L = \(\frac { 2 }{ \pi } \)H; C = \(\frac { { 10 }^{ -4 } }{ \pi } F\); R = 100Ω
V_RMS = 220 V; f = 50Hz
X_L= 2πfl = 2π x 50 x \(\frac { 2 }{ \pi } \) = 200Ω
X_c = \(\frac { 1 }{ 2\pi fC } =\frac { 1 }{ 2\pi \times 50\times \frac { 10^{ -4 } }{ \pi } } 100 \Omega\)
(i) Impedance, Z = \(\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^2 } \)
=\(\sqrt { 100^{ 2 }+(200-100)^{ 2 } } \) = 141.4Ω
(ii) Peak value of current,
I_m = \(\frac { { v }_{ m } }{ Z } =\frac { \sqrt { 2 } V_{ RMS } }{ Z } \)
= \(\frac { \sqrt { 2 } \times 220 }{ 141.4 } \) = 2.2 A
(iii) Power factor of the circuit
\(cos\phi =\frac { R }{ Z } =\frac { 100 }{ 141.4 } \)= 0.707
(iv) Power factor at resonance
\(cos\phi =\frac { R }{ Z } =\frac { R }{ R } \) = 1

f = 50Hz ; V_m = 20V
Instantaneous voltage, υ = V_m sinωt
= V_m sin2πvt
= 20sin(2π x 50)t = 20sin(100 x 3.14)t
υ = 20sin 314t
Time for one cycle, \(T=\frac { 1 }{ f } =\frac { 1 }{ 50 } =0.02s\)
= 20 x 10⁻³ s = 20ms
The wave form is given below

V_p = 12 V; V_s = 240 V
I_s = 50 mA; N_p = 100 turns
\(\frac { { V }_{ s } }{ { V }_{ P } } =\frac { { N }_{ s } }{ { N }_{ p } } =\frac { { I }_{ P } }{ { I }_{ S } } =K\)
Transformation ratio, K = \(\frac{240}{12}=20\)
The number of turns in the secondary
N_S = N_P x K = 100 x 20 = 2000
Primary current,
I_P = K x I_s = 20 x 50 mA = 1 A

(i) Consider a dipole placed in the uniform electric field \(\vec { E } \). A dipole experiences a torque when kept in an uniform electric field \(\vec { E } \).(ii) To rotate the dipole (at constant angular velocity) from its initial angle θ' to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole.

(iii) The work done by the external torque to rotate the dipole from angle θ' to θ at constant angular velocity is
\(W=\int _{ \theta ' }^{ \theta }{ { \tau }_{ ext }d\theta } \quad ...(1)\)
(iv) Since \({ \vec { \tau } }_{ ext }\) is equal and opposite to \({ \vec { \tau } }_{ E }=\vec { p } \times \vec { E } \), We have 
\(|{ \vec { \tau } }_{ ext }|={ |\vec { \tau } }_{ E }|=|\vec { p } \times \vec { E } |\quad \quad \quad ...(2)\)
Substituting equation (2) in equation (1), we get
\(W=\int _{ \theta ' }^{ \theta }{ pEsin\theta d\theta } \)
\(W=pE(cos\theta '-cos\theta )\)
(v) This work done is equal to the potential energy difference between the angular positions θ to θ'.
U(θ) - U(θ') = ∆U= - pE cosθ + pE cosθ' 
If the initial angle is θ' = 90^o and is take as reference point, then U(θ') = pE cos 90^o = 0. The potential energy stored in the system of dipole kept in the uniform electric field is given by
\(U=-pEcos\theta =-\vec { p } .\vec { E } \)  ....(3)
In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.
(vi) The potential energy is a) maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field
b) minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

(a) The electrostatic potential on the surface of the sphere A is V_A = \(\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1 } } \)
The electrostatic potential on the surface of the sphere A is V_B = \(\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 2 } }{ { r }_{ 2 } } \)
Since V_A = V_B. We have
\(\frac { { q }_{ 1 } }{ { r }_{ 1 } } =\frac { { q }_{ 2 } }{ { r }_{ 2 } } \Rightarrow { q }_{ 1 }=\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) { q }_{ 2 }\)
But from the conservation of total charge, Q = q₁ + q₂, we get q₁ = Q – q₂. By substituting this in the above equation,
Q - q₂ = \(\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) { q }_{ 2 }\)
so that q₂ = Q\(\left( \frac { { r }_{ 2 } }{ { r }_{ 1 }+{ r }_{ 2 } } \right) \)
Therefore,
q₂ = 100 x 10^-9 x \(\left( \frac { 2 }{ 10 } \right) \) = 20nC and q₁ = Q - q₂ = 80nC
The electric charge density for sphere A is σ₁ = \(\frac { { q }_{ 1 } }{ 4\pi { r }_{ 1 }^{ 2 } } \)
The electric charge density for sphere B is σ₂ = \(\frac { { q }_{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } \)
Therefore,
σ₁ = \(\frac { 80\times 10^{ -9 } }{ 4 \pi \times 64\times 10^{ -4 } } \) = 0.99 x 10^-6 Cm^-2 and
σ₂ =\(\frac { 20\times 10^{ -9 } }{ 4\pi \times 4\times 10^{ -4 } } \)  = 3.9 x 10^-6 Cm^-2
Note that the surface charge density is greater on the smaller sphere compared to the larger sphere (σ₂ ≈ 4σ₁) which confirms the result \(\frac { { \sigma }_{ 1 } }{ \sigma _{ 2 } } =\frac { { r }_{ 2 } }{ { r }_{ 1 } } \)
The potential on both spheres is the same. So we can calculate the potential on any one of the spheres
V_A=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1 } } =\frac { 9\times 10^{ 9 }\times 80\times { 10 }^{ -9 } }{ 8\times 10^{ -2 } } \)  = 9kV

If the conductor has no net charge, then its motion is the same as usual projectile motion of a mass m which we studied in Kinematics (unit 2, vol-1 XI physics). Here, in this problem, in addition to downward gravitational force, the charge also will experience a downward uniform electrostatic force.
The acceleration of the charged ball due to gravity = -g\(\hat { j } \)
The acceleration of the charged ball due to uniform electric field =\(-\frac { qE }{ m } \hat { j } \)
The total acceleration of charged ball in downward direction \(\vec { a } =-\left( g+\frac { qE }{ m } \right) \vec { j } \)
It is important here to note that the acceleration depends on the mass of the object. Galileo’s conclusion that all objects fall at the same rate towards the Earth is true only in a uniform gravitational field. When a uniform electric field is included, the acceleration of a charged object depends on both mass and charge.
But still the acceleration a = \(\left( g+\frac { qE }{ m } \right) \) is constant throughout the motion. Hence we use kinematic equations to calculate the range, maximum height and time of flight. In fact we can simply replace g by \(g+\frac { qE }{ m } \) in the usual expressions of range, maximum height and time of flight of a projectile.

Note that the time of flight, maximum height, range are all inversely proportional to the acceleration of the object. Since \(\left( g+\frac { qE }{ m } \right) \) >g for charge +q, the quantities T, h_max, and R will decrease when compared to the motion of an object of mass m and zero net charge. Suppose the charge is –q, then \(\left( g-\frac { qE }{ m } \right) \)< g, and the quantities T, h_max and  R will increase. Interestingly the trajectory is still parabolic as shown in the figure.

The proton and the electron attract each other. The magnitude of the electrostatic force between these two particles is given by
\(F_e=\frac { ke^{ 2 } }{ { r }^{ 2 } } =\frac { 9\times 10^{ 9 }\times (1.6\times 10^{ -19 })^{ 2 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
=\(\frac { 9\times 2.56 }{ 28.09 } \) x 10^-7 = 8.2 x 10^-8 N
The gravitational force between the proton and the electron is attractive. The magnitude of the gravitational force between these particles is
F_G = \(\frac { G{ m }_{ e }{ m }_{ p } }{ { r }^{ 2 } } \)
= \(\frac { 6.67\times 10^{ -11 }\times 9.1\times 10^{ -31 }\times 1.6\times 10^{ -27 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
= \(\frac { 97.11 }{ 28.09 } \) x 10^-47 = 3.4 x 10^-47N
The ratio of the two forces \(\frac { { F }_{ e } }{ F_{ G } } =\frac { 8.2\times 10^{ -8 } }{ 3.4\times 10^{ -47 } } \)
= 2.41 x 10³⁹
Note that F_e ≈ 10³⁹ F_G
The electrostatic force between a proton and an electron is enormously greater than the gravitational force between them. Thus the gravitational force is negligible when compared with the electrostatic force in many situations such as for small size objects and in the atomic domain. This is the reason why a charged comb attracts an uncharged piece of paper with greater force even though the piece of paper is attracted downward by the Earth. This given figure is shown in below.

Electrostatic attraction between a comb and pieces of papers

The strength of the magnetic field is one tesla if the unit charge moving in it with unit velocity experiences unit force.
IT = \(\frac { 1Ns }{ Cm } =1\frac { N }{ Am } \)  = 1NA^-1m^-1

If we hold the current carrying conductor in our right hand such that the thumb points in the direction of current flow, then the fingers encircling the wire points in the direction of the magnetic field lines produced.

Gover suggested that the Earth's magnetic field is due to hot rays coming out from the Sun. These rays will heat up the air near equatorial region. Once air becomes hotter, it rises above and will move towards northern and southern hemispheres and get electrified. This may be responsible to magnetize the ferromagnetic materials near the Earth's surface.

An alternating voltage is the voltage which changes polarity at regular intervals of time and the direction of the resulting alternating current also changes accordingly.

The area of cross - section of two wires = 2:1
\(R\alpha \cfrac { \rho l }{ A } oi.eR\alpha \cfrac { 1 }{ R } \)
According to ohm's law
\(I\alpha \cfrac { V }{ R } i.eI\alpha \cfrac { 1 }{ R } \)
The current in the two wires 2 : 1 when the same potential difference is applied across the two wires. When the temperature of a conductor is increased, the amplitude of vibrate of the positive ions in the conductor increases. Consequently, the free electrons collide more frequently with the vibrating ions. This decreases relaxation time \(\tau .\rho \alpha \cfrac { 1 }{ \tau } \). So resistantly increases

The conductivity of a semiconductor increases with rise in temperature. When temperature increases, a large number of covalent bondsbreak. This produces a larger number of charge carriers. So current increases.

Linear mass density of the conductor is = 0.2 g/m
Mass per unit length \(\frac{M}{l}=0.2 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\)
Magnetic field B = 1T.
Acceleration due to gravity, g = 10 ms^-2
Force \(=\frac{m}{l} \times g\)
= 0.2 x 10^-3 x 10 = 0.2 x 10^-2
F = 2 x 10^-3 N   ....(1)
If the coil is placed in the magnetic field then the force acting on the coil is
F= BIl   ....(2)
From the equation (1) and (2) we get
BIl = 2 x 10^-3
∴ 1 x L x I = 2 x 10^-3
∴ I = 2 x 10^-3 A      [∴ l = 1m]
∴ I = 2mA

Change in magnetic field, dB = (8000 - 2000) = 6000 Wb
Change in time, dt = 6s
Radius of the coil, r = 2 x 10^-2 m
Induced emf, e = 44 V,
Area of the coil A =πr² = \(\frac { 22 }{ 7 } \) x (2 x 10^-2)²
= 12.56 x 10^-4 m²
∴ Number of turns in the coil,
\(N=\frac { e }{ A\frac {dB}{dt }} \frac{44}{12.56\times10^{-4} \times(6000/6)}\)
\(N=\frac { 44 \times10 }{12.56 }=3.503\times10=35 \)
∴ Number of turns in the coil = 35

Change in magnetic flux, dф = 4 x 10^-3 Wb
Change in time, dt = 0.4 s
Magnitude of Induced emf \(= |\frac { -d\Phi }{ dt }|=\frac { d\Phi }{ dt }\)
\(=\frac { 4\times 10^{ -3 } }{ 0.4 } \) = 10 x 10^-3 V = 10 mv
∴ Magnitude of induced emf =10 mV

Electric field, E = 6.0 x 10⁶ N C^-1 and magnetic field, B = 0.83 T.
Then.
\(v=\frac { E }{ B } =\frac { { 6.0\times 10 }^{ 6 } }{ 0.83 } =7.23\times { 10 }^{ 6 }{ ms }^{ -1 }\)
When an electron goes with this velocity, it shows null deflection. Since the accelerating potential is 200 V, the electron acquires kinetic energy because of this accelerating potential. Hence,
\(\frac { 1 }{ 2 } mv^{ 2 }=eV \)
\(v=\sqrt { \frac { 2eV }{ m } }\)
Since the mass of the electron, m = 9.1 x 10⁻³¹kg and charge of an electron, \(\left| q \right| =e=1.6\times { 10 }^{ -19 }C.\) The velocity acquired by the electron due to accelerating potential 200 V is
\({ v }_{ 200 }=\sqrt { \frac { 2\left( 1.6\times { 10 }^{ -19 } \right) \left( 200 \right) }{ \left( 9.1\times { 10 }^{ -31 } \right) } } =8.39\times { 10 }^{ 6 }m{ s }^{ -1 }\)
Since the speed v₂₀₀ > v, the electron is deflected towards direction of Lorentz force. So, in order to have null deflection, the potential, we have to supply is
\(v=\frac { { 1mv }^{ 2 } }{ 2\quad e } =\frac { \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 7.23\times { 10 }^{ 6 } \right) ^{ 2 } }{ 2\times \left( 1.6\times { 10 }^{ -19 } \right) } \)
V = 148.65 V

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 

By using superposition principle, the net electric field at point A is
\(\vec { E_{ A } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1A }^{ 2 } } \hat { { r }_{ 1A } } +\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 2A }^{ 2 } } \hat { { r }_{ 2A } } \)
where r_1A and r_2A are the distances of point A from the two charges respectively
\(\vec { E_{ A } } =\frac { 9\times 10^{ 4 }\times 1\times { 10 }^{ -6 } }{ (2\times { 10 }^{ -3 })^{ 2 } } (\hat { j } )+\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ (2\times 10^{ -3 })^{ 2 } } (\hat { i } )\)\(\)
= 2.25 x 10⁹\(\hat { j } \) + 2.25 x 10⁹\(\hat { i } \) = 2.25 x 10⁹ (\((\hat { i } +\hat { j } )\)
The magnitude of electric field
\(|\vec { { E }_{ A } } |=\sqrt { (2.25\times { 10 }^{ 9 })^{ 2 }+(2.25\times 10^{ 9 })^{ 2 } } \)
= 2.25 x \(\sqrt { 2 } \) x 10⁹ NC^-1
The direction of \(\vec { { E }_{ A } } \) is given by \(\frac { \vec { { E }_{ A } } }{ |\vec { { E }_{ A } } | } =\frac { 2.25\times 10^{ 9 }(\hat { i } +\hat { j } ) }{ 2.25\times \sqrt { 2 } \times { 10 }^{ 9 } } =\frac { (\hat { i } +\hat { j } ) }{ \sqrt { 2 } } \), which is the unit vector along OA as shown in the figure.

The acceleration experienced by an electron placed at point A is
\(\vec { a_{ A } } =\frac { \vec { F } }{ m } =\frac { q\vec { { E }_{ A } } }{ m } \)
=\(\frac { (-1.6\times 10^{ -19 })\times (2.25\times 10^{ 9 })(\hat { i } +\hat { j } ) }{ 9.1\times { 10 }^{ -31 } } \)
= -3.95 x 10²⁰ \((\hat { i } +\hat { j } )\)Nkg^-1
The electron is accelerated in a direction exactly opposite to \(\vec { { E }_{ A } } \).

I = 32 A , t = 1 s
Charge of an electron, e = 1.6 x 10^-19 C
The number of electrons flowing per second, n = ?
\(I=\frac { q }{ t } =\frac { ne }{ t } \)
\(n=\frac { It }{ e } \)
\(n=\frac { 32\times 1 }{ 1.6\times { 10 }^{ -19 }C } \)
n = 20 x 10¹⁹ = 2 x 10²⁰ electrons

Cathode rays are stream of negatively charged electron.

r_n ∞ n²
r₁: r₂: r₃ = 1: 4: 9

\(B=\frac { { \mu }_{ ° }NI a^2}{ { 2(a^2+x^2)}^{ \frac { 3 }{ 2 } } } \)
put a = R
and x = R/2, we get,
\(B=\frac { 8N{ \mu }_{ ° }I }{ { 5 }^{ 3 / 2 }R } \)

\(\Phi = \frac{\pi}{3}-\frac{\pi}{3}=0 \)
Power factor = cosФ = cos 0 = 1

Q = q₁ + q₂ = 4 x 10^-2C
\(q_{2f}=Q[\frac{r_2}{r_1+r_2}]\)
= 4 x 10^-2 \([\frac{3}{4}]\)
q_2f = 3 x 10^-2 C

Resistivity ∝ temperature for conductor. so, copper → decreases
Resistivity ∝\(\frac{1}{\text {temperature for semiconductor}}\)
so, germanium → increases.

\(\mathrm{V}=230 \mathrm{~V}  \)
\(P=\frac{V^2}{R} \text { since } \mathrm{V} \text { is same } \mathrm{P} \propto \frac{1}{R}  \)
\(\frac{1}{R_2}=\frac{1}{\frac{R_1}{2}}+\frac{1}{\frac{R_1}{2}}=\frac{2}{R_1}+\frac{2}{R_1}=\frac{2+2}{R_1}  \)
\(\frac{1}{R_2}=\frac{4}{R_1}  \)
\(\therefore R_2=\frac{R_1}{4}  \)
\(R_1=4 R_2  \)
\(\therefore \frac{P_2}{P_1}=\frac{R_1}{R_2}=\frac{4 R_2}{R_2}=4\)

\(R \propto \frac{1}{A}, R \propto \frac{1}{r^2}  \)
\(R_A \propto \frac{1}{r_A^2}, R_B \propto \frac{1}{r_B^2} \)
\(\frac{r_A}{r_B}=\left(\frac{R_B}{R_A}\right)^{1 / 2}=\left(\frac{R_B}{3 R_B}\right)^{1 / 2}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt{3}}\)

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,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)

Wheatstone's bridge:
i) An important application of Kirchhoff's rule is Wheatstone's bridge. It is used to compare Resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected as shown in Figure.
ii) A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.
Applying Kirchhoff's current rule to junction B
I₁ - I_G - I₃ = 0    ..(1)
Applying Kirchhoff's current rule to junction D,
I₂ + I_G - I₄ = 0      ...(2)

Applying Kirchhoff's voltage rule to loop ABDA,
I₁P + I_GG - I₂R = 0      ...(3)
Applying Kirchhoff's voltage rule to loop ABCDA,
I₁P + I₃Q - I₄S - I₂R = 0   ...(4)
(iii) When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (I_G = 0). Substituting I_G = 0 in equation (1), (2) and (3), we get
I₁ = I₃  ..(5)
I₂ = I₄  ..(6)
I₁P = I₂R    ..(7)
Substituting the equation (7) in equation (4),
I₃Q = I₄S  .....(8)
Dividing equation (7) by equation (8), we get
\(\cfrac { P }{ Q } =\cfrac { R }{ S } \)   .....(9)
(iv) This is the bridge balance condition. Only under this condition, galvanometer shows null deflection.