When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric breakdown.

Emf can be produced by changing magnetic flux in any of the following ways:
(i) By changing the magnetic field B
(ii) By changing the area A of the coil and
(iii) By changing the relative orientation θ of the coil with magnetic field.

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.

\(\Phi_B\) = (2t³ + 3t² + 8t + 5)mWb; N = 1;t = 3 s
i) \(ε=\frac { d(N{ \Phi }_{ B }) }{ dt } \)
\(=\frac { d }{ t } \left( { 2t }^{ 3 }+{ 3t }^{ 2 }+8t+5 \right) \times { 10 }^{ -3 }\)
= (6t² + 6t + 8) x 10^-3 V
At t = 3 s,
ε = [( 6 x 9) + (6 x 3) + 8] x 10^-3
= 80 x 10^-3V = 80mV
(ii) As time passes, the magnetic flux linked with the loop increases. According to Lenz’s law, the direction of the induced current should be in a way so as to oppose the flux increase. So, the induced current flows in such a way to produce a magnetic field opposite to the given field. This magnetic field is perpendicularly outwards. Therefore, the induced current flows in anticlockwise direction.

For a collection of n point charges, the electric dipole moment is defined as follows, \(\vec{p}=\stackrel{i=n} \sum _{i=1}q_i\vec{r}_i\) where,  \(\vec{r}_i\) is the position vector of charge q_i from the origin.

It states that the heat developed in an electrical circuit due to the flow of current varies directly as
(i) the square of the current
(ii) the resistance of the circuit and
(iii) the time of flow.

(a) The work done to arrange the charges in the corners of the square is independent of the way they are arranged. We can follow any order.
(i) First, the charge +q is brought to the corner P. This requires no work since no charge is already present, W_P = 0
(ii) Work required to bring the charge –q to the corner Q = (-q) x potential at a point Q due to +q located at a point P
W_Q = -q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ a } =-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^{ 2 } }{ a } \)
(iii) Work required to bring the charge +q to the corner R = q x potential at the point R due to charges at the point P and Q.
W_R = q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( -\frac { q }{ a } +\frac { q }{ \sqrt { 2 } a } \right) \)
= \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }^{ 2 } }{ a } \left( -1+\frac { 1 }{ \sqrt { 2 } } \right) \)
(iv) Work required to bring the fourth charge –q at the position S = q × potential at the point S due the all the three charges at the point P, Q and R.
W_s = - q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ a } +\frac { q }{ a } -\frac { q }{ \sqrt { 2 } a } \right) \)
W_s = \(-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^2 }{ a } \left( 2-\frac { 1 }{ \sqrt { 2 } } \right) \)
(b) Work required to bring the charge q′ to the center of the square = q′ x potential at the center point O due to all the four charges in the four corners.
The potential created by the two +q charges are canceled by the potential created by the -q charges which are located in the opposite corners. Therefore the net electric potential at the center O due to all the charges in the corners is zero.
Hence no work is required to bring any charge to the point O. Physically this implies that if any charge q′ when brought close to O, then it moves to the point O without any external force.

Electric power:
(i) The electric power P is the rate at which the electrical potential energy is delivered.
(ii) The electric power P is the rate at which the work is done.
\( P= \frac{d U}{d t} (or)= \frac{d W}{d t} (or)=VI(or)\frac{V^2}{R}\)
Unit: watt (W)
Electric energy:
(i) The electric energy is the product of power (P) and duration of the time (t) when electric energy is delivered.
(ii) E = Pt
Unit: watt-hour (Wh)

\(\text {emf } e=8 \mathrm{~V}  \)
\(d I=I_1-I_0=2-(-2)=4 \mathrm{~A}  \)
\(\text {dt }=0.05 \mathrm{~s}  \)
\(L=\frac{-e}{d I / d t}=\frac{-8}{4 / 0.05}  \)
\(=\frac{-8 \times 0.05}{4}=\frac{-0.40}{4}  \)
=-0.1 H
-ve sign indicates that self-induced emf always opposes the current w.r.t. time.

\(\vec {E}\) = 10x\(\hat{i},\)  when x = 2 m
\(\vec {E}\) = 10 x 2 x \(\hat{i}\) = 20\(\hat{i}\)
Since, \(E=\frac{-dV}{dx}\therefore V=+20 V\)

The charge + q will be stable between B₁ and B₂ with respect to the displacement.

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

Given: Radius of the toroid r = 15cm = 15 x 10^-2 m
Area of the toroid A = 12cm² = 12 x 10^-4m
No. of turns N = 2000
To find:
Self inductance of the toroid L =?
Formula:
\(L=\frac { { \mu }_{ o }{ N }^{ 2 }{ A } }{ l } =\frac { { \mu }_{ o }{ { N }^{ 2 }A } }{ 2\pi r } \)

= 6.4 x 10^-3 H (or) 6.4 MH

According to Wheatstone bridge \(\cfrac { P }{ Q } =\cfrac { R }{ S } \)
\(\cfrac { R }{ S } =\cfrac { OA }{ OB } =\cfrac { 40 }{ 60 } \)
\(\cfrac { R }{ S } =\cfrac { OA }{ OB } =\cfrac { 40 }{ 60 } \)
If resistance 10 \(\Omega \) connected in series with R, the balance length is 60 cm.
\(\cfrac { R+10 }{ S } =\cfrac { 60 }{ 40 } \Rightarrow 2R+20=3S\)
From (1) & (2) \(\Rightarrow \left[ \cfrac { 4S }{ 3 } +20=3S \right] \)
45 + 60 = 95
55 = 60
\(S=\cfrac { 60 }{ 5 } =12\)
\(2\times \cfrac { 2S }{ 3 } +20=3S\)
\(s=12\Omega \)
From equation (1)
\(R=2\times \cfrac { 12 }{ 3 } =8\Omega \)
\(R=8\Omega \)

\(V=\frac { q }{ 4\pi { \varepsilon }_{ 0 }r } \)
\(F=\frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
= 40.9kV

The line joining B to C is perpendicular to electric field

So potential of B = potential of C
i.e. V_B = V_c
Distance AB = 4 cm
Potential difference
between A & C = E x AB
= 5 x 10³  x (4 x 10^-2)
= 200 volt.
AC² = AB² + B²
AB² = AC² - BC²
= 25 - 9 = 16
AB = 4cm

Working: The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. Let the field magnet be rotated in clockwise direction by some external means (Figure). 
(i) Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin O in the graph between induced emf and time angle.
(ii) When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced emfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Flemin's right hand rule.

(iii) Care has to be taken while applying this rule, the thumb indicates the direction of the motion of the conductor with respect to field. For clockwise rotating poles, the conductor appears to be rotating anticlockwise. Hence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is inwards and for RS outwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.
(iv) For the rotation of 180° from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point B.
(v) The field magnet becomes again parallel to PQRS for 270° rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP. This is represented by point C
(vi) On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature.
(vii) Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle.

Current leads the applied voltage by \(\frac{\pi}{2}\) in a capacitive circuit.
This is the resistance offered by the capacitor, called capacitive reactance (X_c). It measured in ohm.
\({ X }_{ c }=\frac { 1 }{ \omega C } \)
The capacitive reactance (X_c) varies inversely as the frequency. For a steady current, f = 0
∴\({ X }_{ c }=\frac { 1 }{ \omega C } -\frac { 1 }{ 2\pi fC } =\frac { 1 }{ 0 } =\infty \)
Thus a capacitive circuit offers infinite resistance to the steady current.

(i) Carbon resistors consists of a ceramic core, on which a thin layer of crystalline Carbon is deposited. These resistors are inexpensive, stable and compact in size. Color rings are used to indicate the value of the resistance.
(ii) Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth color, silver or gold shows the tolerance of the resistor.

(i) The electric potential at a point P due to a collection of charges q₁, q₂, q₃, ···q_n is equal to sum of the electric potentials due to individual charges.
\({ V }_{ tot }=\frac { k{ q }_{ 1 } }{ { r }_{ 1 } } +\frac { { kq }_{ 2 } }{ { r }_{ 2 } } +\frac { { kq }_{ 3 } }{ { r }_{ 3 } } +...\frac { { kq }_{ n } }{ { r }_{ n } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } { \sum { } }_{ i=1 }^{ n }\frac { { q }_{ i } }{ { r }_{ i } } \)
(ii) where r₁, r₂, r₃ .... r_n are the distances of q₁, q₂, q₃ ..... q_n respectively from P(Figure).