When a current carrying conductor is placed in a magnetic field, the force experienced by the wire is equal to the sum of  Lorentz forces on the individual charge carries in the conductor. Consider a small segment of conductor of length dl, cross-sectional area A and current I as shown in figure. The free electron drift opposite to the direction of current. So the relation between current I and magnitude of drift velocity v_d is
\(I=n e A v_{d}\) ........(1)
If the conductor is kept in a magnetic field \(\vec{B}\) , then average force experienced by the charge (here, electron ) on the conductor is
\(\vec{F}=-e\left(\vec{\nu}_{d} \times \vec{B}\right)\)
Let n be the number of free electrons present in unit volume, therefore
\(\mathrm{n}=\frac{N}{V}\)
where, N is the number of free electrons in the small element of volume V = A dl
Hence Lorentz force \(\overrightarrow{d F}=-e n A d l\left(\overrightarrow{v_{d}} \times \vec{B}\right)\)
Current element,
\(I{\vec d F}=-e n A \vec v_dd l\)
\(d \vec{F}=(I d \vec{l} \times B)\)  ........(2)
The force on a straight current carrying conducting wire of length I placed in a uniform magnetic field is,
\(\vec{F}=(I\vec{l} \times \vec{B})\)
In magnitude,
\(\mathrm{F}=\mathrm{BI} l \sin \theta\)
(a) If the conductor is placed along the direction of the magnetic field, the angle between them is \(\theta\) = 0^o.
∴ F = 0
(b) If the conductor is placed perpendicular to the magnetic field, the angle between them is \(\theta\) = 90^o 
∴ F = BIl. (maximum)

S.No.	Properties	Soft ferromagnetic materials	Hard ferromagnetic materials
i	When external field is removed	Magnetisation disappears	Magnetisation persists
ii	Area of the loop	Small	Large
iii	Retentivity	Low	High
iv	Coercivity	Low	High
v	Susceptibility and magnetic permeability	High	Low
vi	Hysteresis loss	Less	More
vii	Uses	Solenoid core, transformer core and electromagnets	Permanent magnets
viii	Examples	Soft iron, Mumetal, Stalloy etc.	Carbon Steel, Alnico, Lodestone etc

Two long straight parallel current-carrying conductors separated by a distance r are kept in air medium. Let I₁ and I₂ be the electric currents passing through the conductors A and B is same direction (i.e., along z-direction) respectively. The net magnetic field at a distance r due to current I₁ in conductor A is
\(\vec{B}_{1}=\frac{\mu_{o} I_{1}}{2 \pi r}(-\hat{\mathrm{i}})=-\frac{\mu_{o} I_{1}}{2 \pi r} \hat{i}\)

From thumb rule, the direction of magnetic field is perpendicular to the plane of the paper and inwards (arrow into the page ⊗) i.e. along negative \(\vec{i}\)direction
Let us consider a small elemental length dl in conductor B at which the magnetic field \(\vec{B}_1\) present, From equation \(\overrightarrow{d F}=(I \overrightarrow{d l} \times \vec{B})\)
Lorentz force on the element dl of conductor B is
\(\overrightarrow{d F}=\left(I_{2} d \vec{l} \times \vec{B}_{1}\right)=-I_{2} d l \frac{\mu_{o} I_{1}}{2 \pi r}(\hat{k} \times \hat{i})=-\frac{\mu_{o} I_{1} I_{2} d l}{2 \pi r} \hat{j}\)
Therefore the force on dl of wire conductor B is directed towards the conductor A. So the element of length dl in B is attracted towards the conductor A. Hence, the force per unit length of the conductor B due to the current in the conductor A is,
\(\frac{\vec{F}}{l}=-\frac{\mu_{o} I_{1} I_{2}}{2 \pi r} \hat{j}\)
Similarly, the net magnetic induction due to current I₂ (in conductor B) at a distance r in the elemental length dl of conductor A is
\(\vec{B}_{2}=\frac{\mu_{o} I_{2}}{2 \pi r} \hat{i}\)
From the thumb rule direction of magnetic field is perpendicular to the plane of the paper and outwards (arrow out to the page ⊙) i.e., along positive \(\vec{i}\)direction.
Hence the magnetic force at element dl of the conductor A is,
\(\vec{dF} =\left(I_{1} \vec{d} l \times \vec{B}_{2}\right)=I_{1} d l \frac{\mu_{o} I_{2}}{2 \pi r}(\hat{k} \times \hat{i}) \)
\(=\frac{\mu_{o} I_{1} I_{2} d l}{2 \pi r} \hat{j} \)
Therefore the force on dl of conductor A is directed towards the conductor B. So the length dl is attracted towards the conductor B as shown in Figure.

The force acting per unit length of the conductor A due to the conductor B is
\(\frac{\vec{F}}{l}=-\frac{\mu_{0} I_{1} I_{2}}{2 \pi r} \hat{j}\)
Attractive force: The direction of electric current is same.
Repulsive force: The direction of electric current is opposite.

Consider a rectangular loop PQRS carrying current I is placed in a uniform magnetic field \(\vec{B}\). Let a and b be the length and breadth rectangular loop respectively. The unit vector n̂ normal to the plane of the loop makes an angle θ with the magnetic field.
(i) The magnitude of the magnetic force acting on the current-carrying arm PQ is F_PQ = IaBsin \((\frac{π}{2})\)= IaB. The direction of the force is found using right hand cork screw rule and its direction is upwards.

(ii) The magnitude of the force on the arm QR is F_QR = IbB sin \(\left(\frac{\pi}{2}-\theta\right)\) = lbB cos θ and its direction is as shown in the figure.
(iii) The magnitude of the force on the arm RS is F_RS = IaB sin \((\frac{π}{2})\) = IaB and its direction is downwards.
(iv) The magnitude of the force acting on the arm SP is F_SP = lbBsin \(\left(\frac{\pi}{2}+\theta\right)\) = lbB cos θ and its direction is as shown in the figure.
(v) Since the forces F_QR and F_SP are equal, opposite and collinear, they cancel each other. But the forces F_PQ and F_RS which are equal in magnitude and opposite in direction, are not acting along same straight line. Therefore, F_PQ and F_RS constitute a couple which exerts a torque on the loop.
The magnitude of torque acting on the arm PQ about AB is \(\tau_{PQ}=(\frac{b}{2} sin \theta)IaB\) and it points in the direction of AB. The magnitude of the torque acting on the arm RS about AB is \(\tau_{RS}=(\frac{b}{2} sin \theta)IaB\) and points also in the same direction AB as shown in Figure.

The total torque acting on the entire loop about an axis AB is given by,
\(\tau=\left(\frac{b}{2} \sin \theta\right) F_{P Q}+\left(\frac{b}{2} \sin \theta\right) F_{R S}\) 
= Ia (b sinθ) B
\(\tau\) = IAB sinθ along the direction B.
In vector form,
\(\vec{\tau}=(\mathrm{I\overrightarrow{A}}) \times \overrightarrow{\mathrm{B}}\) 
(vi) The above equation can also be written in terms of magnetic dipole moment. 
\(\vec{\tau}={\vec p_m} \times \overrightarrow{\mathrm{B}} \text { where, }{\mathrm{\vec p_m}}={\mathrm{I\overrightarrow A}}\) 
The tendency of the torque is to rotate the loop so as to align its normal vector with the direction of the magnetic field.
(vii) If there are N turns in the rectangular loop, the torque is given by,
\(\tau\) = NIAB sinθ.
Special cases:
a) When θ = 90° or the plane of the loop is parallel to the magnetic field, the torque on the current loop is maximum.
τ_max = IAB
b) When θ = 0°/180° or the plane of the loop is perpendicular to the magnetic field, the torque on the current loop is zero.

Consider a solenoid of length L having N turns. The diameter of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

Consider a rectangular loop abcd. Then from Ampere's circuital law,
\(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } \) = μ₀ I_enclosed = μ₀ x (total current enclosed by Amperian loop)
The left hand side of the equation is
\(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } =\int _{ a }^{ b }{ \vec { B } .\vec { dl } } +\int _{ b }^{ c }{ \vec { B } .\vec { dl } } +\int _{ c }^{ d }{ \vec { B } .\vec { dl } } +\int _{ d }^{ a }{ \vec { B } .\vec { dl } } \)
Elemental lengths bc and da are perpendicular to magnetic field.
\(\therefore \int _{ b }^{ c }{ \vec { B } .\vec { dl } } =\int _{ b }^{ c }{ |\vec { B } ||\vec { dl } | } cos 90^o = 0\)
similarly,
\(\int _{ d }^{ a }{ \vec { B } .\vec { dl } } = 0\) 
Since the magnetic field outside the solenoid is zero,
\(\int _{ c }^{ d }{ \vec { B } .\vec { dl } } =0\) and
\(\int _{ a }^{ b }{ \vec { B } .\vec { dl } } =BL \quad \quad (\because \theta =0^o)\)
Let I be the current passing through the solenoid of N turns, then
\(\int _{ a }^{ b }{ \vec { B } .\vec { dl } }  = BL = μ_o NI ⇒ B = μ_o\frac {NI}{L}\)
The number of turns per unit length is given by \(\frac { N }{ L } \) = n, Then
B = \(\mu_o \frac { nLI }{ L } \) = μ₀nI
Since n is a constant for a given solenoid and μ₀ is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

(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

(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)
\(B_H tan \theta =μ_o\frac{NI}{2R}\)
\(B_H =μ_o\frac{NI}{2R}\frac{1}{tan\theta}\)

Cyclotron:
Device used to accelerate the charged particles to gain large kinetic energy.
Principle:
When a charged particle moves perpendicular to the magnetic field, it experiences magnetic Lorentz force.
Construction:
(i) The particles are allowed to move in between two semi-circular metal containers called Dees (hollow D - shaped objects).
(ii) The uniform magnetic field is controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees.
(iii) Source is kept between two Dees.
(vi) Dees are connected to high frequency alternating potential difference.
Working:
(i) The ion ejected from source is positively charged.
(ii) It is accelerated towards negative potential Dees
(iii) This ion undergoes a circular path.
(iv) At this time, the polarities of the Dees are reversed, so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.

\(\frac { m{ v }^{ 2 } }{ r } \) = qvB
⇒ r = \(\frac { m }{ qB } \)v ........(1)
⇒ r ∝ v
(v) If radius of the circular paths, increases, velocity also increases particles undergo spiral path with increasing radius.
(vi) When the frequency f at which the positive ion ciculates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc. This is called Resonance condition.
From equation, f = \(\frac { qB }{ 2\pi m } \) we have
f_osc = \(\frac { qB }{ 2\pi m } \),
The time period of oscillation is
T = \(\frac { 2\pi m }{ qB } \)
The kinetic energy of the charged particle is,
KE = \(\frac { 1 }{ 2 } mv^{ 2 }=\frac { { q }^{ 2 }B^{ 2 }{ r }^{ 2 } }{ 2m } \)   ........(2)
Limitations:
(i) The speed of ion is limited.
(ii) Electron cannot be accelerated.
(iii) Uncharged particles cannot be accelerated.

i) Let I be current flowing in infinite length of conductor.
ii) Amperian loop is constructed in the form of a circular shape at a distance r from the centre of the conductor.
iii) dl is the line element along the loop.

From the Ampere's law \(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } \) = μ_oI
Hence, the angle between magnetic field vector and line element is zero. Therefore, Here, the angle between magnetic field vector and line element is zero.
\(\oint _{ C }^{ }{ {B dl } } \) = μ_oI
For a circular loop, the circumference is 2πr, which implies,
B\(\int _{ 0 }^{ 2\pi r }{ dl } \) =  μ_oI
\(\vec { B } \).2πr =  μ_oI
B = \(\frac { { \mu }_{ 0 }I }{ 2\pi r } \)
In vector form, the magnetic field is 
\(\vec { B } =\frac { { \mu }_{ 0 }I }{ 2\pi r } \hat { n } \) 
where \(\hat { n } \) is the unit vector along the tangent to the Amperian loop as shown in the Figure.

(i) Consider a bar magnet NS and pole strength q_m and distance 2l. 
(ii) Let C be point along the equatorial line.
(iii) The magnetic field at a point C (lines along the equatorial line) at a distance r from the geometrical center O of the magnet can be computed by keeping unit north pole (q_mC = 1 A m) at C.
\(\vec { { B }_{ N } } =-{ B }_{ N }cos\theta \hat { i } +{ B }_{ N }sin\theta \hat { j } \)    .....(1)
where B_N = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m } }{ r^{ '2 } } \)


The magnetic field at C due to south pole is,
\(\vec { { B }_{ s } } =-{ B }_{ s }cos\theta \hat { i } -{ B }_{ s }sin\theta \hat { j } \)              .....(2)
where B_s = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m } }{ r^{ '2 } } \)
From equations (1) and (2), the net magnetic field at point C due to dipole is \(\vec { { B } } =\vec { { B }_{ N } } +\vec { { B }_{ S } } \).  
\(\vec { { B } } =-({ B }_{ N }+{ B }_{ S })cos\theta \hat { i } \) Since, B_N = B_S
\(\vec { { B } } =-\frac { { 2\mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m } }{ r'^{ 2 } } cos\theta \hat { i } =-\frac { 2{ \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m } }{ ({ r }^{ 2 }+l^{ 2 }) } cos\theta \hat { i } \)       ....(3)
In a right angle triangle NOC, as shown in the figure,
cosθ=\(\frac { adjacent }{ hypotenuse } =\frac { 1 }{ r' } =\frac { 1 }{ ({ r }^{ 2 }+{ l }^{ 2 })^{ \frac { 1 }{ 2 } } } \)          .........(4)
Substituting equation (4) in equation (3) we get
\(\vec { B } =-\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m }\times (2l) }{ ({ r }^{ 2 }+{ l }^{ 2 })^{ \frac { 3 }{ 2 } } } \hat { i } \)          ........(5)
Since, magnitude of magnetic dipole moment is \(|\vec { { p }_{ m } } |\) = p_m = q_m. 2l and substituting in equation (5), we get the magnetic field at a point C is
\( { { \vec B }_{ equatorial } } =-\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { p }_{ m } }{ ({ r }^{ 2 }+{ l }^{ 2 })^{ \frac { 3 }{ 2 } } } \hat { i } \)         ........(6)
If the distance between two poles in a bar magnet are small (looks like short magnet) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r >>l, then,
(r² + l²)^3\2 ≈ r³             ..........(7)
Therefore, using equation (7) in equation (6), we get
\( { { \vec B }_{ equatorial } } =-\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { p }_{ m } }{ r^{ 3 } } \hat { i } \)
In general, the magnetic field at equatorial point is given by
\({ { \vec B }_{ equatorial } } =-\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { \vec p }_{ m } }{ r^{ 3 } } \)   Since p_m\(\hat { i } =\vec { { p }_{ m } } \),    .......(8)

(i) Consider a bar magnet NS whose pole strength is q_m and length is 2l.
(ii) Let C be the point along axis of maget.
(iii) The magnetic field at a point C (lies along the axis of the magnet) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole (q_mc = 1 A m) at C. 

The magnetie field at C due to the north pole is,
\(\vec { { B }_{ N } } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m } }{ (r-l)^{ 2 } } \hat { i } \)   
where (r - I) is the distance between north pole of the bar magnet and unit north pole at C. The magnetic field at C due to the south pole is,
\(\vec { { B }_{ S } } =-\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m } }{ (r+l)^{ 2 } } \hat { i } \)  
where (r + I) is the distance between south pole of the bar magnet and unit north pole at C. The net magnetic field due to magnetic dipole at a point C
\(\vec { B } =\vec { { B }_{ N } } +\vec { B_{ S } } \)
\(\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m } }{ (r-l)^{ 2 } } \hat { i } +\left(- \frac { { \mu }_{ 0 } }{ 4\pi } \frac { { q }_{ m } }{ (r+l)^{ 2 } } \hat { i } \right) \)
\(\vec { B } =\frac { { \mu }_{ 0 }{ q }_{ m } }{ 4\pi } \left( \frac { 1 }{ (r-l)^{ 2 } } -\frac { 1 }{ (r+l)^{ 2 } } \right) \hat { i } \)
\(\vec { B } =\frac { { \mu }_{ 0 }2r }{ 4\pi } \left( \frac { { q }_{ m }.(2l) }{ ({ r }^{ 2 }-{ l }^{ 2 })^{ 2 } } \right) \hat { i } \)
Since, magnitude of magnetic dipole moment is \(|\vec { { p }_{ m } } |\) = p_m = q_m. 2l the magnetic field at a point C can be written as,
\(\vec { { B }_{ axial } } =\frac { { \mu }_{ 0 } }{ 4\pi } \left( \frac { 2rp_{ m } }{ { (r^2-l^2)}^{ 2 } } \right) \hat { i } \)     
If r >> I then, (r² - l²)² ≈ r⁴
\( { { \vec B }_{ axial } } =\frac { { \mu }_{ 0 } 2r}{ 4\pi } \left( \frac { p_{ m } }{ { r }^{ 4 } } \right) \hat { i } =\frac { { \mu }_{ 0 } }{ 4\pi }[ \frac { 2 \vec p_{ m } }{ { r }^{ 3 }}] \)      
∵ \(\vec { { p }_{ m } } =p_{ m }\hat { i } \).

(i) Let 'R' be the radius of a current carrying circular loop.
(ii) I be the current flowing through the wire.
(iii) Let P be a point on the axis of the circular coil at a distance z from its centre 'O'

(iv) Take two diametrically opposite element \(\vec { dl } \) at C and D. According to Biot-Savart's law, the magnetic field at P due to the current element at C is
\(d \vec{B}=\frac{\mu_0}{4 \pi} \frac{I d \vec{l} \times \hat{r}}{r^2}\)
The magnitude of \( { d\vec B } \)is
\(d \vec{B}=\frac{\mu_0}{4 \pi} \frac{I d l \sin \theta}{r^2}=\frac{\mu_0}{4 \pi} \frac{I d l}{r^2}\)
where θ is the angle between  \(I\vec { dl } \) and \(\vec { r } \). Here, θ = 90^o.
\(\vec{B}  =\int d \vec{B}=\int d B \sin \phi \hat{k}  \)
\(\vec{B}  =\frac{\mu_o I}{4 \pi} \int \frac{d l}{r^2} \sin \phi \hat{k}  \)
\(\text {But, } \cos \theta  =\frac{R}{\left(R^2+z^2\right)^{\frac{1}{2}}} \text { (using Pythagoras theorem) }\)
From ΔOCP
\(\sin \phi=\frac{R}{\left(R^2+z^2\right)^{1 / 2}} \text { and } r^2=R^2+z^2.\)
Substituting these in the above equation, we get,
\(\vec{B}=\frac{\mu_0 I}{4 \pi} \frac{R}{\left(R^2+z^2\right)^{3 / 2}} \hat{k}\left(\int d l\right)\)
If we integrate the line element from 0 to 2πR, we get the net magnetic field \(\vec{B}\) at any point P due to the current - carrying circular loop,
\(\vec{B}=\frac{\mu_0 I}{2} \frac{R^2}{\left(R^2+z^2\right)^{3 / 2}} \hat{k}\)
If the circular coil contains N turns, then the magnetic field is
\(\vec{B}=\frac{\mu_0 N I}{2} \frac{R^2}{\left(R^2+z^2\right)^{3 / 2}} \hat{k}\)
The magnetic field at the centre of the coil is,
\(\vec{B}=\frac{\mu_0NI}{2R}\hat k\)  since z= 0

Let YY' be an infinitely long straight conductor carry current I. In order to calculate magnetic field at a point P which is at a distance a from the wire, let us consider a small line element dl (segment AB).
According to Biot Savart law, the magnetic field at a point P due to current element Idl is,
\({ d \vec B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { Idl sin \theta} }{ { r }^{ 2 } }\hat n \).
To apply trigonometry, draw a perpendicular AC to the line BP as shown in Figure.
In triangle ΔABC, \(\sin \theta=\frac{\mathrm{AC}}{\mathrm{AB}}\)
∴ AC = AB sinθ
\(\text { But, } A B  =d l \Rightarrow A C=d l \sin \theta\)
Let dΦ be the angle subtended between AP and BP
ie., \(\angle \mathrm{APB}=\angle \mathrm{APC}=d \phi\)
In a triangle \(\triangle \mathrm{APC}, \sin (d \phi) \simeq A C / A P\)
Since, dΦ is very small, \(\sin (d \phi) \simeq d \phi\)
But,  \(\mathrm{AP}  =r \Rightarrow A C=r d \phi \)
\(\therefore \mathrm{AC} =d l \sin \theta=r d \phi \)
\(\therefore d \vec{B}  =\frac{\mu_0}{4 \pi} \frac{I}{r^2}(r d \phi) \hat{n}=\frac{\mu_0}{4 \pi} \frac{I d \phi}{r} \hat{n}\)
Let Φ be the angle between AP and OP
\(\text {In a } \triangle \mathrm{OPA}, \cos \phi =\frac{\mathrm{OP}}{\mathrm{AP}}=\frac{\mathrm{a}}{\mathrm{r}}  \)
\(r  =\frac{a}{\cos \phi}  \)
\(\text {Now, } d \vec{B}  =\frac{\mu_0}{4 \pi} \frac{I}{a / \cos \phi} d \phi . \hat{n}  \)
\(d \vec{B}  =\frac{\mu_0 I}{4 \pi a} \cos \phi d \phi \hat{n}\)
The total magnetic field at P due to the conductor YY' is
\(\vec { B } = \int _{- \Phi _{ 1 } }^{ { \Phi }_{ 2} }d\vec B =\int _{ -\Phi _{ 1 } }^{ { \Phi }_{ 2 } }\frac { { \mu }_{ 0 }I }{ 4\pi a }{ cos\phi d\phi } \hat { n }\)
\(=\frac { { \mu }_{ 0 }I }{ 4\pi a }[{ sin\phi ]^{\phi_2} _{\phi_-1}} \hat { n }\)
\( \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } (sin{ \Phi }_{ 1 }+sin{ \Phi }_{ 2 })\hat { n } \)
For infinitely long conductor, Φ₁ =  Φ₂ = 90^o
\(\therefore \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } \times 2\hat{n}\Rightarrow\vec { B } =\frac { { \mu }_{ 0 }I }{ 2\pi a } \hat { n } \)
                                                                                                                 

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.

In a right angle triangle OPN let the angle \(\angle\)OPN = \(\theta \)₁ which implies, \({ \varphi }_{ 1 }=\frac { \pi }{ 2 } -{ \theta }_{ 1 }\) and also in a right angle triangle OPM, 
\(\angle\)OPN = \(\theta \)₂ which implies, \({ \varphi }_{ 2 }=\frac { \pi }{ 2 } +{ \theta }_{ 2 }\)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)

i) Consider a bar magnet of length 2l and pole strength q_m
ii) Force experienced by the magnet at each pole is q_m B (equal) in opposite direction. 
iii) So, magnet experiences a torque.

The force experienced by north pole,
\(\vec { { F }_{ N } } ={ q }_{ m }\vec { B } \)             
The force experienced by south pole,
\(\vec { { F }_{ S } } =-{ q }_{ m }\vec { B } \)
∴ The net force acting on the dipole is,
\(\vec { F } =\vec { { F }_{ N } } +\vec { F_{ S } } =\vec { 0 } \)
The moment of force or torque experienced by north and south pole about point O is,
\(\vec { \tau } =\vec { ON } \times \vec { { F }_{ N } } +\vec { OS } \times \vec { { F }_{ S } } \)
\(\vec { \tau } =\vec { ON } \times { q }_{ m }\vec { B } +\vec { OS } \times (-{ q }_{ m }\vec { B } )\)
By using right hand cork screw rule, we conclude that the total torque is pointing into the paper. Since the magnitudes \(|\vec { ON } |=|\vec { OS } |=l\)  and \(|{ q }_{ m }\vec { B } |=|-{ q }_{ m }\vec { B } |\).
The magnitudes of total torque about point O is
ፒ = l x q_mB sinθ + l x q_mB sinθ
ፒ = 2l x q_mB sinθ
ፒ = p_mB sinθ   (∴ q_m x 2l = p_m)
In vector notation, \(\vec { \tau } =\vec { { p }_{ m } } \times \vec { B } \).