According to the Gauss's theorem, the total electric flux through a closed surface is \(\frac{1}{\varepsilon_{0}}\) times the magnitude of net charge enclosed by the surface. Consider a hollow charged sphere.
Take a point inside the sphere where electric field is to be calculated. Draw a Gaussian surface of radius r having point P on its surface.

Now, \(\phi=\oint_{S} \vec{E} \cdot \vec{d} s=E \oint_{S} d s=E .4 \pi r^{2}\)
According to the Gauss's theorem,
\(E .4 \pi r^{2}=\frac{q}{\varepsilon_{0}}(\because q=0)\)
\(\therefore E=0\)

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 \(\vec{p}\))

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
\(\overrightarrow { \tau } =\overrightarrow { m } \times \overrightarrow { B } \)

Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection,
 = 4 mA = 4 x 10^-3 A
Range of the ammeter is 0, which needs to be converted to 6 A.
Current, I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
\(S=\frac{I_{g} G}{I-I_{g}}\)
\(=\frac{4 \times 10^{-3} \times 15}{6-4 \times 10^{-3}}\)
\(S=\frac{6 \times 10^{-2}}{6-0.004}=\frac{0.06}{5.996}\)
\(\approx 0.01 \Omega=10 \mathrm{~m} \Omega\)
Hence, a  10 mΩ shunt resistor is to be connected in parallel with the galvanometer.

(i) 36 x 10^-2 Am²
(ii) 18 x 10^-2Nm

\(\begin{aligned} & \mathrm{E}_i=\lim _{q_0 \rightarrow 0} \frac{\mathbf{F}_i}{q_0}=\lim _{q_0 \rightarrow 0}\left[\frac{1}{q_0}\left(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i q_0}{r_i^2} \hat{\mathbf{r}}_i\right)\right] \\ & \mathrm{E}_i=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i}{r_i^2} \hat{\mathbf{r}}_i \end{aligned}\)..............(i)
If E is eletric ficld at point P due to the system of charges, then by principle of superposition of electric fields, \(\mathrm{E}=\mathrm{E}_1+\mathrm{E}_2+\mathrm{E}_3+\cdots+\mathrm{E}_n=\sum_{i=1}^n \mathrm{E}_i\)
Using Eq. (i), we get \(\mathrm{E}=\sum_{i=1}^n \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i}{r_i^2} \hat{r}_i\)
or   \(\mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_i^2} \hat{\mathbf{r}}_i\)
\(\therefore\) E is a vector quantity.
 

Paper is a dielectric, so when a positively charged glass rod is brought near it, atoms of paper get polarised, with centre of negative charge of atoms coming closer to the glass rod.

Therefore, force of attraction F_a between glass rod and piece of paper becomes greater than the force of repulsion F_r between the glass rod and the piece of paper. This results in attraction of the piece of paper towards the glass rod.

Given, total number of turns, N = 500
Length of solenoid, l = 0.5 m
Current, I = 5 A
Radius, r = 1 cm = 10^-2 m
Here, \(\begin{aligned} \frac{l}{r} & =\frac{0.5}{10^{-2}}=50 \Rightarrow l>>r \\ \end{aligned}\)
\(\begin{aligned} \therefore B & =\mu_0 n I=\frac{\mu_0 N I}{l} \\ \end{aligned}\)
\(\begin{aligned} =4 \pi \times 10^{-7} \times \frac{500}{0.5} \times 5 \end{aligned}\)
= 6.28 \(\times\) 10^-3 T

Lorentz magnetic force on a moving charged particle in the magnetic field is 
\(\overset { \rightarrow }{ F } =q\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) \) 
This force is acting perpendicular to the plane containing \(\overset { \rightarrow }{ v } \) and \(\overset { \rightarrow }{ B } \) , and is directed as given by Right Hand rule. Since work done,
\(\overset { \rightarrow }{ W } =\overset { \rightarrow }{ F } .\overset { \rightarrow }{ s } =Fscos\theta ,\) 
where   is displacement of charged particle in magnetic field. Here, angle \(\theta ={ 90 }^{ o }\) , so 
W = Fs cos 90^o
= 0.

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
 \(=\frac { \left[ { MLT }^{ -2 } \right] }{ \left[ AT \right] \left[ { LT }^{ -1 } \right] } =\left[ { ML }^{ o }{ T }^{ -2 }{ A }^{ -1 } \right] \)

(a) Moving coil galvanometer is an instrument used for detection and measurement of small currents. Moving coil galvanometer works on the principle that when a current-carrying coil is placed in magnetic field it experiences a torque. 
(b) (i) Voltage sensitivity is defined as the deflection produced in the galvanometer when a unit voltage is applied across the two terminals of the galvanometer:
Vs = NBA/KR
(ii) Current sensitivity is defined as the deflection produced in the galvanometer when a limit current flow through the galvanometer:
Is = NBA/K
where N is number of turned in coil, B is the magnetic field, A is the area of the coil, K is the restoring torque per limit twist.

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
\( W =q \int_{r=\infty}^{r} \vec{E} \cdot \overrightarrow{d r}=q \int_{\infty}^{r}\left(\frac{\sigma}{2 \varepsilon_{0}} d r\right) \)
\(=q \cdot \frac{\sigma}{2 \varepsilon_{0}} \int_{r=\infty}^{r} d r \)
\(W =\frac{q \cdot \sigma}{2 \varepsilon_{0}}[r-\infty]=\frac{\infty}{2 \varepsilon_{0}}[r-\infty]=\infty\)

(ii) As, \(\frac{F}{L}=\frac{\mu_0}{4\pi}.\frac{2 I_1 I_2}{r}\)
\(I_1=I_2=I A, r=1 m\)
\(\frac{F}{L}=2\times10^{-7} Nm^{-1}\)
(iii) Here, magnetic field due to the current carrying conductor at a distance d from it is
given by
\(B=\frac{\mu_0}{4\pi} \frac{2I}{d}\)
∴  Force on proton,
F = (e) (v) B sin 90°
⇒ F = evB
\(F=ev(\frac{\mu_0}{4\pi}\frac{2I}{d})\)
\(F=\frac{\mu_0}{4\pi}.\frac{2Iev}{d}\)
The proton is directed perpendicular to straight conductor and away from it.

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) (d)
(ii) (a): Using \(q v B \sin \theta=\frac{m v^{2}}{r}\)
\(r \propto \frac{1}{\sin \theta}\) for the same values of m, v, q and B
\(\therefore \frac{r_{A}}{r_{B}}=\frac{\sin 90^{\circ}}{\sin 30^{\circ}}=2 \text { or } r_{A}=2 r_{B} \text { or } r_{B}<r_{A}\)
(iii) (d): The radius of the helical path of the electron in the uniform magnetic field is
\(r=\frac{m v_{\perp}}{e B}=\frac{m v \sin \theta}{e B}=\frac{\left(2.4 \times 10^{-23} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\right) \times \sin 30^{\circ}}{\left(1.6 \times 10^{-19} \mathrm{C}\right) \times 0.15 \mathrm{~T}}\)
\(=5 \times 10^{-4} \mathrm{~m}=0.5 \times 10^{-3} \mathrm{~m}=0.5 \mathrm{~mm}\)
(iv) (c): Here \(\vec{B}=8.35 \times 10^{-2} \hat{i} \mathrm{~T}\)
\(\vec{v}=2 \times 10^{5} \hat{i}+4 \times 10^{5} \hat{j} \mathrm{~m} / \mathrm{s}, m=1.67 \times 10^{-27} \mathrm{~kg}\)
Pitch of the helix (i.e., the linear distance moved along the magnetic field in one rotation) is given by
Pitch of the helix \(=\frac{2 \pi m v_{\|}}{q B}\)
\(=\frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 2 \times 10^{5}}{1.6 \times 10^{-19} \times 8.35 \times 10^{-2}}=0.157 \mathrm{~m}\)
(v) (b): Period of revolution 
\(T=\frac{2 \pi R}{v \sin \theta} \Rightarrow T=\frac{2 \pi\left(\frac{m v \sin \theta}{q B}\right)}{v \sin \theta} \Rightarrow T=\frac{2 \pi m}{q B}\)
\(\therefore \text { Frequency, } v=\frac{1}{T}=\frac{q B}{2 \pi m}\)

(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

(d) : When current flows through a solenoid, the currents in the various turns of the solenoid are parallel and in the same direction. Since the currents flowing through parallel wires· in the same direction lead to force of attraction between them, the turns of the solenoid will also attract each other and as a result. she solenoid tends to contract