The conductivity of a metallic wire is defined as the degree to which a specified material conduct electricity calculated as the ratio of current density in the material to the electric field which causes the flow of current,

$$\begin{gathered} R = \frac{\rho \mathcal{l}}{A} \\ \rho = \frac{RA}{\mathcal{l}} \\ \frac{1}{\rho } = \frac{\mathcal{l}}{R. A} \\ \sigma = \frac{\mathcal{l}}{R. A} \end{gathered}$$

The SI unit of conductivity is ohm^-1 metre^-1 or mho metre^-1 or siemen metre^-1

$$\begin{gathered} \left|{\overrightarrow{\mathrm{F}}}_1\right| = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(4\mathrm{q}\right)\left(\mathrm{q}\right)}{{\mathcal{l}}^2} \\ {\mathrm{F}}_1 = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(4{\mathrm{q}}^2\right)}{{\mathcal{l}}^2} \\ {\mathrm{F}}_1 = \frac{1}{{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{{\mathcal{l}}^2} \\ {\overrightarrow{\mathrm{F}}}_2| =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(2\mathrm{q}\right)\left(\mathrm{q}\right)}{{\mathcal{l}}^2} \\ {\mathrm{F}}_2 = \frac{1}{2{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{{\mathcal{l}}^2} \\ \mathrm{angle} \mathrm{between} \overrightarrow{{\mathrm{F}}_1} \mathrm{and} \overrightarrow{{\mathrm{F}}_2} \mathrm{is} {120}^{\mathrm{o}} \\ \overrightarrow{\mathrm{F}} = \sqrt{{\mathrm{F}}_1^1 + {\mathrm{F}}_2^2 + 2{\mathrm{F}}_1{\mathrm{F}}_2 \cos {120}^{\mathrm{o}}} \\ {\mathrm{F}}_1 = 2{\mathrm{F}}_2 \\ \mathrm{F} = \sqrt{(2{\mathrm{F}}_2{)}^2 + {\mathrm{F}}_2^2 + 4{\mathrm{F}}_2^2 \cos {120}^0} \\ \mathrm{F} = \sqrt{4{\mathrm{F}}_2^2 + {\mathrm{F}}_2^2 -2{\mathrm{F}}_2^2} \\ \mathrm{F} = \sqrt{3{\mathrm{F}}_2^2} \\ \mathrm{F} = \sqrt{3{\mathrm{F}}_2^2} \\ \mathrm{F} = \sqrt{3}\frac{1}{2{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{{\mathcal{l}}^2} \end{gathered}$$

(b) The amount of work done to separate the charges at infinity will be equal to potential energy.

$$\begin{gathered} \mathrm{U} = \frac{1}{4{\mathrm{\pi \epsilon }}_0 \mathcal{l}}[\mathrm{q} \mathrm{x} (-4\mathrm{q}) + (\mathrm{q} \mathrm{x} 2\mathrm{q}) + (-4\mathrm{q} \mathrm{x} 2\mathrm{q}) \\ \mathrm{U} = \frac{1}{4{\mathrm{\pi \epsilon }}_0 \mathcal{l}} [-4{\mathrm{q}}^2 + 2{\mathrm{q}}^2 -8{\mathrm{q}}^2] \\ \mathrm{U} = \frac{1}{4{\mathrm{\pi \epsilon }}_0 \mathcal{l}} [-10{\mathrm{q}}^2] \\ \mathrm{U} = -\frac{1}{4{\mathrm{\pi \epsilon }}_0 \mathcal{l}} [10{\mathrm{q}}^2 \mathrm{unit} \end{gathered}$$

Potentiometer at open circuit ℓ₁ = 350

R = 9

ℓ₂ = 300

$$\begin{gathered} \mathrm{r} = \mathrm{R} \left(\frac{{\mathrm{I}}_1}{{\mathrm{l}}_2}- 1\right) \\ \mathrm{r} = 9\left(\frac{350}{300}-1\right) \\ = 1.5 \mathrm{\Omega } \end{gathered}$$

There will be two magnetic field one due to current and another due to magnetics.

$$\begin{gathered} \mathrm{B} = {\mathrm{B}}_0 + {\mathrm{B}}_{\mathrm{m}} \\ \mathrm{B} = {\mathrm{\mu }}_0\mathrm{nI} + {\mathrm{\mu }}_0\mathrm{M} \\ \mathrm{B} = {\mathrm{\mu }}_0\mathrm{H} + {\mathrm{\mu }}_0\mathrm{\chi H} \\ \mathrm{B} = {\mathrm{\mu }}_0(1 + \mathrm{\chi })\mathrm{H} \\ \mathrm{B} = {\mathrm{\mu }}_0{\mathrm{\mu }}_{\mathrm{r}}\mathrm{H} \end{gathered}$$

The positive terminals are connected to the offsite ends of the resistor, they send the current in opposite directions.

Hence, the net emf = 200-10 = 190 V

∴ current in the circuit I = E/R

= 190V / 38 Ω

= 5 A

$$\begin{gathered} {\mathrm{F}}_1 = \frac{1}{4{\mathrm{\pi \epsilon }}_0} = \frac{\mathrm{qQ}}{{\mathrm{a}}^2} \\ {\mathrm{F}}_2 = \frac{1}{4{\mathrm{\pi \epsilon }}_0} \frac{\mathrm{qQ}}{{\mathrm{a}}^2} \\ {\mathrm{F}}_3 = \frac{1}{4{\mathrm{\pi \epsilon }}_0} \frac{\mathrm{QQ}}{(\sqrt{2\mathrm{a}{)}^2}} = \frac{1}{4{\mathrm{\pi \epsilon }}_0} \frac{{\mathrm{Q}}^2}{2{\mathrm{a}}^2} \end{gathered}$$

F₁ and F₂ are perpendicular to each other so their resultant will be

$$\begin{gathered} \mathrm{F}\text{'} = \sqrt{{\mathrm{F}}_1^2 + {\mathrm{F}}_2^2 + 2{\mathrm{F}}_1{\mathrm{F}}_2 \cos {90}^{\mathrm{o}}} \\ {\mathrm{F}}_1 = {\mathrm{F}}_2 \\ \mathrm{F}\text{'} = \sqrt{2}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qQ}}{{\mathrm{a}}^2} \end{gathered}$$

F₃ and resultant of F₁ and F₂ will be in the same direction

Net force

F = F' + F₃

$$\begin{gathered} \mathrm{F} = \frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0} \left[\frac{\sqrt{2\mathrm{q}}}{{\mathrm{a}}^2} + \frac{\mathrm{Q}}{2{\mathrm{a}}^2}\right] \\ = \frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{2\sqrt{2\mathrm{q}} + \mathrm{Q}}{2{\mathrm{a}}^2}\right] \\ \mathrm{F} = \frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0} \left[\frac{2\sqrt{2\mathrm{q}} + \mathrm{Q}}{2{\mathrm{a}}^2}\right] \end{gathered}$$

(b) The potential energy of the given system,

$$\begin{gathered} \mathrm{W} = \frac{4\mathrm{KQq}}{\mathrm{a}} +\frac{{\mathrm{Kq}}^2}{\sqrt{2} \mathrm{a}} + \frac{{\mathrm{KQ}}^2}{\sqrt{2} \mathrm{a}} \\ \mathrm{K} = \frac{1}{4{\mathrm{\pi \epsilon }}_0} \end{gathered}$$

Let ratio be x

R_p = x (Resistance of bulb P)

R_q = 2x (Resistance of bulb Q)

We know, P = VI = V.V/R

$$\begin{gathered} \frac{{\mathrm{P}}_{\mathrm{p}}}{{\mathrm{P}}_{\mathrm{q}}} = \frac{{\mathrm{V}}_2}{{\mathrm{R}}_{\mathrm{p}}} \mathrm{x} \frac{{\mathrm{R}}_{\mathrm{q}}}{{\mathrm{V}}_2} \\ (\mathrm{In} \mathrm{series} \mathrm{Potential} \mathrm{will} \mathrm{be} \mathrm{same}) \\ \frac{{\displaystyle {\mathrm{P}}_{\mathrm{p}}}}{{\displaystyle {\mathrm{P}}_{\mathrm{q}}}} =\frac{2\mathrm{x}}{\mathrm{x}} \\ 2:1 \end{gathered}$$

M = 6J/T

θ = 60^o

B = 0.44T

$$\begin{gathered} \mathrm{\tau } = \mathrm{mB} \mathrm{sin\theta } \\ \mathrm{\tau } = 6 \mathrm{x} 0.44 \sin {60}^{\mathrm{o}} \\ = 6 \mathrm{x} 0.44 \mathrm{x} \frac{\sqrt{3}}{2} \\ = 3\sqrt{3} \mathrm{x} 0.44 \\ = 2.836 \\ \mathrm{dw}{\int }_{\mathrm{\theta } ={60}^0}^{{90}^0}\mathrm{\tau }.\mathrm{d\theta } \\ = - \mathrm{mB} [ \cos {90}^0 - \cos {60}^0] \\ = 6 \mathrm{x} 0.44 \mathrm{x} \left[-\frac{1}{2}\right] \\ = 3 \mathrm{x} 0.44 = 1.32 \mathrm{J} \\ \left(\mathrm{ii}\right) \mathrm{dw} = {\int }_{\mathrm{\theta } = {60}^0}^{{180}^0} \mathrm{mB} \sin \mathrm{\theta }.\mathrm{d\theta } \\ = - \mathrm{mB} [ \cos {180}^0 - \cos {60}^0] \\ = 6 \mathrm{x} 0.44 \mathrm{x} \left[-1-\frac{1}{2}\right] \\ = -6\mathrm{x}0.44\left[-\frac{3}{2}\right] \\ = 9 \mathrm{x} 0.44 \\ = 39.6 \mathrm{J} \\ \left(\mathrm{b}\right) \overrightarrow{\mathrm{\tau }} = \overrightarrow{\mathrm{m}} \mathrm{x} \overrightarrow{\mathrm{B}} \\ \mathrm{\tau } = \mathrm{m} \mathrm{Bsin} \mathrm{\theta } \\ = 6 \mathrm{x} 0.44 \mathrm{x} \sin {180}^0 \\ \mathrm{\tau } = 0 \end{gathered}$$

The susceptibility of this material is between 0 and 1 so its a paramagnetic material.

As we know that,

$$\begin{gathered} \mathrm{R} = \frac{\mathrm{\rho }\mathcal{l}}{\mathrm{A}} \\ \mathrm{V} = \mathrm{IR} \\ \mathrm{V} = \frac{\mathrm{I\rho }\mathcal{l}}{\mathrm{A}} \\ \frac{\mathrm{I}}{\mathrm{A}} = \mathrm{j} = \mathrm{current} \mathrm{density} \\ \mathrm{V} = \mathrm{j\rho }\mathcal{l} \\ \mathrm{V} = \mathrm{E}.\mathcal{l} \\ \mathrm{E}.\mathcal{l} = \mathrm{j\rho }\mathcal{l} \\ \mathrm{E} = \mathrm{j\rho } \\ \mathrm{j} =\frac{\mathrm{E}}{\mathrm{\rho }} \\ \frac{1}{\mathrm{\rho }} = \mathrm{\sigma } \\ \mathrm{j} =\mathrm{\sigma E} \\ \mathrm{V} = \mathrm{u} +\mathrm{at} \\ \mathrm{a} = \frac{\mathrm{F}}{\mathrm{m}} \\ \mathrm{a} = \frac{-\mathrm{eE}}{\mathrm{m}} \\ \left|{\mathrm{V}}_{\mathrm{d}}\right| = \frac{\mathrm{eE}}{\mathrm{m}}\mathrm{\tau } \\ \mathrm{I}.∆\mathrm{t} = \mathrm{neA}\left({\mathrm{V}}_{\mathrm{d}}\right)∆\mathrm{t} \\ \mathrm{j} = \mathrm{i}/\mathrm{A} \\ \mathrm{j} =\mathrm{neA}\frac{\mathrm{eE}}{\mathrm{m}}\mathrm{\tau } \\ \mathrm{j} =\frac{{\mathrm{ne}}^2\mathrm{\tau }}{\mathrm{m}}\mathrm{E} \end{gathered}$$