Brewster found experimentally that the reflected and refracted rays are at right angles to each other, when light is incident at polarising angle.
\(\therefore\) i_p + 90^\(\circ\) + r = 180^\(\circ\)
r = 90^\(\circ\) - i_p
From Snell's law, \(\frac{sin \space i}{sin \space r}=\mu\)
\(\therefore \frac{sin \space i_{p}}{sin(90-i_{p})}=\mu \)
\(\frac{sin \space i_{p}}{cos \space i_{p}}=\mu \)
(or) \(\mu\) = tan i_p
Tangent of the polarising angle is numerically equal to the refractive index of the medium.

(i) Bar magnet (magnetic dipole) of dipole moment \(\overrightarrow{p_m}\) is heldat an angle \(\theta\) with the directions of a uniform magnetic field \(\vec{B}\).
(ii) The magnitude of the torque acting on the dipole is \(\left|\overrightarrow{\tau_B}\right|=\left|\overrightarrow{p_m}\right||\vec{B}| \sin \theta\)
(iii) If the dipole is rotated through a very small angular displacement d\(\theta\) against the torque \(\tau_B\) at constant angular velocity, then the work done by external torque \(\left(\vec{\tau}_{e x t}\right)\).
\(\mathrm{dW}=\left|\vec{\tau}_{e x t}\right| d \theta\)
(iv) Since the bar magnet to be moved at constant angular velocity, it implies \(\left|\vec{\tau}_B\right|=\left|\vec{\tau}_{e x t}\right|\)
\(\mathrm{dW}=P_{\mathrm{m}} B \sin \theta d \theta\)
Total work done in rotating the dipole from \(\theta\)' to \(\theta\) is
\(\mathrm{W}=\int_{\theta^{\prime}}^\theta \tau d \theta\)
\(=\int_{\theta^{\prime}}^\theta P_m B \sin \theta d \theta=P_m B(-\cos \theta d \theta)_{\theta^{\prime}}^\theta\)
\(\mathrm{W}=-p_m B\left(\cos \theta-\cos \theta^{\prime}\right)\)
This work done is stored as potential energy
\( \mathrm{U}=-p_m B\left(\cos \theta-\cos \theta^{\prime}\right) \)      ........(i) 
\( If \theta^{\prime}=90^{\circ}, \cos 90=1 \)
\( \mathrm{U}=-p_{\mathrm{n}} B(\cos \theta) \)                 ..............(ii) 
The potential energy stored in a bar magnet in a uniform magnetic field is given by,
\(\mathrm{U}=-\vec{p}_m \cdot \vec{B}\)                .............(iii)

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

\(Y =A \cdot \bar{B}+A B+B C+C A \)
\(=A(\bar{B}+B)+B C+C A[\bar{B}+B=1] \)
\(=A(1)+B C+C A \)
\(=A+C A+B C \)
\(=A(1+C)+B C[1+C=1] \)
= A + BC

λ = 5000 Å = 0.5 x 10^-6 m, 
a = 2.5 μm = 2.5 x 10^-6 m
We know, a sin θ = nλ
For maximum order of diffraction, sin θ = 1
Therefore, 2.5 x 10^-6 = n x 0.5 x 10^-6
\(n=\frac{2.5}{0.5}=5\) 
Maximum order of diffraction = 5.

Advantages of AM :
(i) Easy transmission and reception
(ii) Lesser bandwidth requirements
(iii) Low cost
Limitations of AM :
i) Noise level is high
i) Low efficiency
ii) Small operating range

It provides electrical energy to satellites by means of solar cells. It is used to produce dehydrated fruits, in green houses to keep the plants warm, heat therapy for muscular pain or sprain, TV remote as a signal carrier, to look through haze fog or mist and used in night vision or infrared photography.

The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T
Deflection is 
\(\theta =60°\Rightarrow tan60°=\sqrt { 3 } =1.732\)
\(I=\frac { 2R{ B }_{ H } }{ { \mu }_{ ° }N } tan\theta \)
\(=\frac { 2\times 0.12\times 25\times 1{ 0 }^{ -6 } }{ 4\times 1{ 0 }^{ -7 }\times 3.14\times 100 } \times 1.732=0.82\times 1{ 0 }^{ -1 }A\)
I = 0.082 A

Wings of a morpho butterfly, peacock feathers, lotus leaf surface and sources of parrot fish's bite are some of the natural nano particles.

A = 1, B = 0, C = 1
y = A + B, y = (A + B).C
y = (1 + 0).1 ⇒ y = 1

d' = 2d, β' = β, D' = ?
W.K.T, Fringe width
\(\beta = \frac{D\lambda}{d} \Rightarrow D' = \frac{Dd'}{d}\)
\(D' = \frac{D2d}{d}=2D\)

\(tan \ I=\frac{B_V}{B_H}=1\)
∴ I = 45^o

Magnetic moment,
p'_m = ml
From figure, \(l=\frac{\pi r}{3}\)
\(\therefore r=\frac{3l}{\pi}\)
∴ New magnetic moment,
p'_m = m x r
\(=m\times \frac{3l}{\pi}=\frac{3}{\pi}ml\)
∴ p'_m = \(\frac{3}{\pi}p_m\)

(i) The loop phase shift must be 0^o or integral multiples of 2\(\pi\) .
(ii) The loop gain must be unity \(|A \beta|=1\)
\(\alpha\) \(\rightarrow\) voltage gain of the amplifier
\(\beta \rightarrow\) feed back ratio

NAND and NOR gates are known as universal gates because any other logic gate can be made from NAND or NOR gates.

\(\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=\frac{36}{1}\)
If a₁ and a₂ are their amplitudes,
\(\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=\frac{\left(\mathrm{a}_{1}+\mathrm{a}_{2}\right)^{2}}{\left(\mathrm{a}_{1}-\mathrm{a}_{2}\right)^{2}}=\frac{36}{1}\)  (or) \(\frac{a_{1}+a_{2}}{a_{1}-a_{2}}=\frac{6}{1}\) 
6a₁ - 6a₂  = a₁ + a₂ 
5a₁ - 7a₂ = 5a₁ = 7a₂  
\(\frac{a_{1}}{a_{2}}=\frac{7}{5}\) 
a₁ : a₂ = 7 : 5 

(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.

The energy gap between the valance band and conduction band is called forbidden energy gap

Advantages of Robotics:
(i) Robots are more precise and error free in performing the task.
(ii) Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
Disadvantages of Robotics:
(i) Unemployment problem will increase.
(ii) Robots can perform defined tasks and cannot handle unexpected situations.

(i) The polaroid which polarises the light passing through it is called a polariser.
(ii) The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.

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

When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.

An electromagnetic waves are the waves that are radiated by an accelerated charge which propagates through space as coupled electric and magnetic fields, oscillating perpendicular to each other and to the direction of propagation of the wave.

Condition for bright fringe (or) maxima :
The condition for the point P to have a constructive interference (or) be a bright fringe Is,
Path diference, δ = nλ    Where, n = 0, 1, 2,....
\(\therefore\frac{dy}{D}=n\lambda\)
\(y=n\frac{\lambda D}{d}(or)y_n=n\frac{\lambda D}{d}\)   .....(4)
This is the condition for the point P to have a bright fringe. The distance y_n is the distance or the n^th bright fringe from the point O.
Condition for dark fringe (or) minima:
The condition for the point P to have a destructive interference (or) be a dark fringe is,
Path difference, δ = \((2n-1)\frac{\lambda}{2}\)    Where, n = 1, 2, 3....
\(\therefore\frac{dy}{D}=(2n-1)\frac{\lambda}{2}\)
\(y=\left(\frac{(2n-1)}{2} \frac{\lambda D}{d}\right)(or)\left(\frac{(2 n-1)}{2} \frac{\lambda D}{d}\right) \)    .....(5)
This is the condition for the point P to have a dark fringe. The distance y_n is the distance of the n^th dark fringe from the point O
Bandwidth:
The bandwidth \((\beta)\) is defined as the distance between any two consecutive bright or dark fringes. 
\(\beta=y_{(n+1)}-y_{n}=\left((n+1) \frac{\lambda D}{d}\right)-\left(n \frac{\lambda D}{d}\right) \)
\(\beta=\frac{\lambda D}{d} \)    .....(6)
Bright and Dark tinges are of same width equally spaced on either side of the central bright fringe.

First Theorem :
The complement of the sum of two logical inputs is equal to the product of its complements.
\(\overline { A+B } \) = \(\bar { A } .\bar { B } \)
Proof:
(i) The Boolean equation for NOR gate is Y = \(\overline { A+B } \)
(ii) The Boolean equation for a bubbled AND gate is Y =\(\bar { A } .\bar { B } \)
(iii) Both cases generate same outputs for same inputs. It can be verified using the following truth

(i) From the above truth table, we can conclude \(\overline { A+B } \) = \(\bar { A } .\bar { B } \)
(ii) Thus De Morgan's first theorem is proved.
(iii) Hence, a NOR gate is equal to a bubbled AND gate

Second theorem :
The complement of the product of two is equal to the sum of its complements
\(\overline { A.B } \) = \(\bar { A } +\bar { B } \)
Proof:
(i) The Boolean equation for NAND gate is Y = \(\overline { A.B } \)
(ii) The Boolean equation for bubbled OR gate is Y = \(\bar { A } +\bar { B } \)
(iii) A and B are the inputs and Y is the output. The above two equations produces the same output for the same inputs. It can be verified by using the truth table.

(i) From the above truth table, we can conclude \(\overline { A.B } \) = \(\bar { A } +\bar { B } \)
(ii) Thus, De Morgan's second therom is proved.
(iii) Hence, a NAND gate is equal to a bubbled OR gate.

The static characteristics of the BJT are
(i) Input characteristics
(ii) Output characteristics
(iii) Transfer characteristics.
(i) Input characteristics:
Input characteristics curves give the relationship between the base current (I_B) and base to emitter voltage (v_BE) at constant collector to emitter voltage (v_CE) and are shown in figure

(i) Initially, the collector to emitter voltage (V_CE) is set to a particular value (above 0.7 V to reverse bias the junction).
(ii) Then the base-emitter voltage V_BE, is increased in suitable steps and the corresponding base-current I_B is recorded.
(iii) A graph is plotted with V_BE along the x axis and I_B along the Y - axis
(iv) The procedure is repeated for different values of V_cE
The following observations are made from the graph :
(i) The Curve looks like the forward characteristics of an ordinary p - n junction diode.
(ii) There exists a threshold voltage (or) knee voltage (V_knee) below which the base current (I_b) is very small. This value is 0.7 V for silicon and 0.3 V for germanium transistors. Beyond the knee voltage, the base current increases with the increase in base-emitter voltage.
(iii) It is also noted the increase in V_CE, decreases the I_B. This shifts the curve outward.
(iv) This is because the increase in collector-emitter voltage increases the width of the depletion region which in turn, reduces the effective base width and thereby the base current.
Input resistance :
The ratio of the change in base-emitter voltage (∆V_BE) to the change in base current (∆L_B) at a constant collector-emitter voltage (V_CE) is called the input resistance (r_i)
\(\mathrm{R}_{\mathrm{i}}=\left[\frac{\Delta \mathrm{V}_{\mathrm{BE}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right]_{\dot{\mathrm{V} C \mathrm{E}}}\)
The input impedance is high for a transistor in common emitter configuration.
Output characteristics :
The output characteristics give the relationship between the collector current (I_c) and the collector -emitter voltage (V_CE) at constant input current (I_B)
as shown in figure

(i) Initially I_B is set to a particular voltage. V_CE is increased in suitable steps and I_C is recorded
(ii) A graph is plotted with the V_CE along the x-axis and I_C along the y - axis
(iii) This procedure is repeated for different values I_B
The four important regions ln the output characteristics
(i) Saturation region
(ii) Cut-off region
(iii) Active region
(iv) Breakdown region
Output Resistance :
The ratio of the change in the collector emitter voltage (∆V_CE) to the corresponding change in the collector current (∆l_C) at constant base current (I_B) is called output resistance (r_o).
\(\mathrm{R}_{\mathrm{o}}=\left[\frac{\Delta \mathrm{V}_{\mathrm{BE}}}{\Delta \mathrm{I}_{\mathrm{C}}}\right]_{\mathrm{l}_{\mathrm{B}}}\)
The output impedance for transistor in common emitter configuration is very low.

FuIl wave rectifier :
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier
Construction:

(i) It consists of two p-n junction diodes, a center-tapped transformer, and a load resistor (R₁)
(ii) The centre is usually taken as the ground or zero voltage reference point.
(iii) Due to the centre tap transformer, the output voltage rectified by each diode is only one-half of the total secondary voltage.
Working:
During positive half cycle :
(i) When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G is at zero potential and N is at negative potential.
(ii) This forward biases diode D₁ and reverse biases diode D₂.
(iii) Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁AGC.
During negative half cycle:
(i) When the negative half cycle of the AC input signal passes through the circuit, terminal N becomes positive, C is at zero potential and M is at negative potential.
(ii) This forward biases diode D₂ and reverse biases diode D₁.
(iii) Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂BGC.
(iii) During both positive and negative half cycles of the input signal, the current flows through the load in same direction.

(iv) The output signal corresponding to the input signal is shown in Figure. Though both half cycles of AC input are rectified, the output is still pulsating in nature.
(v) The efficiency (η) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %.

(i) The wavelength of a spectral line can be very accurately determined with the help of a diffraction grating. For that we need to use an instrument called spectrometer.
(ii) The slit of collimator is illuminated by a monochromatic light, whose wavelength is to be determined.
(iii) The telescope is brought in line with collimator to view the image of the slit.
(iv) The given plane transmission grating is then mounted on the prism table with its plane perpendicular to the incident beam of light coming from the collimator.
(v)The telescope is turned to one side until the first order diffraction image of the slit coincides with the vertical cross wire of the eye piece.
(vi) The reading of the position of the telescope is noted.
(vii) Similarly the first order diffraction image on the other side is made to coincide with the vertical cross wire and corresponding reading is noted.
(viii) The difference between two positions gives 2θ. Half of its value gives θ, the diffraction angle for first order maximum as shown in Figure.
The wavelength of light is calculated from the equation.
\(\\ \lambda =\cfrac { sin\theta }{ Nm } \) 

(ix) Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

(i) Let us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY.
(ii) The incident wavefront is AB and the reflected waterfront is A'B'.
(iii) These wavefronts are perpendicular to the incident rays L, M and reflected rays L', M', respectively.

(i) The incident rays, the reflected rays and the normal are in the same plane.
(ii) Angle of incidence, ∠i = ∠NAL = 90°- ∠NAB = ∠BAB'
Angle of reflection ∠r = ∠N'B'M = 90°- ∠N'B'A'= ∠A'B'A
(a) For the two right angle triangles, ∆ABB' and ∆B'A'A, the two right angles, ∠B and ∠A' are equal, (∠B and ∠A' = 90°); the two sides., AA' and BB' are equal, (AA' = BB'); the side AB' is common
(b) Thus the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB' and ∠A'B'A must also be equal.
i = r
Hence, the laws of reflection are proved.

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) 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
\( { { \vec B }_{ axial } } =\frac { { \mu }_{ 0 } 2r}{ 4\pi } \left( \frac { p_{ m } }{ { r }^{ 4 } } \right) \hat { i } =\frac { { \mu }_{ 0 } }{ 4\pi }[ \frac { 2 \vec p_{ m } }{ { r }^{ 3 }}] \)      
∵ \(\vec { { p }_{ m } } =p_{ m }\hat { i } \).

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

Emission spectra:
When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum can be divided into three types:
(i) Continuous emission spectra (or continuous spectra) :
(a) If the light from incandescent lamp (filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours.
(b) Thus, it consists of wavelengths containing all the visible colours ranging from violet to red (in the figure). Examples: spectrum obtained from carbon arc, incandescent solids.
(ii) Line emission spectrum (or line spectrum) :
(a) Suppose light from hot gas is allowed to pass through a prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra consists of  sharp lines of definite wavelengths or frequencies.
(b) Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element  and are different for different elements. Examples: spectra of atomic hydrogen, helium, etc.
(iii) Band emission spectrum (or band spectrum) :
(a) Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces.
(b) This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited.
(c) Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube, etc.