(a) Total resistance in the circuit is,
R_G+3 = 63 Ω. Hence, I = 3 / 63 = 0.048 A.
(b) Resistance of the galvanometer converted to an ammeter is,
\(\frac{R_{G} r_{s}}{R_{G}+r_{s}}=\frac{60 \Omega \times 0.02 \Omega}{(60+0.02) \Omega}=0.02 \Omega\)
Total resistance in the circuit is,
0.02Ω + 3Ω = 3.02Ω . Hence, I = 3 / 3.02 = 0.99 A.
(c) For the ideal ammeter with zero resistance,
I = 3 / 3 = 1.00 A

The capacitive reactance is
\(X_{C}=\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}=212 \Omega\)
The rms current is
\(I=\frac{V}{X_{C}}=\frac{220 \mathrm{~V}}{212 \Omega}=1.04 \mathrm{~A}\)
The peak current is
\(i_{m}=\sqrt{2} I=(1.41)(1.04 A)=1.47 A\)
This current oscillates between +1.47A and -1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

(i) When θ = 0; ፔ = O and p and E are parallel and the dipole is in a position of stable equilibrium.
(ii) When θ = 180°, ፔ = 0 and p and E are anti-parallel and the dipole is in a position of unstable equilibrium.

When a capacitor is charged by a battery, work is done by the charging battery at the expense of its chemical energy. 
This work is stored in the capacitor in the form of electrostatic potential energy. Consider a capacitor of capacitance C. Initial charge on capacitor is zero. Initial potential difference between capacitor plates = zero. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases. Let at any instant when charge on capacitor be q, the potential difference between its plates V = q/C. 

Now work done in giving an additional infinitesimal charge dq to capacitor
dW = Vdq = q/C(dq)
The total work done in giving charge from 0 to Q will be equal to the sum of all such infinitesimal works, which may be obtained by integration. Therefore total work
\(\begin{aligned} W & =\int_0^Q V d q=\int_0^Q \frac{q}{C} d q \\ \end{aligned}\)
\(\begin{aligned} =\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q=\frac{1}{C}\left(\frac{Q^2}{2}-\frac{0}{2}\right)=\frac{Q^2}{2 C} \end{aligned}\)
If V is the final potential difference between capacitor plates, then Q =CV
\(\therefore \quad W=\frac{(C V)^2}{2 C}=\frac{1}{2} C V^2=\frac{1}{2} Q V\)
This work is stored as electrostatic potential energy of capacitor i.e.,
Electrostatic potential energy,
U=Q₂/2C = 1/2(CV₂) = 1/2(QV)
When battery is disconnected (i) Energy stored will be decreased. The energy becomes,
\(U=\frac{Q_0^2}{2 C}=\frac{Q_0^2}{2 K C_0}=\frac{U_0}{K}\)
Thus, energy is reduced to1/K times the initial energy.
(ii) In the presence of dielectric the electric field is reduced to E =E/K= 0.

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.

Given
\({ R }_{ 1 }=1\Omega ,{ R }_{ 2 }=2\Omega ,{ R }_{ 3 }=3\Omega \)
(a) Total resistance of series combination
\({ R }_{ s }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }\)
\({ R }_{ s }\) = 1 + 2 + 3 = \(6\Omega \)

(b) Since E = I (R + r)
I = \(\frac { E }{ { R }_{ s }+0 } =\frac { E }{ { R }_{ s } } =2A\)
\({ V }_{ 1 }={ IR }_{ 1 }=2\times 1=2V\)
\({ V }_{ 2 }={ IR }_{ 2 }=2\times 2=4V\)
\({ V }_{ 3 }={ IR }_{ 3 }=2\times 3=6V\)

Energy stored in 6 μF capacitor = E 

\( \therefore \ \frac{1}{2} C_{1} V_{1}^{2}=E \Rightarrow V_{1}^{2}=\frac{2 E}{C_{1}}=\frac{E}{3} \)
\(\because C_{1} \text { and } C_{2} \text { are in parallel, } \therefore V_{1}^{2}=V_{2}^{2}=\frac{E}{3} \)
Now, \(C_{12}=6+12=18 \mu \mathrm{F}\) 
\( \therefore \text { Charge on } C_{12}=q_{12}=C_{12} . V_{12}=18 \times \sqrt{\frac{E}{3}} \)
\(\therefore \text { Charge on } C_{3}=q_{12}=18 \sqrt{\frac{E}{3}} \)
(a) Now, energy stored in 12 μF capacitor
\(=\frac{1}{2} C_{2} V_{1}^{2}=\frac{1}{2} \times 12 \times \frac{E}{3}=2 E\)
(b) Energy stored in 3 μF capacitor
\(=\frac{q_{3}^{2}}{2 C_{3}}=\frac{1}{2} \times\left(\frac{18}{3}\right)^{2} E=18 E\)
\((c) \because C_{e q}=\left(\frac{18 \times 3}{18+3}\right) \mu \mathrm{F}=\frac{18}{7} \mu \mathrm{F}, Q=18 \sqrt{\frac{E}{3}}\)
Energy drawn from the battery,
\(U=\frac{Q^{2}}{2 C_{e q}}=\frac{18 \times 18 \times E \times 7}{3 \times 18}=42 E\)

(a) I_o = 3.23A
(b) 1.55ms

Given, resistance of heating element at temperature 27°C,   
R₂₇ = 100 \(\Omega\)
Resistance of heating element at temperature t°C,
R_t = 117 \(\Omega\)
\(\alpha\)=1.70 \(\times\)10^-4 °C^-1, t = ?
By using the formula of temperature coefficient of resistance,
\(\alpha=\frac{R_2-R_1}{R_1\left(t_2-t_1\right)}\)   ...(i)
Here, R₁ = R_t, R₁ = R₂₇, t₂ = t and t₁ = 27°C
Such that, \(\alpha=\frac{R_t-R_{27}}{R_{27}(t-27)}\)
 Substituting given values in Eq. (i), we get
\(\begin{aligned} 1.70 \times 10^{-4} & =\frac{117-100}{100(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{17}{100 \times 1.70 \times 10^{-4}} \end{aligned}\)
or t = 1000 + 27 = 1027°C 

As we know that, \(e=\frac{d \phi}{d t}\)
As, \(\phi =5{t}^{3}+4{t}^{2}+2t\)
So, \(e=15{t}^{2}+8t+2\)
For \(t=2s, \ e=15\times {2}^{2}+8\times 2+2\)
= 60 + 16 + 2 = 78 V

\(B_{E}=\frac{\mu_{0} m}{4 \pi r^{3}}=\frac{10^{-7} \times 0.4}{(0.5)^{3}}=\frac{10^{-7} \times 0.4}{0.125}=3.2 \times 10^{-7} \mathrm{~T}\)
From Eq \(B_{A}=\frac{\mu_{0} 2 m}{4 \pi r^{3}}=6.4 \times 10^{-7} \mathrm{~T}\)

\(I=2 \sin (100 \pi t+\pi / 3) A\)
Since, the relation between current and time gives us
\( \frac{2 \pi t}{T}=100 \pi t \\ \therefore T=\frac{2 \pi}{100 \pi}=\frac{1}{50} s \)
\(t=\frac{T}{4}=\frac{1}{50 \times 4}=\frac{1}{200} \mathrm{~s}\)

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.

Force on a current carrying conductor in magnetic field, \(F=IlB{ sin }\theta \) |
 (i) Force F will be minimum if \({ sin }\theta =0\) or \(\theta ={ 0 }^{ o }\) or 180^o ,  i.e.,the linear conductor carrying current is parallel or antiparallel to the direction of magnetic field.
(ii) ) Force F will be maximum if \({ sin }\theta ={ 1 }\) or \(\theta ={ 90 }^{ o }\)  ,  i.e.,the linear conductor carrying current is perpendicular to the direction of uniform magnetic field.

Mobility of charge carriers inside conductor is defined as the magnitude of drift velocity of charge per unit electric field applied.
SI unit of mobility is m²s^-1V^-1 or ms^-1N^-1C.

There are three capacitors cach of capacitance 9 pF.
\(\therefore\) C₁ = C₂ = C₃ = 9 pF
and voltage, V = 120 V
(i) The total capcitance in series combination,
\(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\)
\(\Rightarrow \frac{1}{C_{s}}=\frac{3}{9} \Rightarrow C_{s}=3 pF\)
(ii) Let the charge across the system be q and potentials across C₁, C₂ and C₃ be V₁, V₂ and V₃, respectively.
Charge, q = C_s. V = 3 \(\times\)120 = 360 pC
Potential difference across C₁,
\(V_{1}=\frac{q}{C_{1}}=\frac{360}{9}=40 V\)
Potential difference across C₂,
\(V_{2}=\frac{q}{C_{2}}=\frac{360}{9}=40 V\)
Potential difference across C₃,
\(V_{3}=\frac{q}{C_{3}}=\frac{360}{9}=40 V\)
Thus, the potential difference across each capacitor is 40 V.

Considering the mesh BADB, we have
100I₁ + 15I_g - 60I₂ = 0
or 20I₁ + 3I_g - 12I₂ = 0
Considering the mesh BCDB, we have
10 (I₁ - I_g) - 15I_g - 5 (I₂ + I_g) = 0
10I₁ - 30I_g - 5I₂ = 0
2I₁ - 6I_g - I₂ = 0
Considering the mesh ADCEA,
60I₂ + 5 (I₂ + I_g) = 10
65I₂ + 5I_g = 10
13I₂ + I_g = 2
Multiplying by 10
20I₁ - 60I_g - 10I₂ = 0
From we have
63I_g - 2I₂ = 0
I₂ = 31.5I_g
Substituting the value of I₂ into we get
13 (31.5I_g ) + I_g = 2
410.5 I_g = 2
I_g = 4.87 mA

(a) From Eq, the magnetic energy is
\(U_{B}=\frac{1}{2} L I^{2}\)
\(=\frac{1}{2} L\left(\frac{B}{\mu_{0} n}\right)^{2} \ \left(\text { since } B=\mu_{0} n I, \text { for a solenoid }\right)\)
\(=\frac{1}{2}\left(\mu_{0} n^{2} A l\right)\left(\frac{B}{\mu_{0} n}\right)^{2}\)   [from Eq.]
\(=\frac{1}{2{\mu }_{0}}{B}^{2}Al\)
(b) The magnetic energy per unit volume is,
\({U}_{B}=\frac{{U}_{B}}{V}\)  (where V is volume that contains flux)
\(=\frac{{U}_{B}}{Al}\)
\(=\frac{{B}^{2}}{2{\mu }_{0}}\)
We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor.
\({U}_{E}=\frac{1}{2}{\epsilon }_{0}{E}^{2}\)
In both the cases energy is proportional to the square of the field strength. Equation have been derived for special cases: a solenoid and a parallel plate capacitor, respectively. But they are general and valid for any region of space in which a magnetic field or/and an electric field exist.

(a) The field H is dependent of the material of the core, and is
H = nI = 1000 x 2.0 = 2 x 10³ A/m.
(b) The magnetic field B is given by
B = μ_r μ₀ H
= 400 x 4π x 10^-7 (N/A²) x 2 x 10³ (A/m)
= 1.0 T
(c) Magnetisation is given by
M = (B - μ₀ H)/ μ₀
= (μ_r μ₀ H - μ₀ H) / μ₀ = (μ_r – 1)H = 399 x H
\(\cong\)8 \(\times\)10⁵ A/m
d) The magnetising current I_M is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = μ_r n (I + I_M). Using I = 2A, B = 1 T, we get I_M = 794 A.

Let VL, VR, Vc and V represent the voltage across the inductor, resistor, capacitor and the source respectively. VR is parallel to I. Vc is pi/2 behind I and VL is pi/2 ahead of I.
Clearly,
\(={ i }_{ 0 }^{ 2 }[{ R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 }]\)
\({ i }_{ o }=\frac { { V }_{ 0 } }{ \sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } } } \)
\(Impedence=\frac { { V }_{ 0 } }{ { i }_{ 0 } } =\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } } \)
\(\\ =\sqrt { { R }^{ 2 }+\left( \omega L-\frac { 1 }{ \omega C } \right) ^{ 2 } } \)

The capacitance of a capacitor in the tuning circuit is varied such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal to be received. When this happens, the amplitude of the current becomes maximum in the receiving circuit.

From Kirchhoff's junction rule, we have
\( { I }_{ 1 }={ I }+{ I }_{ 2 }\)   ...........(i)
Applying Kirchhoff's loop rule to outer loop containing 10V cell, we get
\(10=IR+{ 10I }_{ 1 }\) ............(ii)
Applying Kirchhoff's loop rule to outer loop containing 2V cell, we get
\(2={ 5I }_{ 2 }-RI\)
\(2=5\left( { I }_{ 1 }-I \right) -RI\)
\(4={ 10I }_{ 1 }-10I-2RI\)
Subtracting eq (ii) from eq (i), we get
\(6=3RI+10I\)
\(2=I\left( R+\frac { 10 }{ 3 } \right) \)
From Ohm's law, we have
\(V=I\left( R+{ R }_{ off } \right) \)
Comparing eq (iii) and (iv), we get
\({ R }_{ ef }=\frac { 10 }{ 3 } \Omega \)
If \({ E }_{ eff }\) and \({ R }_{ eff }\) are the effective voltage and effective internal resistance of the combination, then the equivalent circuit is shown.

(a): According to Faraday's law of electromagnetic induction, induced emf will be generated in the solenoid because of the relative motion between magnet and solenoid.

(c): Consider a conductor of length I and area of cross section A. Time taken by the free electrons to cross the conductor, t = l/v_d.
Hence, current, \(I=\frac{q}{t}=\frac{A l \times n e}{l / v_{d}}\)
or, \(I=\text { Anev }_{d}\)
or, \(I \propto v_{d}\)
Thus current is directly proportional to drift velocity.

(c): Earth is a good conductor of very large size. When some small charge is given to earth, its potential does not change. Hence potential of earth is assumed to be zero. It is just like sea level which does not alter materially when water is added to it or removed from it. Thus, the potential of all other bodies are measured with reference to the earth. For this, if the connection of a charged body to the ground by a metallic conductor would cause electrons to flow to that body from ground, the body is at positive potential. Conversely, is also true. In either case the conductor is neutralized and brought to zero potential. In fact the atmosphere does possess significant electric field.

(i) (a) :Magnetic field due to a long current carrying wire at r
\(B=\frac{\mu_{0}}{2 \pi} \frac{I}{r}\)
(ii) (d): Let I' be the current in region r < R
Then, \(I^{\prime}=\frac{I}{\pi R^{2}} \pi\left(r^{2}\right) \text { or } I^{\prime}=\frac{I r^{2}}{R^{2}}\)
So, magnetic. field \(B=\frac{\mu_{0} I^{\prime}}{2 \pi r}=\frac{\mu_{0} I r^{2}}{2 \pi R^{2} r}=\frac{\mu_{0} I r}{2 \pi R^{2}}\)
(iii) (a): Magnetic field due to a long straight wire of radius a carrying current I at a point distant r from the centre of the wire is given as follows

\(B=\frac{\mu_{0} I r}{2 \pi a^{2}} \quad \text { for } \quad r<a\)
\(B=\frac{\mu_{0} I}{2 \pi a} \quad \text { for } r=a\)
\(B=\frac{\mu_{0} I}{2 \pi r} \quad \text { for } r>a\)
The variation of magnetic field B with distance r from the centre of wire is shown in the figure.
(iv) (d): Let the magnetic fields due to a long straight wire of radius R carrying a steady current I at a distance r from the centre of the wire are
\(B_{1}=\frac{\mu_{0} I r}{2 \pi R^{2}} \quad(\text { For } r<R)\)
and \(B_{2}=\frac{\mu_{0} I}{2 \pi R} \quad(\text { For } r>R)\)
So, the magnetic field at \(r=\frac{R}{2} \text { is } B_{1}=\frac{\mu_{0} I}{2 \pi R^{2}}\left(\frac{R}{2}\right)=\frac{\mu_{0} I}{4 \pi R}\)
and at \(r=2 R \text { is } B_{2}=\frac{\mu_{0} I}{2 \pi(2 R)}=\frac{\mu_{0} I}{4 \pi R}\)
\(\therefore\) Their corresponding ratio is \(\frac{B_{1}}{B_{2}}=\frac{\left(\mu_{0} I / 4 \pi R\right)}{\left(\mu_{0} I / 4 \pi R\right)}=1\)
(v) (c)

(I) (b): \((S+x)=\frac{Q}{P} R\)
\(x=\frac{Q}{P} R-S=\frac{6}{5} \times 10-5=7 \Omega\)
(ii) (a): A Wheatstone bridge consisting of four arms of resistance P, Q, R, S is most sensitive when all the resistances are equal.
(iii) (a) : When metal wire is heated, its resistance increases R₁ increases,L₁ increases.
The null point shift to the right.
(iv) (c): The percentage error in measuring resistance with a metre bridge can be minimized by adjusting the balancing point near the middle of the bridge i.e. close to 50 cm
(v) (c): \(\frac{P}{Q}=\frac{l_{1}}{100-l_{1}} \text { or } \frac{2}{3}=\frac{l_{1}}{100-l_{1}}\)
\(\text { or } \quad 5 l_{1}=200 \text { or } l_{1}=40 \mathrm{~cm}\)