10^–2 m
Since q_mA = q_mB = q_m, we have
9 x 10^-3 = 10^-7 x \(\frac { { q }_{ m }^{ 2 } }{ { \left( 10\times { 10 }^{ -2 } \right) }^{ 2 } } \Rightarrow { q }_{ m }\) = 30NT^-1 

B_H = 0.15 G and B_V = 0.26 G
tan I = \(\frac { 0.26 }{ 0.15 } \Rightarrow I=ta{ n }^{ -1 }(1.732)=60°\)
The resultant magnetic field of the Earth is
\(B=\sqrt { { B }_{ H }^{ 2 }+{ B }_{ V }^{ 2 } } =0.3G\)

Uniform magnetic field B = 0.8 T
Angle of orientation θ = 30°
Torque, ፒ = 0.2 Nm.
(i) We know that torque ፒ = P_mB sinθ
\(0.2=p_\mathrm{m} \times 0.8 \times \sin 30^{\circ} \)
\(0.2=p_\mathrm{m} \times 0.8 \times \frac{1}{2} \)
0.2 = 0.4 p_m
\(p_m=\frac{0.2}{0.4}=0.5 \mathrm{Am}^{2}\)
(ii) Work done by the applied force to move the magnet from stable to unstable position.
\(\mathrm{W} =-p_\mathrm{m}B\left[\cos \theta_{2}-\cos \theta_{1}\right] \)
\(\mathrm{W} =-p_\mathrm{m}B\left(\cos 180^{\circ}-\cos 0^{\circ}\right) \)
\(\mathrm{W} =-p_\mathrm{m}B(-1-1)=2 \mathrm{p_mB} \)
\(\mathrm{W} =2 \times 0.5 \times 0.8=\mathbf{0 . 8} \mathbf{J} \)
Work done by the applied magnetic field are in opposite direction
\(\mathbf{W}_{\text {mag }}=-0.8 \mathrm{J}\)

Charge of an electron q = –1.60 × 10^–19 C ⇒ |q| = 1 60 x 10^-19 C
Magnitude of magnetic field B = 0.500 T
Mass of the electron, m = 9.11 × 10^–31 kg
Radius of the orbit, r = 2.50 mm = 2.50 × 10^–3 m
Speed of the electron,  V = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

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

\(tan \ I=\frac{B_V}{B_H}=1\)
∴ I = 45^o

\(\vec { B } =\frac { \beta }{ \sqrt { 3 } } =(\hat { i } +\hat { j } +\hat { k } )T\)
Using an equation,
Lorentz force, \(\vec{F}=Il\hat{j}\times\vec B\)
We can get,
Lorentz force \(F=\sqrt { \frac { 2 }{ { 3 } } } \beta Il\)

Dipole moment, \(\vec{p}_m=n\times I\times\vec{A}\)
\(\vec{p}_m\) = 50 x 3 x 3.14 x 25 x 10^-4 ≈ 1.2 A m²

(i) When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.
(ii) Let B be the magnetic field produced by passing current through the coil of the tangent galvanometer and B_H  be the horizontal component of earth's magnetic field.
(iii) Under the action of two magnetic fields, the needle comes to rest making angle with B_H , such that
B = B_H  tan \(\theta\)    .........(1)
Where B ⇒ magnetic field produced by current
B_H ⇒ horizontal component of earth's magnetic field
Construction:
(i) Copper coil of wire wound on a non-magnetic circular frame such as brass or wood. Compass box is kept at centre.
(ii) This compass box consists of pivoted magnet and aluminum pointer.
(iii) This compass box is having circular scale graduated with four quadrants.
Working:
(i) Two magnetic fields are perpendicular to each other.
(ii) Magnetic induction due to the current in the coil acting to normal to the plane of the coil.
(iii) Magnetic induction at the centre of the coil,
\(B=μ_o\frac{NI}{2R}\)   .....(2)
Sub. eqn.(1) in eqn. (2)
\(B_H tan \theta =μ_o\frac{NI}{2R}\)
\(B_H =μ_o\frac{NI}{2R}\frac{1}{tan\theta}\)

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.

Capacitor

Magnetic dipole moment

(b) Assertion and Reason are true but Reason is the false explanation of the Assertion.

II only

both are correct