(i) Torque acting on the loop is given by
\(\tau=M B \sin \theta\)
As the angle between the magnetic field vector and the dipole moment vector is zero.
\(\tau=M B \sin (0)=0 \mathrm{Nm}\)
(ii) Magnitude of force is given by
\(F=\frac{2 \mu_{0} I_{1} I_{2} l}{4 \pi}\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right]\)
where \( l=10 \times 10^{-2} \mathrm{~m}, I_{1}=2 \mathrm{~A}, I_{2}=5 \mathrm{~A} r_{1}=1 \times 10^{-2} \mathrm{~m}, r_{2}=5 \times 10^{-2} \mathrm{~m} \)
\(F=\left[2 \times 10^{-7} \times 2 \times 5 \times 10^{-1}\right] \quad\left[\frac{1}{10^{-2}}-\frac{1}{5 \times 10^{-2}}\right] \)
\( F =20 \times 10^{-8}\left[1-\frac{1}{5}\right] \times \frac{1}{10^{-2}} \)
\(=\frac{20 \times 10^{-6} \times 4}{5}=16 \times 10^{-6} \mathrm{~N}\)
The net force is attractive because the arm of the loop carrying current in the same direction as the Direction of current in the wire is nearer.

Let us consider a circular loop of radius a with centre C. Let the plane of the coil be
perpendicular to the plane of the paper and current I be flowing in the direction as shown in the figure. Suppose P is any point on the axis at a direction from the centre.

Now, consider a current element Idl on top (L) where current comes out of paper normally, whereas at bottom (M) enters into the plane of paper normally.
\(\because LP \bot Idl\)
Also \(MP \bot Idl\)
\(\because LP=MP=\sqrt{r^{2}+a^{2}}\)
The magnetic field at point P due to current element Idl. According to Biot-Savart's law,
dB = \(\frac{\mu_0}{4\pi}.\frac{Idl sin 90^{0}}{(r^{2}+a^{2})}\)
where, a = radius of circular loop,
r = distance of point P from centre along the axis. The direction of dB is perpendicular to LP and along PQ, where \(PQ \bot LP\)  Similarly, the same magnitude of magnetic field is obtained due to current element Idl at the bottom and direction is along PQ', where \(PQ^{'} \bot MP\).
Now, resolving dB due to current clement at Land M dB cos \(\phi\) components balance each other and net magnetic field is given by
B = \(\oint dB sin \phi = \oint \frac{\mu_0 }{4\pi} (\frac{Idl}{r{2}+a^{2}}).\frac{a}{\sqrt{r^{2}+a^{2}}}\)
\([\therefore In \Delta PCL, sin \phi = \frac{a}{\sqrt{r^{2}+a^{2}}}]\)
\(=\frac{\mu_0}{4\pi} \frac{Ia}{(r^{2}+a^{2})^{3/2}} \oint dl=\frac{\mu_0}{4\pi}\frac{Ia}{(r^{2}+a^{2})^{3/2}}(2\pi a)\)
or  \(B=\frac{\mu_0Ia^{2}}{2(r^{2}+a^{2})^{3/2}}\)
For N turns, B = \(\frac{\mu_0 N Ia^{2}}{2(r^{2}+a^{2})^{3/2}}\) Tesla.
The diagram of magnetic field lines due to a circular wire carrying current I is.

(i) Magnetic field due to straight segments is
\(B=\int { \frac { \mu _{ \circ } }{ 4\pi } } .\frac { Id1\times r }{ { r }^{ 3 } } \)

For point O, dI and r for each element of straight segments PQ and RS are parallel.
Therefore, dI x r = 0
Thus, magnetic field due to straight segments is zero.
(ii) Magnetic field at center O due to semi-circular arc
\(=\frac { Magnetic\ field\ at\ center\ of \ circular\ coil }{ 2 } \)
 \(=\frac{1}{2}\left(\frac{\mu_0 I}{2 r}\right)=\frac{\mu_0 I}{4 r}=\frac{\left(4 \pi \times 10^{-7}\right) \times 10}{4 \times 2 \times 10^{-2}}\)
[Given, I = 10 A and r = 2.0 cm = \(2\times { 10 }^{ -2 }\)m]
\(=5\pi \times { 10 }^{ -5 }T\)

Capacitance, C = 30μF = 30 × 10⁻⁶F
Inductance, L = 27 mH = 27 × 10⁻³ H
Angular frequency is given as:

\({ \omega }_{ r }=\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 27\times { 10 }^{ -3 }\times 30\times { 10 }^{ -6 } } } =\frac { { 10 }^{ 4 } }{ 9 } =1.1\times { 10 }^{ 3 }rad/s\)
Hence, the angular frequency of free oscillations of the circuit is 1.11 × 10³ rad/s.

44 kV, 0.01 amp.

\(Given \ r=0.1m,I=5A.\)
\(\\ \therefore \ d B=\frac { { \mu }_{ 0 }I }{ 2\pi r } =\frac { 4\pi \times { 10 }^{ -7 }\times 5 }{ 2\pi \times 0.1 }\)
\(={ 10 }^{ -5 }J\)

I = 0.3 A

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) The force is due to the interaction between magnetic fields due to currents in two linear parallel conductors.
(ii) Here, \({ l }_{ 1 }=2cm=2\times { 10 }^{ -2 }m;{ l }_{ 2 }=20\times { 10 }^{ -2 }m;{ I }_{ 1 }=2.0A;{ I }_{ 2 }=5.0A;r=4\times { 10 }^{ -2 }m\)
Since action and reaction are equal and opposite so the magnitude of repulsive force on 20 cm long wire = the magnitude of repulsive force on 2 cm long wire
= \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2{ I }_{ 1 }{ I }_{ 2 } }{ r } { l }_{ 1 }={ 10 }^{ -7 }\times \frac { 2\times 2\times 5\times (2\times { 10 }^{ -2 }) }{ 4\times { 10 }^{ -2 } }\)
\( ={ 10 }^{ -6 }\)
This study shows that when the current through two parallel conductors flow in the same direction, they attract each other and vice versa. It implies that when the thoughts and actions of two business partners are aligned in the same direction, their behavior is cohesive and they succeed. If there thoughts and actions are opposing, the partnership is likely to collapse. The same thing is true for husband and wife. For a successful family life, the coherence of thoughts and action is a must. 

Resistance of two line wires, R = 20\(\times\)2\(\times\)0.4 = 16\(\Omega \)
For the transformer, E_p = 3000 V, E_s = 220 V and P = 600 kW = 6\(\times\)10⁵ W 
\(\therefore \) Current carried by line wires,  \({ I }_{ v }=\frac { P }{ { E }_{ p } } =\frac { 6\times { 10 }^{ 5 } }{ 3000 } =200A\)
Power loss in the form of heat = \({ I }_{ v }^{ 2 }\times R=\left( 200 \right) ^{ 2 }\times 16=640000W=640kW\)
\(Power\ supplied \ by\ plant\ =\ power\ demand\ +\ power\ loss\)
\(=\ 600+640=1240kW\)

(a) 1.0 J, yes
(b) \(\omega \) = 10³ rad s^-1, v = 159 Hz
(c) 1.0 J

The quality factor (Q) of resonance in series L-C-R circuit is defined as the ratio of voltage drop across inductor (or capacitor) to the applied voltage,
\(\text { i.e., } \mathrm{Q}=\frac{V_L}{V_R}=\frac{I_0 X_L}{I_0 R}=\frac{\omega_0 L}{R}=\frac{I}{\omega_0 C R}\)
It is an indicator of sharpness of the resonance. Quality factor has no unit.

\(\text { Average power, } \vec{P}=V_{r m s} \times i_{r m s} \times \cos \phi\)
\( \text { Where } \phi=90^{\circ} \\ \therefore \vec{P}=V_{r m s} \times i_{r m s} \times \cos 90^{\circ} \\ =0 \)

From the graph, I₁ = 1 A, I₂ = -2 A and I₃ = 1A
\(\begin{aligned} I_{\mathrm{rms}} & =\sqrt{\frac{I_1^2+I_2^2+I_3^2}{3}}=\sqrt{\frac{1^2+(-2)^2+1^2}{3}} \\ \end{aligned}\)
\(\begin{aligned} =\sqrt{\frac{6}{3}}=\sqrt{2}=1.414 \mathrm{~A} \end{aligned}\)

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

\({ M }_{ 1 }=I\left( { \pi r }^{ 2 } \right) \) 
\({ M }_{ 2 }=\frac { I }{ 2 } \left( { \pi 4r }^{ 2 } \right) =I\left( { 2\pi r }^{ 2 } \right) \quad \therefore \frac { { M }_{ 1 } }{ { M }_{ 2 } } =\frac { 1 }{ 2 } \) 
The direction of \(\overset { \rightarrow }{ { M }_{ 1 } } \) is opposite to the direction of \(\overset { \rightarrow }{ { M }_{ 2 } } \)

Fraction of current passing through galvanometer 
\(=\frac { { i }_{ g } }{ i } =\frac { S }{ G+S } =\frac { 5 }{ 50+5 } =\frac { 1 }{ 11 } \) 
Fraction of current passing through shunt
\(=\frac { { i }_{ s } }{ i } =\frac { i-{ i }_{ g } }{ i } =1-\frac { { i }_{ g } }{ i } =1-\frac { 1 }{ 11 } =\frac { 10 }{ 11 } \)

The force on moving charged particle in a magnetic field, \(\overset { \rightarrow }{ F } =q\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) \)  is always perpendicular to \(\overset { \rightarrow }{ v } \) and \(\overset { \rightarrow }{ B }.\) Thus work done = FS cos 90^o = 0.
This means that a uniform magnetic field can do not work on charged particle although it can change its direction of motion. Hence the kinetic energy of charged particle in a magnetic field remains unchanged.

(i) The phase difference berween E and I is \(\frac{\pi}{2}\). the voltage E leads the current I by phase angle \(\frac{\pi}{2}\).
(ii) The phasor diagram is given by

(iii) L = 3 H, E_rms = 150 V, f = 50 Hz
\(I_0=\frac{E_0}{X_L}=\frac{\sqrt{2} E_{\mathrm{mms}}}{2 \pi f L}=\frac{\sqrt{2} \times 150}{2 \pi \times 50 \times 3}=0.225 \mathrm{~A}\)
(iv) f = 200 Hz, L = 15 mH = 15 \(\times\)10^-3 H
\(\therefore\) Inductive reactance, X_L = \(\omega\)L = 2\(\pi\)fL
= 2 \(\pi\) \(\times\)200 \(\times\)15 \(\times\)10^-3 = 18.84 \(\Omega\)

(i) (a): For stationary electron, \(\vec{v}=0\)
\(\therefore\) Force on the electron is \(\vec{F}_{m}=-e(\vec{v} \times \vec{B})=0\)
(ii) (d): Force on the proton \(\vec{F}_{B}=e(\vec{v} \times \vec{B})\)
Since, \(\vec{v}\) is parallel to \(\vec{B}\)
\(\therefore \quad \vec{F}_{B} \doteq 0\)
Hence proton will continue to move with velocity v along the axis of solenoid.
(iii) (b): Magnetic force on the charged particle q is 
\(\vec{F}_{m}=q(\vec{v} \times \vec{B}) \text { or } F_{m}=q v B \sin \theta\)
where \(\theta\) is the angle between \(\vec{v} \text { and } \vec{B}\)
Out of the given cases, only in case (b) it will experience the force while in other cases it will experience no force
(iv) (a) : \(\vec{F}=q(\vec{v} \times \vec{B})\)
\(=q[(2 \hat{i} \times(\hat{i}+2 \hat{j}+3 \hat{k})]=(4 q) \hat{k}-(6 q) \hat{j}\)
(v) (c): When an electric charge is moving both electric and magnetic fields are produced, whereas a static charge produces only electric field.

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

(a): Transformer works on the principle of mutual induction i.e., if two coils are inductively coupled and when current or magnetic flux is changed through one of the two coils, then induced e.m.f. is produced in the other coil. So whenever there is change in current or magnetic flux, only then e.m.f. is induced. But in case of D.C. current or voltage, e.m.f. is not induced because it remain constant throughout and never changes its direction and magnitude. Therefore transformer cannot work when D.C. is applied.

(c): When charged particle enters the uniform field they makes angle \(\theta\) with the field. Then its path is decided by combined effect of two component of velocity. \(v \cos \theta\) parallel to the field. Due to the parallel
field the charge will follow a linear path and due to the perpendicular component \((v \sin \theta)\) of the field will be circular. This results in a helical path whose axis is parallel to the parallel component of the field

(b): The wires are parallel to each other but the direction of current in it is in same direction so they attract each other. If the current in the wires is in opposite direction then wires repel each other. When the currents are in opposite directions, the magnetic forces are reversed and the wires repels each other