It is an arrangement of eletric field (E) and magnetic field (B) perpendicular to each other. When charged particles enter that region, particles with a certain velocity can pass through that region.
v = E/B
The speed is independent of charge and mass.

An ammeter is connected in series with the circuit because the purpose of the ammeter is to measure the current through the circuit. Since the ammeter is a low impedance device connecting it in parallel with the circuit would cause a short circuit, damaging the ammeter and the circuit.

(i) Stretch out forefinger, the middle finger and the thumb of the left hand such that they are in three mutually perpendicular directions.
(ii) If the forefinger points in the direction of magnetic field, the middle finger in the direction of the electric current, then thumb will point in the direction of the force experienced by the conductor.

Resonance condition happens, when the frequency f at which the positive ion circulates in the magtetic field must be equal to the constant frequency of the electrical oscillator f_osc.
 \(\mathrm{f}_{\mathrm{osc}}=\frac{\mathrm{q} \mathrm{B}}{2 \pi \mathrm{m}}\)

(i) Magnetic declination is the angle between magnetic meridian at a point and geographical meridian.
(ii) Magnetic inclination or dip at a point is defined as the angle subtended by the Earth's total magnetic field \(\vec{B}\) with the horizontal direction in the magnetic meridian.

One ampere is defined as that constant current when it is passed through each of the two infinitely long parallel straight conductors kept side by side parallely at a distance of one meter apart in vacuum causes each conductor to experience a force of 2 x 10^-7 newton per meter length of conductor.

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.

Magnetic permeability is the measure of ability of the material to allow the passage of magnetic field lines through it.

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 \(\vec { dl } \).
\(d\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { I\vec { dl } \times \hat { r } }{ { r }^{ 2 } } \)

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

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.

Lorentz deduced the force experienced by a moving charge in the magnetic field \(\overrightarrow{F_{m}}\)
\(\overrightarrow{F_{m}}=q(\vec{v} \times \vec{B})\)   ........(1)
In magnitude, 
\(\mathrm{F}_{\mathrm{m}}=\mathrm{qvB} \sin \theta\)   ...(2)
The equations (1) and (2) imply
i) \(\overrightarrow{F_{m}}\) directly proportional to the magnetic field \(\overrightarrow{B}\)
ii) \(\overrightarrow{F_{m}}\) is directly proportional to the velocity \(\overrightarrow{v}\) of the moving charge
iii) \(\overrightarrow{F_{m}}\) is directly proportional to sine of the angle between the velocity and magnetic field
iv) \(\overrightarrow{F_{m}}\) is directly proportional to the magnitude of the charge q
v) The direction of \(\overrightarrow{F_{m}}\) is always perpendicular to \(\overrightarrow{v}\) and \(\overrightarrow{B}\) as \(\overrightarrow{F_{m}}\) is the cross product of \(\overrightarrow{v}\) and \(\overrightarrow{B}\)
vi) The direction of \(\overrightarrow{F_{m}}\) on negative charge is opposite to the direction of \(\overrightarrow{F_{m}}\) on positive charge provided other factors are identical.
vii) If velocity \(\overrightarrow{v}\) of the charge q is along magnetic field \(\overrightarrow{B}\) then, \(\overrightarrow{F_{m}}\) is zero.

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.

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.

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
\(\therefore \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } \times 2\hat{n}\Rightarrow\vec { B } =\frac { { \mu }_{ 0 }I }{ 2\pi a } \hat { n } \)
                                                                                                                 

(i) Galvanometer to an Ammeter

Ammeter is an instrument used to measure current flowing in the electrical circuit. 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.
A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer. This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of ammeter depends on the values of the shunt resistance
Let I be the current passing through the circuit as shown in Figure. When current I reaches the junction A, it divides into two components. Let I_g be the current passing through the galvanometer of resistance R₈ through a path AGE and the remaining current (l - I_g) passes along the path ACDE through shunt resistance S. The value of shunt resistance is so adjusted that current I_g produces full scale deflection in the galvanometer. The potential difference across galvanometer is same as the potential difference across shunt resistance.
\(\mathrm{V}_{\text {galvanometer }} =\mathrm{V}_{\text {shunt }} \)
\(\Rightarrow \mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}} =\left(\mathrm{I}-\mathrm{I}_{g}\right) \mathrm{S} \)
\(\mathrm{S} =\frac{I_{g}}{\left(I-I_{g}\right)} R_{g} \mathrm{Or} \)
\(\mathrm{I}_{\mathrm{g}}=\frac{S}{S+R_{g}} I \Rightarrow I_{g} \propto I\)
Since, the deflection in the galvanometer is proportional to the current passing through it.
\(\theta=\frac{1}{G} I_{g} \Rightarrow \theta \propto I_{g} \Rightarrow \theta \propto I\)
So, the deflection in the galvanometer measures the current I passing through the circuit (ammeter).
Shunt resistance is connected in parallel to galvanometer. Therefore, resistance of ammeter can be determined by computing the effective resistance, which is
\(\frac{1}{R_{e f f}}=\frac{1}{R_{g}}+\frac{1}{S} \Rightarrow R_{e f f}=\frac{R_{g} S}{R_{g}+S}=R_{a}\)
Since the shunt resistance is very low resistance and the ration is also small. This means R_g is also small, i.e., the resistance offered by the ammeter is small. So, when we connect ammeter in series, the, ammeter will not change the resistance appreciably and also the current in the circuit. For an ideal ammeter, the resistance must be equal to zero. Hence, the reading in ammeter is always lesser than the actual current in the circuit. Let I_ideal be current measured from ideal ammeter and I_actual be the actual current measured in the circuit by the ammeter. Then, the percentage error in measuring a curent 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
A voltmeter is an instrument used to measure potential difference across any two points in the electrical circuits. It should not draw any current ftom the circuit otherwise the value of potential difference to be measured will change.
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.
A galvanometer is converted into a voltmeter by connecting high resistance  R_h in series with galvanometer as shown in Figure. The scale is now calibrated in volt and the raoge of voltmeter depends on the values of the resistance connected in series i.e., the value of resistance is so adjusted that only current Is produces full scale deflection in the galvanometer.
Let R_g be the resistance of galvanometer and I_g be the current with which the galvanometer produces full scale deflection. 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.

\(\mathrm{I}=\mathrm{I}_{\mathrm{g}} \)
\(\mathrm{I}=I_{g} \Rightarrow I_{g}=\frac{\text { potential difference }}{\text { totalresistance }} \)
Since the galvanometer and high resistance are connected in series, the total resistance or effective resistance gives the resistance of voltmeter. The voltmeter resistance is
\(R_{v} =R_{g}+R_{h} \)
\(I_{g} =\frac{V}{R_{g}+R_{h}} \)
\(\Rightarrow R_{h} =\frac{V}{I_{g}}-R_{g} \)
Note that \(I_{g} \propto V\)

Biot-Savart's law states that the magnitude of magnetic field \(d\vec { B } \) at a point P at a distance 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 (say,θ) between \(\vec { dl } \) and \(\hat { r } \).
(iv) inversely as the square of the distance between the point P and length element \(\vec { dl } \).
\(d\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { I\vec { dl } \times \hat { r } }{ { r }^{ 2 } } \)