(a) Drift velocity: The average velocity gained by free electrons, when a unit electric field is applied across the conductor.
\(I=neAv_d\)
\(=neA\frac{eI}{m}\tau\)
\(\therefore \) Current density \(J=\frac{I}{A}=\frac{ne^2 Be}{m}\)
(a) P = I²R
Current flowing through the resistance 2\(\Omega\)
\(l=\sqrt{\frac{200}{2}}=10A\)
∴ Potential drop across the 2\(\Omega\) resistors = 20V Therefore potential across parallel combination of 40\(\Omega\) and 10\(\Omega\) = 80V
Current through 5\(\Omega\); I = \(\frac{80}{10}A=8A\)
∴  Power dissipated in the 5\(\Omega\) resistor = (8)² x 5W
= 320W

(i) Average power delivered by an AC circuit is
\({ P }_{ av }={ V }_{ rms }{ I }_{ rms }cos\phi \)
where is minimum, the power delivered is minimum and hence, power dissipated will be maximum for the circuit.

Here,
 R = \(5.0 \ \Omega\);
r₁ = 5 x 10^-2 m;
D = 2.0 x 10^-3 m;
\(\rho= 2.0 \times 10^{-7} \ \Omega m\).
R = \(\frac{\rho l}{\pi D^2/4}\)
or l = \(\frac{R \pi D^2}{4 \rho}\)
Let n be the number of turns in the coil. Then total length of the wire used, l = \(2 \pi r_1 n\)
or n = \(\frac{l}{2 \pi r_1}=\frac{R \pi D^2}{4 \rho \times 2 \pi r_1}=\frac{RD^2}{8 \rho r_1}\)
 = \(\frac{5.0 \times (2.0 \times 10^{-3})^2}{8 \times (2.0 \times 10^{-7})\times (5 \times 10^{-2})^2}\)
 = 250

Here, I = 20 A, B₂ = 3 x 10^-4 T, 
r = 2.0 x 10^-2 m
Magnetic field due to a straight wire carrying current is 
\({ B }_{ 1 }=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ r } ={ 10 }^{ -7 }\times \frac { 2\times 20 }{ 2.0\times { 10 }^{ -2 } } =2\times { 10 }^{ -4 }\quad T\) 
This magnetic field will act perpendicular to magnetic field B₂(= 3 x 10^-4T). Therefore, the magnitude of the resultant magnetic field
\(B=\sqrt { { B }_{ 1 }^{ 2 }+{ B }_{ 2 }^{ 2 } } =\sqrt { { \left( 2\times { 10 }^{ -4 } \right) }^{ 2 }+{ \left( 3\times { 10 }^{ -4 } \right) }^{ 2 } } \) 
= 3.6 x 10^-4 T

\(Here, q=\pm20\mu C=\pm20\times10^{-6}C\)
\(2a=1 cm=10^{-2}m, r=50cm={1\over 2}m\)
As 2a<<r, therefore, intensity on equatorial line of short dipole is
\(E={1\over 4\pi\epsilon_o}{P\over r^3}={q\times 2a\over 4\pi\epsilon_or^3}\)
\(={9\times 10^9\times 20\times10^{-6}\times10^{-2}\over(1/2)^3}\)
\(E=1.44\times 10^4N/C\)

\(Here, \ { E }_{ v }=200V,\ v=50hz,\ R=100\Omega ,\ { V }_{ R }=65V,\ { V }_{ C }=415V,\ { V }_{ L }=204V\)
(i) If Iv is current in the circuit, then
\({ V }_{ R }={ I }_{ V }\times R; \ 65={ I }_{ V }\times 100, \ { I }_{ V }=0.65A\)
\( (ii) \ { V }_{ L }={ I }_{ V }{ X }_{ L }; \ { X }_{ L }=\frac { { V }_{ L } }{ { I }_{ v } } =\frac { 204 }{ 0.65 } =313.85\Omega\)
\({ X }_{ L }=\omega L=2\pi vL=313.85\)
\(L=\frac { 313.85 }{ 2\pi v } =\frac { 313.85 }{ 3.14\times 50 } =1.0H\)
\( (iii) \ { V }_{ C }={ I }_{ v }{ X }_{ C }, \ { X }_{ C }=\frac { { V }_{ C } }{ { I }_{ v } } =\frac { 415 }{ 0.65 } =638.5\Omega \)
\({ X }_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } ; \ C=\frac { 1 }{ 2\pi v \ { X }_{ c } } =\frac { 1 }{ 2\times 3.14\times 50\times 638.5 } =4.99\times { 10 }^{ -6 }F\)
(iv) Let C′ be the capacitance that would produce resonance with L= 1.0H, then
\( v=\frac { 1 }{ 2\pi \sqrt { LC' } } ; \ C'=\frac { 1 }{ 4{ \pi }^{ 2 }{ v }^{ 2 }L } =\frac { 1 }{ 4\times \left( 3.14 \right) ^{ 2 }\times \left( 50 \right) ^{ 2 }\times 1 } =10.1\times { }^{ }F=10.1\mu F\)

\(Here, \ R=50cm=0.5m,\)
\(r=5cm=5\times { 10 }^{ -2 }m\)
\(x=50cm=0.5m, \ { I }_{ 1 }=3A, \ { I }_{ 2 }=2A, \ M=?\)
\(\\ { M }^{ \ \ }=\frac { { \mu }_{ 0 } }{ 5\pi } \times \frac { 2{ \pi }^{ 2 }{ R }^{ 2 }{ r }^{ 2 } }{ \left( { R }^{ 2 }+{ x }^{ 2 } \right) ^{ 3/2 } } =\frac { { 10 }^{ -7 }\times 2\left( 3.14 \right) ^{ 2 }\left( { 0.5 } \right) ^{ 2 }\left( { 5\times 10 }^{ -2 } \right) ^{ 2 } }{ \left[ { 0.5 }^{ 2 }+{ 0.5 }^{ 2 } \right] ^{ 3/2 } } \)
\(M=3.48\times { 10 }^{ -9 }H\)

Uses of electromagnetic waves
Following are some important uses of electromagnetic waves:
1. Radio waves. Radio waves are electromagnetic waves and are used in radio and television communication systems.
2. Microwaves. Microwaves are used in radar and other communication systems.
3. Infrared radiations. Infrared rays are used in:
(a) solar water heater and solar cooker.
(b) weather forecasting
(c) taking photographs during fog, smoke etc.
(d) dehydrating fruits.
(e) the treatment of muscular strain.
(f) greenhouse to keep the plants warm.
4. Ultraviolet rays.  ultraviolet rays are used:
(a) for checking mineral samples by making use of its property of causing fluorescence and also used for the study of molecular structure.
(b) for sterilizing the surgical instruments because U.V.-rays destroy bacteria.
(c) in food preservations
(d) in detection of invisible writing 
5. X-rays. X-rays are used:
(a) in surgery
(b) in radiotherapy.
(c) in medical diagnosis to detect the fracture in bones etc.
(d) in detective departments to detect gold, diamond etc concealed in bags etc without opening them.
(e) in scientific research to study the crystal structure etc.
6. \(\gamma \)-rays. Gamma rays are used to get information of structure of atomic nucleus.

Given, potential difference = 230 V
Initial current at 27°C = I_27°C = 3.2 A
Final current at t°C = I_t°C = 2.8 A
Room temperature = 27°C
Temperature coefficient of resistance, \(\alpha=1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)
Resistance at 27°C, R_27°C\(=\frac{V}{I_{27^{\circ} \mathrm{C}}}=\frac{230}{3.2}=\frac{2300}{32} \Omega\)
Resistance at t°C, R_t°C = \(\frac{V}{I_{t^{\circ} \mathrm{C}}}=\frac{230}{2.8}=\frac{2300}{28} \Omega\)
Temperature coefficient of resistance
\(\begin{aligned} \alpha & =\frac{R_t-R_{27}}{R_{27}(t-27)} \end{aligned}\)
\(\begin{aligned} \Rightarrow 1.70 \times 10^{-4} & =\frac{\frac{2300}{28}-\frac{2300}{32}}{\frac{2300}{32}(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{82.143-71.875}{71.875 \times 1.70 \times 10^{-4}}=840.347 \end{aligned}\)
or t = 840.3 + 27 = 867.3 °C
Thus, the steady temperature of heating element is 867.3 °C

The circuit diagram of the potentiometer, is as shown here

The potential drop, V, across a length 1of a uniform wire, is proportional to the length 1of the wire.(or \(V\propto l\) for a 
uniform wire)
Derivation: From the figure, connect points 1 and 3 together with balance point at \({ N }_{ 1 }\) point where A\({ N }_{ 1 }\) = \({ l }_{ 1 }\) Now connect points 2 and 3 together with balance point at point \({ N }_{ 2 }\) where A\({ N }_{ 2 }\) = \({ l }_{ 2 }\), Next let the points 2 and 3 be connected together.Let the balance point be at the point N₂ where We then have and 
\({ \varepsilon }_{ 1 }=k{ l }_{ 1 }\)
\({ \varepsilon }_{ 2 }=k{ l }_{ 2 }\)
\(\frac { { \varepsilon }_{ 1 } }{ { \varepsilon }_{ 2 } } =\frac { { l }_{ 1 } }{ { l }_{ 2 } } \)

The expression in vector form is given by F = q(vｘB). 
The direction of the magnetic force is in the direction of (v ｘ B), i.e. perpendicular to the plane containing v and B.

As, A and B are points on the equatorial plane of dipole
V_A = V_B = 0
Net potential = V_A + V_B = 0
Work done, W=\(\frac{V}{q},\) as V = 0, W = 0
So, the work done by the process will be zero.

According to principle of superposition of electric fields, E (electric field) at a point due to system of three charges,
E  = \({{1}\over{4\pi{\epsilon}_{0}}}=\left[{{q}\over{{ r }_{ { 1 }^{ p } }^{ 2 }}}\hat{r}_{{1}_{p}} +{{{q}_{2}}\over{{ r }_{ { 2 }^{ p } }^{ 2 }}}\hat{r}_{{2}_{p}}+{{{q}_{3}}\over{{ r }_{ { 3 }^{ p } }^{ 2 }}}\hat{r}_{{3}_{p}} \right]\)
= \({ { q }\over{ 4\pi{\epsilon}_{0} } }\times{ { 9 }\over{ 8 } }\) [ using \({S}_{\infty}={ { a }\over{ 1 -r } }\) ] = \({ { 1 }\over{ 4\pi{\epsilon}_{0} } }.{{9q}\over{8}}{NC}^{-1}\)

Examples of Molecular solid : Solid \(\mathrm{SO}_2, \mathrm{NH}_3, I_2\)
Examples of Ionic solid : NaCI,ZnS,CuCI

When the switch is thrown from the OFF position (open circuit) to the ON position (closed circuit), then neither B nor A and the angle between B and A does not change.Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently, no current will flow in the circuit.

It can be justified by placing a magnetic needle around current carrying wire, which shows deflection of needle.

Drift velocity is the average velocity with which the free electrons get drifted towards the positive end of the conductor under the influence of an external electric field applied.
The relation between current I and drift velocity  v_d is I = nAev_d
Where n is the number density of electrons in a conductor of area of cross-section A and e is the charge on an electron. 

Principle of Cyclotron The cyclotron works on the principle that a positively charged ion 'can be accelerated to high kinetic energy by making it pass again and again smaller value of same oscillating electrical field by making use of strong perpendicular magnetic field. Also, the frequency of charge particle must be equal to the frequency of oscillating electrical field. Expression for cyclotron frequency, \(f=\frac{qB}{2\pi m}\)

Working Let initially positively charged is accelerated towards D2 and enter into it. Now, the charged particle experiences magnetic Lorentz force due to a strong normal magnetic field. It performs circular motion. The time taken by the charge particle to complete half revolution is equal to half of time period of AC oscillator between two dees.
The charge d particle again accelerated towards D₁ as D₂ acquires positive and d negative polarity. Thus, the charge particle is brought again and again in the small region of oscillating electrical field by strong normal magnetic field. The charged particle repeatedly passes through oscillating electrical field. It traversed on spiral path and finally having radius of its circular path becomes equal to the radius of dees and finally  comes out through window W and strikes to the target.

Two charges q_A  and q_B  are located at points A(0, 0 -15cm) and B(0, 0, 15 cm) on Z-axis. They form an electric dipole.

Total charge, q = q_A + q_B  = 2.5 x 10^-7 -2.5 x 10^-7
q = 0
Also AB = 15 + 15 = 30 cm or AB = 30 x 10^-2 m
Electric dipole moment, 
p = Either charge x BA 
= 2.5 x 10^-7 x (30 x 10^-2) (-k)
= -7.5 x 10^-8 k C-m

(i) In combination 1, net emf of combination is E = E₁ + E ₂, whereas for combination 2 net emf is E₂ - E₁
E1 - E₂ = Kl₁
where, K = potential gradient
l₁ = 400 cm
For combination2, E₂ - E₁ = Kl₂
where, l₂ = 240 cm
\(\therefore \ \ \frac { E1+E2 }{ E1-E2 } =\frac { Kl1 }{ Kl2 } =\frac { 400 }{ 240 } =\frac { 5 }{ 3 } \Longrightarrow \frac { E1+E2 }{ E2-E1 } =\frac { 5 }{ 3 } \)
Applying componendo and dividendo theorem, we get
\(\therefore \ \ \frac { E1+E2 }{ E1-E2 } =\frac { Kl1 }{ Kl2 } =\frac { 400 }{ 240 } =\frac { 5 }{ 3 } \Longrightarrow \frac { E1+E2 }{ E2-E1 } =\frac { 5 }{ 3 } \)
\(\frac { E2 }{ E1 } =\frac { 8 }{ 2 } \Longrightarrow \frac { E1 }{ E2 } =\frac { 1 }{ 4 } \)        ..... (iii)
(ii)  \(\therefore \frac { E1 }{ E2 } =\frac { 1 }{ 4 } \Rightarrow E1=E(say)\)  
Then, E₂  = 4E
\(\Rightarrow\) E₁+ E₂ = K x 400\(\Rightarrow\) 5E = K X 400
\(K=\frac { 5E }{ 400 } =\frac { E }{ 80 }\)
Now, balancing length for E₁ is l₁
  \(\therefore \ \ E2=Kl1\Rightarrow E=\frac { E }{ 80 } Xl1\Rightarrow l1=80cm\)