Using right hand rule, current flows upwards.

\(\overset { \rightarrow }{ B } =\mu _{ ° }(\overset { \rightarrow }{ H+ } \overset { \rightarrow }{ M } )\) 
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 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 

The magnetic field at point O is zero. Because the upper and lower current carrying conductors are identical and so the magnetic fields caused by them at centre O will be equal and opposite.

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.

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

Given magnetic moment 0.5 J T^-1 and distance r = 0.1 m
(a) When the point lies on the axial line of the bar magnet, the magnetic field for short magnet is given by
\({ { \vec B }_{ axial } } =\frac { { \mu }_{ ° } }{ 4\pi } \left( \frac { 2{ p }_{ m } }{ { r }^{ 3 } } \right) \hat { i } \)
\({ { \vec B }_{ axial } } =1{ 0 }^{ -7 }\times \left( \frac { 2\times 0.5 }{ { \left( 0.1 \right) }^{ 3 } } \right) =1\times { 10 }^{ -4 }\hat { i } \ T\)
Hence, the magnitude of the magnetic field along axial is B_axial = 1 x 10^-4 T and direction is towards South to North.
(b) When the point lies on the normal bisector (equatorial) line of the bar magnet, the magnetic field for short magnet is given by
\({ {\vec B }_{ equatorial } } =-\frac { { \mu }_{ ° } }{ 4\pi } \frac { { p }_{ m } }{ { r }^{ 3 } } \hat { i } \)
\({ {\vec B }_{ equatorial } } =-1{ 0 }^{ -7 }\left( \frac { 0.5 }{ { \left( 0.1 \right) }^{ 3 } } \right) \hat { i } =-0.5\times 1{ 0 }^{ -4 }\hat { i } \ T \)
Hence, the magnitude of the magnetic field along axial is B_equatorial = 0.5 x 10^-4 T and direction is towards North to South.
Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of B_axial and B_equatorial are opposite.

(i) Galvanometer to an Ammeter:

(i) Ammeter is an instrument used to measure current flowing in the electrical circuit.
(ii) The Ammeter must offer low resistance such that it will not change the current passing through it. So, ammeter is connected in series to measure the circuit current.
(iii) A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
(iv) Let I be the current passing through the circuit. When current I reaches the junction A, it divides into two components.
a) I_g → Current passing through the galvanometer
b) I - I_g → Current passing through the shunt resistance.
(v) The potential difference across the galvanometer is same as the potential difference across the shunt resistance.
\(\mathrm{V}_{\text {galvanometer }} =\mathrm{V}_{\text {shunt }} \)
\(\Rightarrow \mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}} =\left(\mathrm{I}-\mathrm{I}_{g}\right) \mathrm{S} \)
\(\mathrm{S} =\frac{I_{g}}{\left(I-I_{g}\right)} R_{g} \)  (or)
\(\mathrm{I}_{\mathrm{g}}=\frac{S}{S+R_{g}} I \Rightarrow I_{g} \propto I\)
Since, the deflection in the galvanometer is proportional to the current passing through it.
\(\theta=\frac{1}{G} I_{g} \Rightarrow \theta \propto I_{g} \Rightarrow \theta \propto I\)
Where, R_g → Galvanometer resistance, S → Shunt resistance.
Since shunt resistance is connected in parallel to galvanometer,
Effective resistance,\(\frac{1}{R_{e f f}}=\frac{1}{R_{g}}+\frac{1}{S} \Rightarrow R_{e f f}=\frac{R_{g} S}{R_{g}+S}=R_{a}\)
R_a ⇒ low resistance. An ideal ammeter has zero resistance.
The percentage error in measuring a current through an ammeter is,
\(\frac{\Delta I}{I} \times 100 \%=\frac{I_{i d e a l}-I_{a c t u a l}}{I_{a c t u a l}} \times 100 \%\)
(ii) Galvanometer to a voltmeter:
i) A voltmeter is an instrument used to measure potential difference across any two points in the electrical circuits. 
ii) Voltmeter must have high resistance and when it is connected in parallel, it will rot draw appreciable current so that it will indicate the true potential difference.
iii) A galvanometer is converted into a voltmeter by connecting high resistance  R_h in series with galvanometer.
iv) Let R_g be the resistance of galvanometer and I_g be the current with which the galvanometer produces full scale deflection.
v) Since the galvanometer is connected in series with high resistance, the current in the electrical circuit is same as the current passing through the galvanometer.

\(\mathrm{I}=\mathrm{I}_{\mathrm{g}} \)
\(\mathrm{I}=I_{g} \Rightarrow I_{g}=\frac{\text { potential difference }}{\text { total resistance }} \)
Since the galvanometer and high resistance are connected in series, the voltmeter resistance is,
\(R_{v} =R_{g}+R_{h} \)
Therefore,
\(I_{g} =\frac{V}{R_{g}+R_{h}} \)
\(\Rightarrow R_{h} =\frac{V}{I_{g}}-R_{g} \)
Note that \(I_{g} \propto V\)
R_h is very large. An ideal voltmeter has infinite resistance

Principle : When a current carrying loop is placed in a uniform magnetic field it experiences a torque.
Construction : A moving coil galvanometer consists of a rectangular coil PQRS of insulated thin copper wire. The coil contains a large number of turns wound over a light metallic frame. A cylindrical soft-iron core is placed symmetrically inside the coil as shown in Figure. The rectangular coil is suspended freely between two pole pieces of a horse-shoe magnet.

The upper end of the rectangular coil is attached to one end of fine strip of phosphor bronze and the lower end of the coil is connected to a hair spring which is also made up of phosphor bronze. In a fine suspension strip, a small plane mirror is attached in order to measure the deflection of the coil with the help of lamp and scale arrangement. The other end of the mirror is connected to a torsion head. In order to pass electric current through the galvanometer, the suspension strip and the spring S are connected to terminals.
Working : Consider a single turn of the rectangular coil PQRS whose length be l and breadth b. PQ = RS = l and QR = SP = b.
Let I be the electric current flowing through the rectangular coil PQRS as shown in Figure. The horse-shoe magnet has hemi - spherical magnetic poles which produces a radial magnetic field. Due to this radial field, the sides QR and SP are always parallel to the magnetic field B and experience no force. The sides PQ and RS are always parallel to the magnetic field and experience force in opposite directions. Due to this, torque is produced.

For single turn, the deflection torque is,
て = bF = bBIl = (lb)BI
て = ABI
since, area of the coil A = lb
For coil with N turns, we get
て = NABI               ........(1)
Due to this deflecting torque, the coil gets twisted and restoring torque (also known as restoring couple) is developed. Hence the moment of restoring couple is proportional to the amount of twist θ. Thus
て = Kθ      ............(2)
where K is the restoring couple per unit twist or torsional constant of the spring.
At equilibrium, the deflection couple is equal to the restoring couple. Therefore by comparing equations (1) and (2), we get,
NABI = Kθ
⇒ I =\(\frac { K }{ NAB } \) θ      ...........(3)
(or) I = Gθ
where G = \(\frac { K }{ NAB } \) is called galvanometer constant or current reduction factor of the galvanometer.
Since, suspended moving coil galvanometer is very sensitive, we have to handle with high care while doing experiments. Most of the galvanometer we use are pointer type moving coil galvanometer.