Here, q = 25 x 10^-9C, 2a = 6m, r = 4 m,
p = q(2a)= 25 x 10^-9 x 6 =1.5 x 10^-7 c-m
(i) Now, \(E_{\text {axial }}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 p r}{\left(r^{2}-a^{2}\right)^{2}}\)
\(=\frac{9 \times 10^{9} \times 2 \times 1.5 \times 10^{-7} \times 4}{\left(4^{2}-3^{2}\right)^{2}}=\frac{2700 \times 4}{49}\)
= 220.4 NC ^-1
(ii) \(\therefore E_{\text {equatorial }}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{p}{\left(r^{2}+a^{2}\right)^{3 / 2}}\)
\(=\frac{9 \times 10^{9} \times 1.5 \times 10^{-7}}{\left(4^{2}+3^{2}\right)^{3 / 2}}=\frac{1350}{125}=\) 10.8 NC^-1

(a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge
(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Here, n = 100, r = 8 cm = 8 \(\times\)10^-2m and I = 0.40 A
\(\therefore\) Magnetic field B at the centre,
\(\begin{aligned} B=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi I n}{r} & =\frac{10^{-7} \times 2 \times 3.14 \times 0.40 \times 100}{8 \times 10^{-2}} \\ \end{aligned}\)
\(\begin{aligned} =3.1 \times 10^{-4} \mathrm{~T} \end{aligned}\)

Magnetic flux density at a point in a magnetic field means magnetic field induction at that point. It is defined as the force experienced by a unit charge while moving with a unit velocity, perpendicular to the direction of magnetic field at that point. Force experienced by the charged particle having charge q moving with velocity \(\overset { \rightarrow }{ v } \) through a magnetic field \(\overset { \rightarrow }{ B } \) is given by
\( \left| \overset { \rightarrow }{ F } \right| =q\left| \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right| =qvBsin\theta\)  
\(or \ B=\frac { F }{ qvsin\theta } \)
The SI unit of B is tesla, where 1 tesla is the magnetic flux density at a point if 1 coulomb charge while moving with a velocity of 1 ms^-1, perpendicular to a magnetic field experiences a force of 1 N at that point.
The dimensional formula of B
 \(=\frac { \left[ { MLT }^{ -2 } \right] }{ \left[ AT \right] \left[ { LT }^{ -1 } \right] } =\left[ { ML }^{ o }{ T }^{ -2 }{ A }^{ -1 } \right] \)

Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection,  I_g = 3 mA = 3 x 10^-3 A
Range of the voltmeter is 0, which needs to be converted to 18 V.
therefore, V = 18 V
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
\(R=\frac{V}{I_{\mathrm{g}}}-\mathrm{G}\)
\(=\frac{18}{3 \times 10^{-3}}-12=6000-12=5988 \Omega\)
Hence, a resistor of resistance 5998 Ω is to be connected in series with the galvanometer.

Electric field intensity due to a thin infinite plane sheet charge: Consider a thin infinite sheet of charge with uniform surface charge density a. To calculate electric field at a point P distant r from the sheet we imagine a symmetrical Gaussian surface in such a way that the point charge lies on it. Here we assume a cylinder of cross-sectional area A and length 2r with its axis perpendicular to the sheet.

Flux through the curved surface of the cylinder,
\(\phi_{1}=\int \vec{E} \cdot \overrightarrow{d s}=0 \left(\because \theta=90^{\circ}\right)\)
Total flux through plane faces of the cylinder,
\(\phi_{2}=2 \int \vec{E} \cdot \overrightarrow{d s}=2 E A \left(\because \theta=0^{\circ}\right)\)
Net flux through the Gaussian surface is
\(\phi=\phi_{1}+\phi_{2}=2 E A\)
Net charge enclosed by the Gaussian surface is
\(Q=\sigma A\)
According to the Gauss's theorem, \(\phi=\frac{Q}{\varepsilon_{0}}\)
\(\therefore \phi=\frac{\sigma A}{\varepsilon_{0}}\)
From equations (i) and (ii), we get
\(2 E A=\frac{\sigma A}{\varepsilon_{0}} \Rightarrow E=\frac{\sigma}{2 \varepsilon_{0}}\)
(b) 

Work done in bringing a charge from ∞ to given point P is given by
\( W =q \int_{r=\infty}^{r} \vec{E} \cdot \overrightarrow{d r}=q \int_{\infty}^{r}\left(\frac{\sigma}{2 \varepsilon_{0}} d r\right) \)
\(=q \cdot \frac{\sigma}{2 \varepsilon_{0}} \int_{r=\infty}^{r} d r \)
\(W =\frac{q \cdot \sigma}{2 \varepsilon_{0}}[r-\infty]=\frac{\infty}{2 \varepsilon_{0}}[r-\infty]=\infty\)

Statement for Biot-Savart Law: The magnitude of magnetic field \(d\overrightarrow { B } \) due to current element is directly proportional to the current I, the elements length \(\left| dl \right| \) and inversely proportional to the square of the distance r of the field point. Its direction is perpendicular to the plane containing \(\overrightarrow { dl } \)and \(\overrightarrow{r}\).
\(d\overrightarrow B\alpha\frac{I\overrightarrow dl \times \overrightarrow r}{r^{ 3}}\)
Or \(d\overrightarrow B=\frac {\mu_0}{4 \pi}\frac{I\overrightarrow dl \times \overrightarrow r}{r^{ 3}}\)

The magnetic field due to \(\overrightarrow {dl}\) is given by Biot Savart law as
\(dB=\frac {\mu_0}{4\pi}.\frac { I\left| \overrightarrow { dl } \times \overrightarrow { r } \right| }{ { r }^{ 3 } } \)
Now dB_x= Db Cos \(\theta\) = \(\frac { \mu _{ 0 } }{ 4\pi } .\frac { Idl }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) } \cos { \theta } \)
\(=\frac { \mu _{ 0 } }{ 4\pi } .\frac { Idl }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) } \frac { R }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 1/2 } } \)
So, \({ B }_{ x }=\int { dB_{ s }=\frac { { \mu }_{ 0 } }{ 4\pi } } \frac { IR }{ \left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \int { dl } \)
\(=\frac { { \mu }_{ 0 }IR^{ 2 } }{ 2\left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \)
(The y-components, of the field, add up to zero,due to symmetry)
\(\therefore\)Magnetic field at P due to a circular loop
\(=B={ B }_{ x }\overrightarrow { i } =\frac { { \mu }_{ 0 }IR^{ 2 } }{ 2\left( { x }^{ 2 }+{ R }^{ 2 } \right) ^{ 3/2 } } \overrightarrow { i } \)
Explanation: A circular current loop produces magnetic field and its magnetic moment is the product of current and its area \(\overrightarrow M=\overrightarrow {LA}\)

(i) A galvanometer of range I_g and resistance G₁ can be converted into
(a) a voltmeter of range V, by connecting a high resistance R in series with galvanometer whose value is given by
\(R=\frac { V }{ { I }_{ g } } -G\)
(b) an ammeter of range I, by connecting a very low resistance (shunt) in parallel with galvanometer whose value is given by
\(S=\frac { { I }_{ g }G }{ I-{ I }_{ g } } -G\)
Thus, the nature of force is attractive.
When direction of flow of current is in opposite direction, the nature of force becomes repulsive.

Current sensitivity, \({ I }_{ s }=\frac { NAB }{ k } \) and
Voltage sensitivity, \(V_{ s }=\frac { NAB }{ kR } \)
Since, the resistance of the coil may vary, it implies an increase in current sensitivity may not necessarily increase voltage sensitivity.
Thus, the trajectory of both the particles will be same.

\(\text { Given: } G=100 \Omega ; I_{\mathrm{g}}=10^{-3} \mathrm{~A} ; I=1 \mathrm{~A} ; S=?\)
\(\therefore \ S=\frac{1_{g} G}{I-I_{g}}=\frac{10^{-3} \times 100}{1-10^{-3}}=0.1 \Omega\)

Electric dipole moment is a measurement of the strength of electric dipole. It is given by \(\vec{p}=q(\overrightarrow{2 a})\) cm, where \(\vec{p}\) is the electric dipole moment and 2a is the separation between the charges. It is a vector quantity directed from negative to positive charge on the line joining them.

Let the dipole be made of two equal and opposite charges +q and -q, separated by 2a. Consider a point P at a distance r from the mid-point. Field at P due to each charge will be of equal magnitude \(\left|\vec{E}_{\pm q}\right|=\frac{k q}{\left(r^{2}+a^{2}\right)}\) pointing as shown.
Resolving electric fields due to two charges. We can see that Y-axis components get cancelled out.

∴ Net field at P,E = 2E_q cos θ
\( E= \frac{2 k q}{\left(r^{2}+a^{2}\right)} \cdot \frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}} \)
\(= \frac{2 a q k}{\left(r^{2}+a^{2}\right)^{3 / 2}}=\frac{k p}{\left(r^{2}+a^{2}\right)^{3 / 2}} \)
\(\left(\because \cos \theta=\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\right)\)
(pointing anti-parallel to dipole moment) 
If r >> a, i.e. a² can be neglected in comparison to r².
\(\therefore \ E=\frac{k p}{r^{3}}\) (anti-parallel to \(\vec{p}\))

Given, p = 4 \(\times\)10^-9 C-m, E = 5 \(\times\) 10⁴,
\(\theta\) = 30°
\(\therefore\) \(\tau\) = pE sin \(\theta\)
= 4 \(\times\)10^-9 \(\times\)5 \(\times\)10⁴ \(\times\)sin 30°
\(=4\times 10^{-9}\times 5\times 10^{4}\times \frac{1}{2}\)    [\(\because\)sin 30° =\(\frac{1}{2}\)]
= 10 \(\times\)10^-5 = 10^-4 N-m

The force, on a wire of length I,carrying a current I, in a magneticfield \(\overrightarrow { B } \) is given by \(\overrightarrow { F } =\left( \overrightarrow { l } \times \overrightarrow { B } \right) \) For a rectangular loop, places as shown, in a magnetic field \(\overrightarrow { B } \)

|Force on arm BCI = IForce on arm DAI = l/Bsin \(\alpha\)
Where \(\alpha\)=angle between side BC and \(\overrightarrow { B } \)
These two forces add up to zero as they are collinear (along the axis of the coil) and act in opposite directions
|Force on arm ABI = IForce on arm CDI = IbB
These two equal and opposite forces are not collinear. The perpendicular distance between their lines of action is, as shown
\(2\times\frac{a}{2}sin\theta=a sin\theta\)
Torque acting on the coil, has a magnitude \(\tau \) where
\(\tau \)=(lbB)x(a sin \(\theta\))=IAB sin \(\theta\) (A=ab=Area of the coil)
In vector form, \(\overrightarrow { \tau } =\overrightarrow { A } \times \overrightarrow { B } \)
But \(I\overrightarrow { A } \)=\(\overrightarrow { m } \), as given
\(\overrightarrow { \tau } =\overrightarrow { m } \times \overrightarrow { B } \)

(i) The required expression for the face is
F = q (v x B) = q vB sinθ
Expression for magnetic field is,
\(\overrightarrow F=q(\overrightarrow v \times \overrightarrow B)\)
If \(\overrightarrow v\) and \(\overrightarrow B\) both are normal to each other, then
F = qvB.
Now if q = 1, v = 1, then F = B
Hence, the magnetic field at a point is equal to the force which a unit charge experiences when it enters with unit velocity in a direction normal to the direction of the magnetic field at that point. In S.I system, unit of magnetic field is weber/metre^-2 (Wbm^-2)or tesla (T) or newton per ampere per metre (NA^-1m^-1).
(ii) Consider the segment of a conductor given in below figure.

Let the number of electrons per unit volume of the conductor is n; the drift speed of electron inside the conductor is V_d and the magnetic field is B.
Then, Lorentz magnetic force, f = -e(V_d х B)
∴ Force on all the mobile electrons of the conductor F = f.nAl = -nAle(V_d х B)
or F = l(l х B) = IlB sinθ
This is the expression for the force experienced by current carrying wire.

(c): Since \(d B \propto \sin \theta\) where \(\theta\) is angle between the direction of the flow of current and the line joining the elementary portion to the observation point which is zero in this case, so the magnetic field is also zero (because \(\sin \theta\) is equal to zero).

(c): A body becomes negatively charged only when some electrons are transferred to the body i.e. the body gains some electrons. Hence its mass increases slightly. Mass of a body decreases only when body gives some electrons to some other body.