B_H = B (i.e the horizontal component of earth's magnetic field B_H = B)
I = 60° ; (i.e the horizontal component of earth's magnetic field B_H = B)
B_H = BE cos I
(B_E - Net earth's magnetic field)
B = BE cos 60° ⇒ BE = \(\frac { B }{ cos{ 60 }^{ 0 } } \) ⇒ BE = 2B
At equator I = 0  (∵ cos0° = 1)
B_H = 2B. cos0° = 2B
B_H = 2B

It is defined as the ability of the materials to retain the magnetism in them even magnetising field vanishes.

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

The magnetic field due to current in the upper semicircle and lower semicircle of the circular coil are equal in magnitude but opposite in direction. Hence, the net magnetic field at the center of the loop (at point O) is zero \(\overset { \rightarrow }{ B } =\overset { \rightarrow }{ 0 } \).

\(\overset { \rightarrow }{ B } =\mu _{ ° }(\overset { \rightarrow }{ H+ } \overset { \rightarrow }{ M } )\) 
But from equation (3.33), in vector form,
\(\overset { \rightarrow }{ M } ={ x }_{ m }\overset { \rightarrow }{ H } \)
Hence, \(\overset { \rightarrow }{ B } =\mu _{ ° }({ x }_{ m }+1)\overset { \rightarrow }{ H } \Rightarrow \overset { \rightarrow }{ B } =\mu \overset { \rightarrow }{ H } \)
where, \(\mu =\mu _{ ° }({ x }_{ m }+1)\Rightarrow { x }_{ m }+1=\frac { \mu }{ \mu _{ ° } } =\mu _{ r }\)
\(\Rightarrow { x }_{ m }=\mu _{ r }-1\)

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.

When an electric charge q is moving with velocity \(\vec { v } \) in the magnetic field \(\vec { B } \), it experiences a force, called magnetic force \(\vec { { F }_{ m } } \). After careful experiments, Lorentz deduced the force experienced by a moving charge in the magnetic field \(\vec { { F }_{ m } } \).
\(\vec { { F }_{ m } } =q(\vec { v } \times \vec { B } )\)             ...........(1)
In magnitude, F_m = qvB sinθ         .......(2)
The equations (1) and equation (2) imply
(i) \(\vec { { F }_{ m } } \) is directly proportional to the magnetic field \(\vec { B } \).
(ii) \(\vec { { F }_{ m } } \) is directly proportional to the velocity \(\vec { v } \).
(iii) \(\vec { { F }_{ m } } \) is directly proportional to sine of the angle between the velocity and magnetic field.
(iv) \(\vec { { F }_{ m } } \) is directly proportional to the magnitude of the charge q.
(v) The-direction of \(\vec { { F }_{ m } } \) is always perpendicular to \(\vec { v } \) and B as \(\vec { { F }_{ m } } \) in the cross product of \(\vec { v } \) and \(\vec { B } \).

(vi) The direction of \(\vec { { F }_{ m } } \) on a negative charge is opposite to the direction of \(\vec { { F }_{ m } } \) on positive charge provided other factors are identified as shown in Figure.
(vii) If the velocity \(\vec { v } \) of the charge, q is along the magnetic field \(\vec { B } \) then, \(\vec { { F }_{ m } } \) is zero.

The magnetic field from the center of a circular loop of radius R along the axis is given by
\(\vec { B } =\frac { { \mu }_{ 0 }I }{ 2 } \frac { { R }^{ 2 } }{ ({ R }^{ 2 }+{ z }^{ 2 })^{ \frac { 3 }{ 2 } } } \hat { k } \)
At larger distance z >> R, therefore R² + z² ≈ z²,
we have
\(\vec { B } =\frac { { \mu }_{ 0 }I }{ 2 } \frac { { R }^{ 2 } }{ { z }^{ 3 } } \hat { k } \)               ........(1)
Let A be the area of the circular loop A = πR². So rewriting the equation (1) in terms of the area of the loop, we have
\(\vec { B } =\frac { { \mu }_{ 0 }I }{ 4\pi } \frac { { R }^{ 2 } }{ { z }^{ 3 } } \hat { k } \)
\(\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2IA }{ { z }^{ 3 } } \hat { k } \)              .......(2)
Comparing equation (2) with equation (1) dimensionally, we get
p_m = IA
where P_m is called a magnetic dipole moment. In vector notation,
\(\vec { { p }_{ m } } =I\vec { A } \)       .........(3)
This implies that a current - carrying circular loop behaves as a magnetic dipole of the magnetic moment \(\vec { { p }_{ m } } \) So, the magnetic dipole moment of any current loop is equal to the product of the current and area of the loop.

The magnitude of the reverse magnetising field for which the residual magnetism of the material vanishes is called its coercivity.

Since isotopes are singly ionized, they have equal charge which is equal to the charge of an electron, q = - 1.6 x 10^-19 C. Mass of uranium \(_{ 92 }^{ 235 }{ U and _{ 92 }^{ 238 }{ U } }\) are 3.90 x 10^-25 kg and 3.95 x 10^-25 kg respectively. Magnetic field applied, B = 0.500 T. Velocity of the electron is 1.00 x 10⁵ m s^-1, then 
(a) the radius of the path of \(_{ 92 }^{ 235 }{ U }\) is r₂₃₅
\({ r }_{ 235 }=\frac { { m }_{ 235 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =48.8\times { 10 }^{ -2 }m\)
r₂₃₅ = 48.8cm 
The diameter of the semi-circle due to \(_{ 92 }^{ 235 }{ U\ \ is \ \ { d }_{ 235 }=2{ r }_{ 235 } }\) = 97.6 cm
The radius of the path of _{\(_{ 92 }^{ 238 }{ U\ is\ 2{ r }_{ 238 }\ then}\)}
\({ r }_{ 238 }=\frac { { m }_{ 238 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =49.4\times { 10 }^{ -2 }m\)
r₂₃₈ = 49.4 cm
The diameter of the semi-circle due to  \(^{ 238 }_{92}{ U\ is \ 2{ r }_{ 238 } \ =98.8 \ cm}\)
Therefore the separation distance between the isotopes is \(\triangle d={ d }_{ 238 }-{ d }_{ 235 }=1.2 \ cm\)
(b) The time taken by each isotope to complete one semi-circular path are
\({ t }_{ 235 }=\frac { \text{ magnitude of the displacement} }{ velocity } \)
\(=\frac { 97.6\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.76\times { 10 }^{ -6 }s=9.76\mu s\)
\({ t }_{ 238 }=\frac { \text{magnitude of the displacement }}{ velocity } \)
\(=\frac { 98.8\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.88\times { 10 }^{ -6 }s=9.88\mu s\) 

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

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

Current I₁ = 2A ; I₂ = 5A
Two wires are separated by a distance a = 1 cm
= 1 x 10^-2m
The force between two parallel wires per unit length
F =?
F = \(\frac { { \mu }_{ 0 } }{ 2\pi } .\frac { { I }_{ 1 }{ I }_{ 2 } }{ a } \)
= 2 x 10^-7 x \(\frac { 2\times 5 }{ 1\times { 10 }^{ -2 } } \)
F = 20 x 10^-5 N
This force is repulsive F = 20 x 10^-5 N.

In the general case, the unit normal vector \(\hat { n } \) and magnetic field \(\vec { B } \) is with an angle 8 as shown in Figure.

(a) The force on section PQ
\(\vec { i } =a\hat { j } \) and \(\vec { B } =B\hat { i } \)
\(\vec { { F }_{ PQ } } =\vec { Il } \times \vec { B } =IaB(\hat { j } \times \hat { j } )=-IaB\hat { k } \)
Since the unit vector normal to the plane \(\hat { n } \) is along the direction of .\(\vec { k } \)
(b) The force on section QR
\(\vec { l } =bcos\left( \frac { \pi }{ 2 } -\theta \right) \hat { i } -sin\left( \frac { \pi }{ 2 } -\theta \right) \hat { k } \)
\(\vec { { F }_{ QR } } =\vec { Il } \times \vec { B } =-IbB\left( \frac { \pi }{ 2 } -\theta \right) \hat { j } \)
\(\vec { { F }_{ QR } } =-IbBcos\theta \hat { j } \)
(c) The force on section RS
\(\vec { l } =a\hat { j } \) and \(\vec { B } =B\hat { i } \)
\(\vec { { F }_{ RS } } =\vec { Il\times \vec { B } } =IaB(\hat { j } \times \hat { j } )=-IaB\hat { k } \)
Since the unit vector normal to the plane is along the direction of \(\hat { k } \).
(d) The force on section SP
\(\vec { l } =bcos\left( \frac { \pi }{ 2 } -\theta \right) \hat { i } +sin\left( \frac { \pi }{ 2 } +\theta \right) \hat { k } \quad \vec { B } =B\hat { i } \)
\(\vec { { F }_{ SP } } =\vec { Il } \times \vec { B } =IbBsin\left( \frac { \pi }{ 2 } -\theta \right) \hat { j } \)
\(\vec { { F }_{ SP } } =-IbBcos\theta \hat { j } \)
The net force on the rectangular loop is
\(\vec { { F }_{ net } } =\vec { { F }_{ PQ } } +\vec { { F }_{ QR } } +\vec { { F }_{ RS } } +\vec { { F }_{ SP } } \)
\({ F }_{ net }=IaB\hat { k } -IbBcos\theta \hat { j } -IaB\hat { k } +IbBcos\theta \hat { j } \)
\(\vec { { F }_{ net } } =\vec { 0 } \)
Hence, the net force on the rectangular loop in this configuration is also zero. Notice that the force on section QR and SP is not zero here. But, they have equal and opposite effects, but we assume that the loop to be rigid, so no deformation. So, no torque was produced by these two sections.
Even though the forces PQ and RS also are equal and opposite, they are not collinear. So these two forces constitute a couple as shown in Figure (a). Hence the net torque produced by these two forces about the axis of the rectangular loop is given by
\(\vec { { \tau }_{ net } } =baBIsin\theta \hat { k } =ABIsin\theta \hat { k } \)

\(\vec { OA } =\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (-\hat { i } )+\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (-\hat { k } )\)
=\(\frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } )\)
\(\vec { OB } =\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (\hat { i } )+\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (\hat { k } )\)
=\(\frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } )\)
\(\vec { OA } \times \vec { { F }_{ PQ } } =\left\{ \frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } \right\} \times \left\{ IaB\hat { k } \right\} \)
= \(\frac { 1 }{ 2 } IabBsin\theta \hat { j } \)
\(\vec { OA } \times \vec { { F }_{ RS } } =\left\{ \frac { b }{ 2 } (sin\theta \hat { i } +cos\theta \hat { k } \right\} \times \left\{ -IaB\hat { k } \right\} \)
= \(\frac { 1 }{ 2 } IabBsin\theta \hat { j } \)
The net torque \(\vec { \tau _{ net } } =IaBsin\theta \hat { j } \)         ..........(1)
Note that the net torque is in the positive y-direction which tends to rotate the loop in a clockwise direction about the y axis. If the current is passed in the other way (P⟶S⟶R⟶Q⟶P), then total torque will point in the negative y-direction which tends to rotate the loop in an anticlockwise direction about the y-axis.
Another important point is to note that the torque is less in this case compared to the earlier case (where the \(\hat { n } \) is perpendicular to the magnetic field \(\vec { B } \)). It is because the perpendicular distance is reduced between the forces \(\vec { { F }_{ PQ } } \) and \(\vec { { F }_{ RS } } \) in this case.
The equation (1) can also be rewritten in terms of magnetic dipole moment \(\vec { { p }_{ m } } =I\vec { A } =Iab\hat { n } \)
\(\vec { \tau _{ net } } =\vec { p } \times \vec { B } \).