(i) 36 x 10^-2 Am²
(ii) 18 x 10^-2Nm

Charge on \(\alpha \) -particle, q=+2e, where e is the charge of an electron. Thus
q = 2 x 1.6 x 10^-19 C.
r = 2 x 1.6 x 10^-19 C.
r = 2 x 10^-10 m; v = 2 x 10⁶ ms^-1
Current \(I=\frac { q }{ T } =\frac { q }{ 2\pi r/v } =\frac { qv }{ 2\pi r } \) 
Magnetic field at the centre of circular path
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi I }{ 2r } =\frac { { \mu }_{ o }I }{ 2r } =\frac { { \mu }_{ o } }{ 2r } \left( \frac { qv }{ 2\pi r } \right) =\frac { { \mu }_{ o } }{ 4\pi } \frac { qv }{ { r }^{ 2 } }\)
\(=\frac { { 10 }^{ -7 }\times \left( 2\times 1.6\times { 10 }^{ -19 } \right) \times \left( 2\times { 10 }^{ 6 } \right) }{ { \left( 2\times { 10 }^{ -10 } \right) }^{ 2 } } =1.6T\)

Here, \(N=100,\ A=3{ m }^{ 2 },\ \omega =60/sec.\)
\(B=0.04T,\ R=500\Omega ,\ { I }_{ 0 }=?,\ { P }_{ 0 }=?\)
\({ I }_{ 0 }=\frac { { E }_{ 0 } }{ R } =\frac { NAB\omega }{ R } =\frac { 100\times 3\times 0.04\times 60 }{ 500 } \)
\(=1.44A\)
Max. power dissipation \(={ E }_{ \upsilon }{ I }_{ \upsilon }\)
\(=\frac { { E }_{ 0 } }{ \sqrt { 2 } } \frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { { I }_{ 0 }^{ 2 }R }{ 2 } =\frac { { \left( 1.44 \right) }^{ 2 }\times 500 }{ 2 } \)
\(=518.4 \ W\)

Figure shows the variation of I_m with \(\omega \) in a L-C-R series circuit for two values of resistance R₁ and R₂(R₁>R₂).

The condition for resources in the L-C-R circuit is,
X_L = X_C
\({ \omega }_{ 0 }L\ =\ \frac { 1 }{ { \omega }_{ 0 }C } \Rightarrow { \omega }_{ 0 }^{ 2 }=\frac { I }{ LC } \Rightarrow { \omega }_{ 0 }=\frac { 1 }{ \sqrt { LC } } \)
We see that, the current amplitude is maximum at the resonance frequency \({ \omega }_{ 0 }\) .
 Since, \({ I }_{ m }=\frac { { V }_{ m } }{ R } \) ar resonance, the current amplitude for case R₂
 is sharper to that for case R₁.
Quality factor or simply the Q-factor of a resonant L-C-R circuit is defined as the ratio of voltage drop across the capacitor (or inductor) to that of applied voltage.
It is given by \(Q=\frac { 1 }{ R } \sqrt { \frac { L }{ C } } \)
The Q-factor determines the sharpness of the resonance curve and if the resonance is less sharp, not only the maximum current becomes less, but also the circuit will be closed to the resonance for a larger range \(\triangle \omega \) frequencies and the tuning of the circuit will not be good. So, less sharp the resonance, less in the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be loss in energy of the circuit.

Given, L = 100 mH = 100 x 10^-3 H,
C = 2\(\mu\)F = 2 x 10^-6 F, \(\omega\) = 1000 and R = 400 \(\Omega\)
For phase difference,
tan \(\phi={{\left( \omega L-{{1}\over{\omega C}} \right)}\over{R}}\)
[where, \(\phi\) is phase difference between current and voltage]
\(\therefore\)  \(\omega\) = 1000 \(\Rightarrow\) \(\omega\)L = 1000 x 100 x 10^-3 = 100 \(\Omega\)
\({{1}\over{\omega C}}={{1}\over{1000\times2\times{10}^{-6}}}={{1}\over{2\times{10}^{-3}}}=500\Omega\)
\(\Rightarrow\)  \(tan\ \phi={{100-500}\over{400}}={{-400}\over{400}}=-1\)
\(\Rightarrow\)  \(\phi\ ={tan}^{-1}(-1)\Rightarrow\ \phi=135°\)
Since, \(\omega L< {{1}\over{\omega C}}\)  or X_L < X_C
Therefore, current is leading in phase by a phase angle 135°.
For unit power factor cos \(\phi\) = 1
\(\Rightarrow{{R}\over{\sqrt{{R}^{2}+\left(\omega L-{{1}\over{\omega C'}} \right)}}}=1\)
where C' is the total capacitance.
\(\Rightarrow\)  \({R}^{2}+\left( \omega L-{{1}\over{\omega C'}} \right)^{2}={R}^{2}\)
\(\Rightarrow\)  \(\omega L={{1}\over{\omega C'}}\)
\(\Rightarrow\)  \(\omega L=100={{1}\over{\omega C}}={{1}\over{1000}}c'\)
C' = \({{1}\over{{10}^{5}}}={10}^{-5}\) F = \(10 \mu F\)
Additional capacitance C'required in parallel
= \(C'-C=10\mu F-2\mu F=8\mu F\)

(a) Current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it.
∴ Current sensitivity, \(I_{s}=\frac{\theta}{I}=\frac{N B A}{k}\)
Its SI unit is radian/ampere.
(b) For a balanced wheatstone bridge, there will be no deflection in the galvanometer.
Hence, for the first given figure 
\(\frac{4}{R_{1}}=\frac{6}{9} \Rightarrow R_{1}=6 \Omega\)
If the bridge is balanced, then on interchanging the position of the galvanometer and the battery, there is no effect on the balance of the bridge. Hence for the 2nd given figure,
\( \frac{6}{12}=\frac{R_{2}}{8} \)
\(\Rightarrow \ R_{2}=4 \Omega \)
Therefore, the required ratio is calculated as
\(\frac{R_{1}}{R_{2}}=\frac{6}{4}=\frac{3}{2}=3: 2\)

Potential energy of the configuration arises due to potential energy of one dipole (say B), in the magnetic field due to the other (A). Use the result that.
\(B_{A}=\frac{\mu_{0} M_{A}}{4 \pi r^{3}}\) (on the normal bisector)
\(B_{A}=\frac{\mu_{0} 2 M_{A}}{4 \pi r^{3}}\) (on the axis)
Equilibrium is stable when M_B is parallel to BA and unstable when M_B is anti-parallel to B_A.
(a) AB₁ and AB₂
(b) (i) AB₃, AB₆ (stable), (ii) AB₅, AB₄ (unstable)
(c) AB₆ (Maximum negative)

Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection,  I_g = 3 mA = 3 x 10^-3 A
Range of the voltmeter is 0, which needs to be converted to 18 V.
therefore, V = 18 V
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:
\(R=\frac{V}{I_{\mathrm{g}}}-\mathrm{G}\)
\(=\frac{18}{3 \times 10^{-3}}-12=6000-12=5988 \Omega\)
Hence, a resistor of resistance 5998 Ω is to be connected in series with the galvanometer.

Gauss's law correlates surface integral of electric field \(\overset { \rightarrow }{ E } \)  with charge q over a surface [i.e. \(\oint _{ s }^{ }{ \overset { \rightarrow }{ E } .\overset { \rightarrow }{ ds } } =\frac { q }{ { \in }_{ o } } \) ], whereas Ampere's law correlates the line integral of magnetic field \(\overset { \rightarrow }{ B } \)  with current over a closed path [i.e. \(\oint { \overset { \rightarrow }{ B } } .\overset { \rightarrow }{ dl } ={ \mu }_{ o }I\) ].

When an electron is moving along the direction of the magnetic field, then angle θ between velocity vector \(\vec v\) and magnetic field \(\vec B\) is zero degree, i.e.,θ=0o 
Lorentz magnetic force 
 F=−evBsin0o=0.

\(t=\pi / \omega=\frac{1}{2}\left(\frac{2 \pi}{\omega}\right)=\frac{T}{2} \text { and } t=\frac{2 \pi}{\omega}=T\)
\(I_{a v}=-\frac{2 I_0}{\pi}\)

Dia: copper, water, mercury, hydrogen. para: aluminum, oxygen.

\(|\mathrm{dB}|=\frac{\mu_{0}}{4 \pi} \frac{I \mathrm{~d} l \sin \theta}{r^{2}}\)
dl = Δx = 10⁻²m , I = 10 A, r = 0.5 m = y, \(\mu_{0} / 4 \pi=10^{-7} \frac{\mathrm{T} \mathrm{m}}{\mathrm{A}}\) θ = 90° ; sin θ = 1
\(|\mathrm{dB}|=\frac{10^{-7} \times 10 \times 10^{-2}}{25 \times 10^{-2}}=4 \times 10^{-8} \mathrm{~T}\)
The direction of the field is in the +z-direction. This is so since
\(\mathrm{d} \mathbf{l} \times \mathbf{r}=\Delta x \hat{\mathbf{i}} \times y \hat{\mathbf{j}}=y \Delta x(\hat{\mathbf{i}} \times \hat{\mathbf{j}})=y \Delta x \hat{\mathbf{k}}\)
We remind you of the following cyclic property of cross-products
\(\hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}} ; \hat{\mathbf{j}} \times \hat{\mathbf{k}}=\hat{\mathbf{i}} ; \hat{\mathbf{k}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}}\)
Note that the field is small in magnitude.

(i) Decreases From X to Z 
(ii) Decreases From X to Z
In case of  X Quality factor Is More , Impedance, Decreases From X to Z.

Using Eq. we find
r = m v / (qB) = 9 x 10^–31 kg x 3 x 10⁷ m s^–1 / ( 1.6 x 10^–19 C x 6 x 10^–4 T )
= 28 x 10^–2 m = 28 cm
ν = v / (2 \(\pi\)r) = 17 x 10⁶ s^–1 = 17 x 10⁶ Hz = 17 MHz.
E = (½ ) mv² = (½ ) 9 x 10^–31 kg x 9 x 10¹⁴ m²/s² = 40.5 x 10^–17 J
\(\approx \) 4 x 10^–16 J = 2.5 keV.

Power is the rate of doing work. In A.C current and voltage change at every moment. Since power is the product of instantaneous voltage and current, therefore it changes at every moment.
To determine average power in a circuit containing pure resistor, determine total energy spent in one cycle and divide it by the period.
Instantaneous power  \(=EI=\left[ { E }_{ 0 }sin\omega t \right] \left[ { I }_{ 0 }sin\omega t \right] \)
Instantaneous work done in small interval dt is
\( \ dW=EI \ dt={ E }_{ 0 }{ I }_{ 0 }{ sin }^{ 2 }\omega t \ dt\)
\(W=\int _{ 0 }^{ T }{ { E }_{ 0 }{ I }_{ 0 }{ sin }^{ 2 }\omega t \ dt }\)
\(W={ E }_{ 0 }{ I }_{ 0 }\int _{ 0 }^{ T }{ { \left( \frac { 1-cos2\omega t }{ 2 } \right) }dt } \)
\(=\left[ \because \int _{ 0 }^{ T }{ dt } -\int _{ 0 }^{ T }{ cos2\omega t } \right]\)
\(or \ \ W=\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } \left[ T-0 \right] =\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } T\)
\( \left[ \because \int _{ 0 }^{ T }{ dt } -\int _{ 0 }^{ T }{ cos2\omega t } \right] =0\)
\(or \ \ { P }_{ av }=\frac { W }{ T } =\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } =\frac { { E }_{ 0 } }{ \sqrt { 2 } } .\frac { { I }_{ 0 } }{ \sqrt { 2 } } \)
\(\ {P }_{ av }={ E }_{ v }{ I }_{ v }\)
\(\therefore\) Average power over a complete cycle of a.c through a resistor is the product of virtual voltage and virtual current.

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.

Given, current, I = 4A
Number of turns per unit length, n = 600
Relative permeability, \(\mu \)_r = 5000
Since, magnetic intensity, H = nI = 600 x 4 = 2400Am^-1
Since,   \(\mu\)_r = 1 + X_m
\(\Rightarrow\)  X_m = \(\mu\)_r - 1
= 5000 -1 = 4999 \(\approx \) 5000
Here, X_m = magnetic susceptibility.
Intensity of magnetisation can be given as
I = X_mH = 5000 x 2400
= 1.2 x 10⁷ Am^-1
Therefore, magnetic field, B = \(\mu\)_r \(\mu\)₀ H
= 5000 x (4\(\pi\)  x 10^-7) x 2400
= 15 T

Given, R₁ = 10 \(\Omega\), N₁ = 30, A₁ = 3.6 \(\times\)10^-3m²,
B₁ = 0.25 T, R₂ = 14 \(\Omega\), N₂ = 42.
A₂ = 1.8 \(\times\)10^-3 m², B₂ = 0.50 T
k₁ = k₂ (spring constants are smae)    ...(i)
(i) Using the formula of current sensitivity, \(I=\frac{N A B}{k}\)
\(\therefore \quad \frac{I_{S_2}}{I_{S_1}}=\frac{N_2 B_2 A_2 k_1}{N_1 B_1 A_1 k_2}=\frac{42 \times 0.50 \times 1.8 \times 10^{-3}}{30 \times 0.25 \times 3.6 \times 10^{-3}}\)
= 1.4     [from Eq. (i)]
(ii) Using the formula of voltage sensitivity,
\(\begin{aligned} V & =\frac{N A B}{k R} \end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{V_{S_2}}{V_{S_1}} & =\frac{N_2 B_2 A_2 k_1 R_1}{k_2 R_2 N_1 B_1 A_1} \\ \end{aligned}\)
\(\begin{aligned} =\frac{42 \times 0.50 \times 1.8 \times 10^{-3} \times 10}{14 \times 30 \times 0.25 \times 3.6 \times 10^{-3}} \end{aligned}\)
= 1             [from Eq. (i)]

(i) (d): A galvanometer can be converted into a voltmeter of given range by connecting a suitable high resistance R in series of galvanometer, which is given by 
\(R=\frac{V}{I_{g}}-G=\frac{100}{10 \times 10^{-3}}-25=10000-25=9975 \Omega\)
(ii) (c): An ideal voltmeter should have a very high resistance.
(iii) (c): Resistance of voltmeter \(=\frac{25}{25 \times 10^{-3}}=1000 \Omega\)
\(\therefore \quad X=1000-10=990 \Omega\)
(iv) (d): To convert a moving coil galvanometer into a voltmeter, it is connected with a high resistance in series. The voltmeter is connected in parallel to measure the potential difference. As the resistance is high, the voltmeter itself does not consume current.
(v) (d): The resistance of an ideal voltmeter is infinity.

(I) (b): Inductance, L = 80 mH = 80 x 10^-3 H
Capacitance, C = \(60 \mu \mathrm{F}=60 \times 10^{-6} \mathrm{~F}, V=230 \mathrm{~V}\)
Frequency, u = 50 Hz
\(\omega=2 \pi v=100 \pi \mathrm{rad} \mathrm{s}^{-1}\)
Peak voltage \(V_{0}=V \sqrt{2}=230 \sqrt{2} \mathrm{~V}\)
Maximum current is given by, \(I_{0}=\frac{V_{0}}{\left(\omega L-\frac{1}{\omega C}\right)}\)
\(I_{0}=\frac{230 \sqrt{2}}{\left(100 \pi \times 80 \times 10^{-3}-\frac{1}{100 \pi \times 60 \times 10^{-6}}\right)}\)
Amplitude of maximum current, I_o = 11.63 A
(ii) (c): rms value of current,
\(I=\frac{I_{0}}{\sqrt{2}}=\frac{-11.63}{\sqrt{2}}=-8.23 \mathrm{~A}\)
Negative sign appears as \(\omega L<\frac{1}{\omega C}\)
(iii) (a): Average power consumed by the inductor is zero because of phase difference of \(\frac{\pi}{2}\) between voltage and current through inductor.
(iv) (b): Average power consumed by the capacitor is zero because of phase difference of \(\frac{\pi}{2}\) between voltage and current through capacitor
(v) (a)

(a): The magnetic field at the centre of circular coil is given by
\(B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi n I}{a}\)
So if current through coil is doubled then magnetic field is \(B^{\prime}=2 B\)
The magnetic field also get doubled. The magnetic field is directly proportional to the current in conductor

(c): The transmission is done at high voltage due to which current through the wire is reduced By reduction in current corresponding dissipation of energy is also reduced \(\left(\text { as } H \propto I^{2} R\right)\) If transmissioh is done at low voltage then we have to use thick wire-in order to reduce the dissipation of energy. This increase the cost of transmission lines wires. In order to reduce both energy dissipation and cost of transmission wire, transmission is done at high voltage by using step-up transformers.

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