The figure (a) indicates a diamagnetic material.
The figure (b) indicates a ferromagnetic material.
The figure (c) indicates a paramagnetic material.

(i) As there is no electric field inside thin charged metallic spherical shell,
the force on the charge \(\frac{Q}{2}\) is zero.
Force on charge 2Q at point A.
\( F =E \times 2 Q=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{3 Q}{2 x^{2}}\right) \cdot 2 Q \)
\(=\frac{1}{4 \pi \varepsilon_{0}} \frac{3 Q^{2}}{x^{2}}\)
(ii) Electric flux through the shell,
\(\phi_{E}=\frac{Q}{2 \varepsilon_{0}}\)

(a) The lines of force due to a positive charge placed near a metal plate are as shown in the figure.

(b) In the region II between the plates \(\vec{E}_{A} \text { and } \vec{E}_{B}\) are opposite to each other.

\(\text {As } \sigma_{1}>\sigma_{2},\left|\vec{E}_{A}\right|>\left|\vec{E}_{B}\right|\)
and resultant field \(=E_{A}-E_{B}=\frac{\sigma_{1}}{2 \varepsilon_{0}}-\frac{\sigma_{2}}{2 \varepsilon_{0}}\) 
\(\therefore \ \overrightarrow{E_{I I}}=\frac{1}{2 \varepsilon_{0}}\left(\sigma_{1}+\sigma_{2}\right) \text { from } A \text { to } B\)
In the region III, both \(\vec{E}_{A} \text { and } \vec{E}_{B}\) are supporting each other.
∴ \(\overrightarrow{E_{I I I}}=\frac{1}{2 \varepsilon_{0}}\left(\sigma_{1}+\sigma_{2}\right) \text { away from } B .\)

 This law gives the relationship between the total flux passing through any closed surface and the net charge enclosed within the surface. Gauss's law states that the total flux through a closed surface is \({{1}\over{{\epsilon}_{0}}}\) times to the net charge enclosed by the closed surface mathematically, \({\phi}_{E}={\phi}_{s}E.dS={{q}\over{{\epsilon}_{0}}}\).
Electric field intensity due to an infinitely long uniformly charged wire at point P at distance r from it is obtained as follows: Consider a thin cylindrical Gaussian surface S with charged wire on its axis and point P on its surface, then net electric flux through surface S is

\(\phi=\oint_{s}E.dS=\overset {\int{Eds\ \cos \ 90°} }{ Upper\ plan\ face } + \overset { \int{Eds\ cos\ 90°} }{ Caved\ surface }+\overset { \int{Eds\ cos\ 90°} }{ Lower\ plane\ face }\)
\(\phi\) = 0 + EA + 0 or  \(\phi\) = E.2\(\pi rl\)
But by Gauss's theorem,
\(\phi={{q}\over{{\epsilon}_{0}}}={{\lambda \ l}\over{{\epsilon}_{0}}}\)
Where, q is the charge on length l of wire enclosed by cylindrical surface S and \(\lambda\) is uniform linear charge density of wire.
\(\therefore\)    \(E\times2\pi r l={{\lambda\ l}\over{{\epsilon}_{0}}}\Rightarrow\ E={{\lambda}\over{3\pi{\epsilon}_{0}r}}\)
Thus, electric field of a line charge is inversely proportional to distance directed normal to the surface of charged wire.
Let us consider a long straight wire carrying + q charge on its length / and linear charge density \(\lambda\) Cm.
\(\therefore\)   \(\lambda={{q}\over{l}}\)
\(\Rightarrow\) q = \(\lambda\ l\)        ...(i)
Let electric Iiclu intensity is [Q be: obtained .H a distance r from it. Since, magnitude of E due to long charged wire is same at every point which lie at the same distance from the wire. So, Gaussian surface will be a cylinder of radius r and length / such that wire lies along the axis as shown in figure.
\(\because\) Angle between E and dSis 90° at caps, whereas 0° at any point on curved surface of a cylinder. Now, applying Gauss' theorem.
\(\int{E.dS={{q}\over{{\epsilon}_{0}}}}\)
\(\oint_{CSA}E.dS+\oint_{CSA}E.dS={{\lambda\ l}\over{{\epsilon }_{0}}}\)  [FromEq. (i)]
 [CSA = Close Surface Area]
\(\int_{CSA}E\ dScos\ 90°+\int_{CSA}E\ dS\ cos\ 0°={{\lambda\ l}\over{{\epsilon}_{0}}}\)
(S₁ and S₃ are caps and S₂ represents CSA)
\(0+\int_{CSA}E\ dS={{\lambda l}\over{{\epsilon}_{0}}}\)      \([\because\ cos 90°=0]\)
\(E \oint_{CSA}\ dS={{\lambda l}\over{\epsilon_0}}\)
[ \(\because\) E is a constant at every point at CSA]
\(E\times2\pi r l={{\lambda l}\over{{\epsilon}_{0}}}\)
E = \({ {\lambda l }\over{ 2\pi{\epsilon}_{0} r l} }\)
\(\Rightarrow\)  E = \({ {\lambda }\over{2\pi{ \epsilon}_{ 0} r} }\)

Electric field due to a uniformly charged thin spherical shell :


(i) When point P lies outside the spherical shell :Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.
Let \(\overrightarrow{E}\) be the electric field at point P, then the electric flux through area element is \(\overrightarrow{ds}\) given by
\(d\phi = \overrightarrow{E}.\overrightarrow{\Delta S}\)
 Since \(\Delta\)S is also along normal to the surface,
\(d\phi = E ds\)
\(\therefore\) Total electric flux through the Gaussian surface is given by,
\(\phi = \oint_s Eds = E\oint_s ds\)
 Now  \(\oint_s ds = 4\pi r^2\)
\(\phi = E \times 4\pi r^2\)................(i)
Since the charge enclosed by the Gaussian surface is q₁ according to Gauss theorem,
\(\phi = \frac{q}{\epsilon_0}\)   .......................(ii)
From equations (i) and (ii),we obtain
\(E\times4\pi r^2 = \frac{q}{\epsilon_0}\)
\(E = \frac{1}{4\pi\epsilon_0}\frac{q}{r_2}\) ( for ( r > R ) )
(ii) When point P lies inside the spherical shell : In such a case, the Gaussian surface encloses no charge. According to Gauss law,
 \(E\times4\pi r^2 = 0\)
i.e., E = 0 (r < R)
Graph showing the variation of electric field as a function of r :

Given, I = 10 A, N = 100, l = 40 cm, b = 20 cm
B = 5 T, \(\theta\)= 60°
A = l \(\times\)b = 40 \(\times\)20 = 800 cm² = 8 \(\times\)10^-2m²
\(\begin{array}{r} \because \quad \tau=N B I A \sin \theta=100 \times 5 \times 10 \times 8 \times 10^{-2} \times \frac{\sqrt{3}}{2} \\ \end{array}\)
\(\begin{array}{r} {\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]} \end{array}\)
= 346.41 N-m

Electric energy consumed per second = 100 J
Heat produced = 80% = \(\frac{80}{100}\times 100\)
= 80 J = \(\frac{80}{4.2}\)
= 19.05 cal

392.8 volt

(a)  \(B=\frac { { \mu }_{ 0 }NI }{ 2\pi r } ;\)
\(A=12{ cm }^{ 2 }=12\times { 10 }^{ -4 }{ m }^{ 2 },\)     
\(r=15cm=0.15m\)
\(\therefore \)     \(\phi =\frac { { \mu }_{ 0 }NI }{ 2\pi r } A,\)
Total flux  \(N\phi =\frac { { \mu }_{ 0 }A{ N }^{ 2 } }{ 2\pi r } I\)
Since   \(\phi =LI\)
\(\therefore \)   \(L=\frac { { \mu }_{ 0 }A{ N }^{ 2 } }{ 2\pi r } \)
\(=\frac { 4\pi \times { 10 }^{ -7 }\times 12\times { 10 }^{ -4 }\times { \left( 1200 \right) }^{ 2 } }{ 2\pi \times 0.05 } \)
\(=2.3mH\)
\(\left| e \right| =\frac { d }{ dt } \left( { N }_{ 2 }{ \phi }_{ 2 } \right) \)
\(={ N }_{ 2 }\frac { { \mu }_{ 0 }{ N }_{ 1 }A }{ 2\pi r } \frac { d{ I }_{ 1 } }{ dt } \)

\(=\frac { 4\pi \times { 10 }^{ -7 }\times 300\times 1200\times 12\times { 10 }^{ -4 }\times 2 }{ 2\pi \times 0.15\times 0.05 } \)
\(\left| e \right| =0.023 \ V\)

The ratio of length to radius of the solenoid is quite large (about 35). Therefore, to estimate B approximately, we can use the exact result for a closely wound infinitely long solenoid.
(a) At the centre or near it,
 \(B={ \mu }_{ 0 }nI\) 
Where n is the number of turns per unit length.
Note 1. the radius of the wire does not enter this equation. Therefore, to get n, simply multiply number of turns per layer and divide the product by the length of the solenoid.
 \(n=\frac { 350\times 4 }{ 0.50 } =2800{ m }^{ -1 }\)
Now  \(I=6.0A,\)  
and  \({ \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }T{ mA }^{ -1 },\)
Which gives \(B=2.1\times { 10 }^{ -2 }T\) 
Note 2. This estimate of B is for both on and off the axis, since for an infinitely long solenoid, the internal field near the centre is uniform over the entire cross-section.
(b) At the end of the solenoid,
\(B=\frac { { \mu }_{ 0 }nI }{ 2 }\)
\(=1.05\times { 10 }^{ -2 }T\)
(c) The outside field near the centre of long solenoid is negligible to the internal field.

Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T₂ = 100°C
Resistance of the silver wire at T₂, R₂ = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as 
\(\alpha=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}\)
\(=\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1}\)
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.

Given: \(C=55 \times 10^{-12} \mathrm{~F}, V=5.3 \mathrm{~V}\) 
\( \because \ Q =C V=n e \)
\(\therefore n=\frac{55 \times 10^{-12} \times 5.3}{1.6 \times 10^{-19}}=1.8 \times 10^{9} \text { electrons } \)

If length of wire is 2l , then magnetic moment
M = m x 2l = 4\(\pi\) A-m²  [given]
As wire is bent in the form of semicircle, effective distance between the ends is 2r.
So, new dipole moment
\(M^{\prime}=m \times 2 r=m \times 2 \times \frac{2 l}{\pi}=\frac{2}{\pi}(m \times 2 l) \quad[\because \pi r=2 l]\)
\(=\frac{2}{\pi} M=\frac{2}{\pi} 4 \pi\)
= 8  A - m²

It is also attractive in nature.

r = mv/qB .........(i)
Also P = mv = \(\sqrt { 2mE } \)..........(ii)
By equ (i) and equ (ii)
As the radius is \(r\frac { \sqrt { 2mE } }{ qB } \) proportional to square root of kinetic energy, so if the kinetic energy is halved the radius become \(\sqrt { 1 } /2\) times of its initial value 

The magnitude of the emf induced in the circuit due to magnetic flux depends on the time rate of change of magnetic flux through the circuit.
\(|\varepsilon|=\frac{\Delta \phi}{\Delta t}\)

Drift velocity,
\(v_d=\frac{1}{n A \varepsilon}=\frac{I}{n\left(\pi D^2 / 4\right) e}, \text { i.e., } v_d \propto \frac{1}{n}\)
Since the wires are connected in series, current I through both is same. Therefore,
\(\frac{\left(v_d\right)_x}{\left(v_d\right)_y}=\frac{n_y}{n_x}=\frac{1}{2}\)(vd)x(vd)y=nynx=12

Given, C = 200 \(\mu\)F, d = 5 mm, t = 5 mm, V = 100 V
\(\text { (i) } C=\frac{\varepsilon_0 A}{d} \Rightarrow A=\frac{C d}{\varepsilon_0}\)
\(A=\frac{200 \times 10^{-6} \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}=112.99 \times 10^3 \mathrm{~m}^2\)
When d' = 2d, then C' = \(\frac{\varepsilon_0 A}{2 d-t+\frac{t}{K}}\)
\(=\frac{8.85 \times 10^{-12} \times 112.99 \times 10^3}{\left(10-5+\frac{5}{10}\right) \times 10^{-3}}\)
\(=181.8 \times 10^{-6}=181.8 \mu \mathrm{F}\)
(i) Charge on capacitor, q = C₀V₀
= 200 \(\times\) 10^-6 \(\times\) 100
= 2 \(\times\) 10^-2 C
\(\begin{aligned} \Rightarrow \quad C_0 V_0 & =C^{\prime} V^{\prime} \end{aligned}\)
\(\begin{aligned} \text { or } \quad V^{\prime} & =\frac{C_0 V_0}{C^{\prime}}=\frac{2 \times 10^{-2}}{181.8 \times 10^{-6}}=110 \mathrm{~V} \end{aligned}\)
\(\begin{aligned} E_0 & =\frac{V_0}{d}=\frac{100}{5 \times 10^{-3}}=20 \times 10^3 \mathrm{~V} / \mathrm{m} \\ \end{aligned}\)
\(\begin{aligned} E^{\prime} & =\frac{V^{\prime}}{2 d}=\frac{110}{10 \times 10^{-3}}=11 \times 10^3 \mathrm{~V} / \mathrm{m} \end{aligned}\)
(iii) \(\begin{gathered} \bar{U}=\frac{1}{2} \varepsilon_0 E_0^2=\frac{1}{2} \times 8.85 \times 10^{-12} \times\left(20 \times 10^3\right)^2 \\ \end{gathered}\)
\(\begin{aligned} =1770 \times 10^{-6} \mathrm{~J} / \mathrm{m}^3 \end{aligned}\)
\(\begin{aligned} (\bar{U})^{\prime} & =\frac{1}{2} \times \varepsilon_0\left(E^{\prime}\right)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 8.85 \times 10^{-12} \times\left(11 \times 10^3\right)^2 \\ \end{aligned}\)
\(\begin{aligned} =535.42 \times 10^{-6} \mathrm{~J} / \mathrm{m}^3 \end{aligned}\)

The given loop can be shown below as

The force acting on the arms MN and PO of the given loop are equal, mutually opposite and collinear, hence they balance each other.
Force on arm PM,
\(F_1=\frac{\mu_0 I_1 I_2 L}{2 \pi L}=\frac{\mu_0 I_1 I_2}{2 \pi}\), attractive in nature ...(i)
Force on arm NO,
\(F_2=\frac{\mu_0 I_1 I_2 L}{2 \pi(2 L)}=\frac{\mu_0 I_1 I_2}{4 \pi},\) , repulsive in nature ...(ii)
From Eqs. (i) and (ii), we get
F = F = F₁ - F₂
\(=\frac{\mu_0 I_1 I_2}{2 \pi}-\frac{\mu_0 I_1 I_2}{4 \pi}\)
\(=\frac{\mu_0 I_1 I_2}{4 \pi}\)attractive in nature                            ...(iii)
So, from Eqs. (ii) and (iii), we can conclude that the magnitude of the force on side NO of the loop is \(F=\left(\frac{\mu_0 I_1 I_2}{4 \pi}\right)\) when the net force F is towards the wire.

Let E and r be the emf and internal resistance of each cell.
Case I When the cells are in series.
Total emf of cells = 14E
Total resistance of circuit = 82.6 + 14r
∴ Current in the circuit is given by
\(\frac{14 E}{82.6+14 r}=0.25 \mathrm{~A}\)
Case II When the cells are in parallel.
Total emf of cells = E
Total resistance of circuit = 0.053 \(+\frac{r}{14}\)
∴ Current in the circuit is given by
\(\frac{E}{0.053+\frac{r}{14}}=25 \mathrm{~A}\)
Dividing Eq. (i) by Eq. (ii), we get
\(14 \frac{\left(0.053+\frac{r}{14}\right)}{(82.6+14 r)}=10^{-2}\)
\(\Rightarrow \ 14 \times \frac{14 \times 0.053+r}{14} \times 10^{2}=82.6+14 r\)
⇒ 53 x 14 + 100r = 82.6 + 14r
Solving, we get
r = 0.097Ω ≈ 0.1 Ω
Substituting the value of r in Eq. (i), we get
E = 1.5V

Let the net electric field be zero at point P at a distance x from charge +4\(\mu\)C, then
\(\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \times 10^{-6}}{x^{2}}-\frac{1 \times 10^{-6} \times 1}{4 \pi \varepsilon_{0} \times(2-x)^{2}}=0\)
\(\Rightarrow \ \frac{4}{x^{2}}=\frac{1}{(2-x)^{2}}\)
\(\Rightarrow \ \frac{2}{x}=\frac{1}{2-x}\)
\(\Rightarrow\)  x = 4-2x
\(\Rightarrow \ x=\frac{4}{3} \mathrm{~m}\)

(a) Owing to therandom thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.
(b) The induced dipole moment in a diamagnetic substance is always opposite to the magnetising field. Hence, the internal motion of the atoms (which is related to the temperature) does not affect the diamagnetism of a material.
(c) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.
(d)The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field and vice versa. 
(e)The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point. 
(f)The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation. 

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(a) (a) Both A and R are true and R is the correct explanation of A

(i) (c) :The mutual inductance ofthe system is \(M=\mu_{0} n_{1} n_{2} \pi a^{2} L\)
(ii) (c): Here \(\phi_{B}=2 \times 10^{-2} \mathrm{~Wb}, I=0.01 \mathrm{~A}\)
As \(\phi_{B}=M I\)
\(\therefore\) Mutual inductance between two coils is \(M=\frac{\phi_{B}}{I}=\frac{2 \times 10^{-2} \mathrm{~Wb}}{0.01 \mathrm{~A}}=2 \mathrm{H}\)
(iii) (b) : Mutual inductance of coils \(M=\frac{\mu_{0} \mu_{r} N_{1} N_{2} A}{l}\)
It is clear that mutual inductance of coils can be increased by increasing the number of turns in the coils.
(iv) (a) : We know that the mutual inductance depends (directly proportional) on the permeability of the medium surrounding the coils. When the permeability of the medium is increased by inserting a sheet of iron, then the mutual inductance between the coils also increases.
(v) (c)

(i) (b)
(ii) (d): Magnetic permeability - Henry m^-1
(iii) (d): Given, L= 3 cm, A = 2 cm², M = 3 A m²
.Intensity of magnetisation \(=\frac{M}{l A}=\frac{3}{3 \times 10^{-2} \times 2 \times 10^{-4}}\)
\(=\frac{1}{2 \times 10^{-6}}=0.5 \times 10^{6}=5 \times 10^{5} \mathrm{~A} / \mathrm{m}\)
(iv) (b): Here, n = 500 turns/m
\(I=1 \mathrm{~A}, \mu_{-}=500\)
Magnetic intensity \(H=n I=500 \mathrm{~m}^{-1} \times 1 \mathrm{~A}=500 \mathrm{~A} \mathrm{~m}^{-1}\)
As \(\mu_{r}=1+\chi \quad \text { or } \chi=\left(\mu_{r}-1\right)\)
Magnetisation, M = XH
\(=\left(\mu_{r}-1\right) H=(500-1) \times 500 \mathrm{~A} \mathrm{~m}^{-1}\)
\(=2.495 \times 10^{5} \mathrm{~A} \mathrm{~m}^{-1} \approx 2.5 \times 10^{5} \mathrm{~A} \mathrm{~m}^{-1}\)
(v) (a): Relative permeability of iron \(\mu_{r}=6000\)
Magnetic susceptibility \(\chi_{m}=\mu_{r}-1=5999\)