Magnetic field is the region or space around every magnet within which its influence will be felt by keeping another magnet in that region.
\(\vec { B } =\frac{1}{q_m}\vec{E}\)
Its unit is N A^-1m^-1

The component of Earth's magnetic field along the horizontal direction in the magnetic meridian is called horizontal component of Earth's magnetic field, denoted by B_H.

Magnetic susceptibility is defined as the ratio of the intensity of magnetisation (\(\vec { M } \)) induced in the material due to the magnetising field |\(\vec H\)|.
\( \chi _{ m }=\frac { |\vec { M } | }{ |\vec { H } | } \).

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

Velocity of the particle is \(\vec { v } =v\hat { j } \)
(a) Magnetic field is along positive y-direction, this implies \(\vec B=B\hat { j } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { j } )=\vec 0\)
So, no force acts on the particle when it moves along the direction of magnetic field.
(b) Since the magnetic field points in positive z - direction, this implies, \(\vec { B } =B\hat { k } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { k } )=qvB\vec i \)
Therefore, the magnitude of the Lorentz force is qvB and direction is along positive x - direction.
(c) Magnetic field is in zy - plane and making an angle θ with the velocity of the particle, which implies \( {\vec B } =Bcos\theta \hat { j } +Bsin\theta \hat { k } \)

From Lorentz force,
\({ \vec F }_{ m }=q(v\hat { j } )\times (Bcos\theta \hat { j } +Bsin\theta \hat k)\) 
\(=qvBsin\theta \hat { i } \)

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.

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

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

The like currents i.e current in both the wire are in the same direction attracts each other. The force is repulsive when the current flows in opposite direction through the wires.
F = \(\frac { { \mu }_{ 0 }{ I }_{ 1 }{ I }_{ 2 }dl }{ 2\pi r } \)
i.e \(F\alpha \frac { 1 }{ r } \)

As the wire of the loop carrying the opposite current is near so the net force acting on the loop is repulsive.

The required resistance has the least value for ammeter and maximum value in the case of a voltmeter.
(i) The shunt resistance required to convert a galvanometer to an ammeter has the value
S = \(\frac { { I }_{ g } }{ I-{ I }_{ g } } \) x G
The shunt required for the milliammeter be a higher value.
(ii) Similarly, The voltmeter should have high resistance, the value of the required resistance should be highest in the case of the voltmeter. This is connected in series with the coil of the galvanometer.

Net force an electron moving in the combined electric field \(\vec { E } \) and a magnetic field \(\vec { B } \) is
\(\vec { F } =-e[\vec { E } +\vec { v } \times \vec { B } ]\)
Since electron moves undeflected then \(\vec { F } \) = 0
\(\vec { E } +(\vec { v } \times \vec { B } )\) = 0
\(|\vec { E } |=(|\vec { v } |\times |\vec { B } |)\Rightarrow |\vec { v } |=\frac { |\vec { E } | }{ |\vec { B } | } \). 

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 \(\vec { { p }_{ m } } =I\vec { A } =Iab\hat { n } \)
\(\vec { \tau _{ net } } =\vec { p } \times \vec { B } \).