In stable equilibrium, the dipole moment vector and the magnetic field vector are in same direction.

It is given that a charged particle oscillates about its mean equilibrium position and the frequency of charged particle is 10⁹  Hz.
The frequency of an electromagnetic wave produced by the oscillator will be same to the value of frequency of a charged particle that is oscillating about its mean position.
Thus, the value of the frequency of an electromagnetic wave produced by oscillator is 10⁹ Hz.

The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v x B is along the z-axis (screw rule or right-hand thumb rule). So,
(a) for electron it will be along –z axis.
(b) for a positive charge (proton) the force is along + z axis.

Self-inductance is the property of a coil by virtue of which, the coil opposes any change in the strength of current flowing through it by inducing an emf in itself. Its SI unit is henry (H).

(i) Conductor
(ii) Semiconductor
In conductors, average relaxation time decreases with increase in temperature, resulting in an increase in resistivity.
In semiconductors, the increase in number density (with increase in temperature) is more than the decrease in relaxation time; the net result is, therefore, a decrease in resistivity.

\(Here, \ l=10cm=0.1 \ m; \ M={ 10 }^{ 6 } \ A/m\)
\(As \ M=\frac { { I }_{ M }(magnetisation \ current) }{ l(length) } \)
\(\\ { I }_{ M }=M\times l={ 10 }^{ 6 }\times 0.1={ 10 }^{ 5 }\)

Given,
Capacitance of the capacitor, C = 12pF = 12 x 10^-12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
\(\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 12 \times 10^{-12} \mathrm{\times}(50)^{2} \mathrm{~J}=1.5 \times 10^{-8} \mathrm{~J}\)
Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10^-8 J. was disconnected.

Magnetic field strength, B = 6.5 x 10⁻⁴ T
Charge of the electron, e = 1.6 x 10⁻¹⁹ C
Mass of the electron, m_e = 9.1 x 10⁻³¹ kg
Velocity of the electron, v = 4.8 x 10⁶ m/s
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = ν
Angular frequency of the electron = ω = 2πν
Velocity of the electron is related to the angular frequency as:
v = rω
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
\(e v B=\frac{m v^{2}}{r}\)
\(e B=\frac{m}{r}(r \omega)=\frac{m}{r}(2 \pi r v)\)
\(v=\frac{B e}{2 \pi m}\)
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as:
\(V=\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}}\)
= 18.2 x 10⁶ Hz
\(\approx\) 18 MHz
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

Current flowing through various branches of the circuit is represented in the given figure.

I₁ = Current flowing through the outer circuit
I₂ = Current flowing through branch AB
I₃ = Current flowing through branch AD
I₂ - I₄ = Current flowing through branch BC
I₃ + I₄ = Current flowing through branch CD
I₄ = Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
10I₂ + 5I₄ - 5I₃ = 0
2I₂ + I₄ - I₃ = 0
I₃ = 2I₂ + I₄ … (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I₂ - I₄) - 10(I₃ + I₄) - 5I₄ = 0
5I₂ + 5I_{4 -} 10I₃ - 10I₄ - 5I₄ = 0
5I₂ - 10I₃ - 20I₄ = 0
I₂ = 2I₃ + 4I₄ … (2)
For the closed circuit ABCFEA, potential is zero i.e.,
-10 + 10 (I₁) + 10(I₂) + 5(I₂ - I₄) = 0
10 = 15I₂ + 10I₁ - 5I₄
3I₂ + 2I₁ - I₄ = 2 … (3)
From equations (1) and (2), we obtain
I₃ = 2(2I₃ + 4I₄) + I₄
I₃ = 4I₃ + 8I₄ + I4
- 3I₃ = 9I₄
- 3I₄ = + I₃ … (4)
Putting equation (4) in equation (1), we obtain
I₃ = 2I₂ + I₄
- 4I₄ = 2I₂
I₂ = - 2I₄ … (5)
It is evident from the given figure that,
I₁ = I₃ + I₂ … (6)
Putting equation (6) in equation (1), we obtain
3I₂ +2(I₃ + I₂) - I₄ = 2
5I₂ + 2I₃ - I₄ = 2 … (7)
Putting equations (4) and (5) in equation (7), we obtain
5(- 2 I₄) + 2(- 3 I₄) - I₄ = 2
- 10I₄ - 6I₄ - I₄ = 2
17I₄ = - 2
\(I_{4}=-\frac{2}{17} A\)
Equation (4) reduces to
I₃ = - 3(I₄)
\(=-3\left(-\frac{2}{17}\right)=\frac{6}{17} A\)
\(I_{2}=-2\left(I_{4}\right)\)
\(=-2\left(-\frac{2}{17}\right)=\frac{4}{17} A\)
\(I_{2}-I_{4}=\frac{4}{17}-\left(-\frac{2}{17}\right)=\frac{6}{17} A\)
\(I_{3}+I_{4}=\frac{6}{17}+\left(-\frac{2}{17}\right)=\frac{4}{17} A\)
\(I_{1}=I_{3}+I_{2}\)
\(=\frac{6}{17}+\frac{4}{17}=\frac{10}{17} A\)
Therefore, current in branch \(A B=\frac{4}{17} A\)
In branch BC \(=\frac{6}{17} A\)
In branch CD = \(-\frac{4}{17} A\)
In branch AD = \(\frac{6}{17} A\)
In branch BD = \(\left(-\frac{2}{17}\right) A\)
Total current = \(\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17} A\)

(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps:
(i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.
(ii) Work needed to bring –q to B when +q is at A. This is given by (charge at B) x (electrostatic potential at B due to charge +q at A)
\(=-q \times\left(\frac{q}{4 \pi \varepsilon_{0} d}\right)=-\frac{q^{2}}{4 \pi \varepsilon_{0} d}\)
(iii) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) x (potential at C due to charges at A and B)
\(=+q\left(\frac{+q}{4 \pi \varepsilon_{0} d \sqrt{2}}+\frac{-q}{4 \pi \varepsilon_{0} d}\right)\)
\(=\frac{-q^{2}}{4 \pi \varepsilon_{0} d}\left(1-\frac{1}{\sqrt{2}}\right)\)
(iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) x (potential at D due to charges at A, B and C)
\(=-q\left(\frac{+q}{4 \pi \varepsilon_{0} d}+\frac{-q}{4 \pi \varepsilon_{0} d \sqrt{2}}+\frac{q}{4 \pi \varepsilon_{0} d}\right)\)
\(=\frac{-q^{2}}{4 \pi \varepsilon_{0} d}\left(2-\frac{1}{\sqrt{2}}\right)\)
Add the work done in steps (i), (ii), (iii) and (iv). The total work required is
\(=\frac{-q^{2}}{4 \pi \varepsilon_{0} d}\left\{(0)+(1)+\left(1-\frac{1}{\sqrt{2}}\right)+\left(2-\frac{1}{\sqrt{2}}\right)\right\}\)

\(=\frac{-q^{2}}{4 \pi \varepsilon_{0} d}(4-\sqrt{2})\)
The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
(Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)
(b) The extra work necessary to bring a charge q₀ to the point E when the four charges are at A, B, C and D is q₀ x (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by thatdue to B and D. Hence no work is required to bring any charge to  point E.

a) Given: The peak voltage of supply is 100 V.
The rms voltage is give as,
v m = 2 ×V
Where, the peak value of supply voltage is v m and its rms value is V.
By substituting the given values in the above equation, we get
300= 2 ×V V= 300 2 =212.1 V
Thus, the value of rms voltage is 212.1V.
b) Given: The rms current in an ac circuit is 10 A.
The peak current in the circuit is given as,i m = 2 ×I
Where, the peak current in an ac circuit is i m and its rms value is I.
By substituting the given values in the above equation, we get
i m = 2 ×10 =14.1 A
Thus, the value of peak current in the given ac circuit is 14.1 A.

Here, number of turns per unit length,
\(n=\frac{N}{l}=15\) turns/cm = 1500 turns/m
A = 2.0 cm² = 2 \(\times\)10^-4m²
\(\begin{aligned} \therefore \frac{d I}{d t}=\frac{4-2}{0.1} \text { or } \frac{d I}{d t}=20 \mathrm{As}^{-1} \\ \end{aligned}\)
\(\begin{aligned} \therefore|e|=\frac{d \phi}{d t}=\frac{d}{d t}(B A) \quad\left[\because B=\frac{\mu_0 N I}{l}\right] \\ \end{aligned}\)
\(= \frac{A d}{d t}\left(\mu_0 \frac{N I}{l}\right)=A \mu_0\left(\frac{N}{l}\right) \frac{d l}{d t}\)
= (2 \(\times\) 10^-4) \(\times\)4 \(\pi\)\(\times\)10^-7 \(\times\)1500 \(\times\)20 V
= 7.5 \(\times\)10^-6 V

Energy of photon, \(\mathrm{E}=\mathrm{hv}\)
This implies,\(\mathrm{E}=\mathrm{h} \frac{\mathrm{c}}{\lambda}\)
Where, \(\mathrm{h}=6.62 \times 10^{-34} \mathrm{js}\)
\( \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\)
If wave length \(\lambda\) is in meter and energy is in joule then, we will divide E by \(1.6 \times 10^{-19}\) to convert into eV (Electron volt).
\(\therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda \times 1.6 \times 10^{-19}} \mathrm{eV}\)
(1) For y - rays wave length ranges from  to less that \(10^{-14} \mathrm{~m}\)
Therefore,  \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \times 10^3 \mathrm{eV} \approx 10^4 \mathrm{eV}\)
Thus, \( \lambda=10^{-10} \mathrm{~m}, \text { energy }=10^4 \mathrm{eV} \text { and }\) \( \lambda=10^{-14} \mathrm{~m} \text {, energy }=10^8 \mathrm{eV}\)
Energy of y - rays ranges between 10⁴ to \(10^8 \mathrm{eV}\)
(2) For X - rays wave length ranges from \(10^{-8} \mathrm{~m}\) to \(10^{-7} \mathrm{~m}\) For \(\lambda=10^{-8}\)
Therefore, \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-8} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \approx 10^2 \mathrm{eV}\)
\( \lambda=10^{-13} \mathrm{~m} \text {, energy }=10^7 \mathrm{eV}\)
(3) For violet radiation \(\lambda\) ranges from \(4 \times 10^{-7}\) to \(6 \times 10^{-10}\)
Therefore, for \(\lambda=4 \times 10^{-7}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV} =3.1 \mathrm{eV} \approx 10^{10} \mathrm{eV}\)
\(\lambda=6 \times 10^{-10} \mathrm{~m} \text {, Energy }=10^3 \mathrm{eV}\)
Energy of ultraviolet radiation vary between \(10^{10}\) to \(10^3 \mathrm{eV}\).
(4) For visible radiations wave length range from \(4 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-7} \mathrm{~m}\)
Therefore,
For \(\lambda=4 \times 10^{-7} \mathrm{~m}\), and Energy \(=10^{10} \mathrm{eV}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{7 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{VV} \)
\(=1.77 \mathrm{eV} \approx 10^{\circ} \mathrm{eV}\)
(5) For infrared radiation $\lambda$ range from \(7 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-14} \mathrm{~m}\)
Therefore, \(\lambda=7 \times 10^{-7} \text {, energy }=10^{\circ} \mathrm{eV}\)
For \(\lambda=7 \times 10^{-4} \text {, energy }=\frac{1}{1000} \text { times }\)
the other order of  \(10^{-3}\)eV
(6) For micro waves $\lambda$ ranges from 1 mm to 0.3 m
For \(\lambda=1 \mathrm{~mm}\) or \(10^{-3}\)
energy is equal to \( \text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-3} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-3} \mathrm{eV} \approx 10^{-3} \mathrm{eV}\)
For \(\lambda=0.3 \mathrm{~m} \text {, Energy }=4.1 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV} \text {. }\)
(7) For Radio waves $\lambda$ ranges from 1 m to few km For $\lambda=1 \mathrm{~m}$
For λ=1m
Energy is equal to
\(=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^0 \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV}\)
Energy for λ of the order of few km≈10⁻⁶eV
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source

(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

Initial current, I₁ = 5.0 A
Final current, I₂ = 0.0 A
Change in current dl = I^-1- I^-2 = 5A
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
\(e=L\frac{di}{dt}\)
\(L=\frac{e}{\frac{di}{dt}}\)
\(=\frac{200}{\frac{5}{0.1}}=4H\)
Hence, the self induction of the coil is 4 H.

The capacitance of the capacitor in the circuit is C = 60 μ F or 60 x 10 ^-6 F
The source voltage is V = 110 V
The frequency of the source is ν = 60 Hz
The angular frequency can be calculated using the following relation,
ω = 2πν
The capacitive reactance in the circuit is calculated as follows:
\(X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi \nu C}=\frac{1}{2 \pi \times 60 \times 60 \times 10^{-6}} \Omega\)
Now, the RMS value of the current is determined as follows:
\(I=\frac{V}{X_{C}}=\frac{220}{2 \pi \times 60 \times 60 \times 10^{-6}}=2.49 A\)
Therefore, the RMS current is 2.49 A.

B_H = Bcos \(\delta \), B_v = Bsin \(\delta \)
\({ Also } \tan \delta=\frac{B_{V}}{B_{H}}\)

Given, v = 2.3 x 10⁶ ms ^-1 ,
r = 0.53 \(\overset{o}{A}\) = 0.53 x 10^-10 m
Equivalent current, \( I=\frac{e}{T}=\frac{e}{\frac{2 \pi r}{v}}=\frac{e v}{2 \pi r}\)
\(=\frac{1.6 \times 10^{-19} \times 2.3 \times 10^{6}}{2 \times 3.14 \times 0.53 \times 10^{-10}}\)
\(\therefore\) Magnetic moment,
M = IA = I (\(\pi r\)²) = 1.105 x 10^-3 x 3.14 x (0.53 x 10^-10)²
= 9.75 x 10^-24 A - m²

(i) By Gauss' theorem, total electric flux through closed Gaussian surface is given by
\(\phi =\frac { q }{ { \varepsilon }_{ 0 } } \)
\(\therefore\)  \(q=\phi { \varepsilon }_{ 0 }\) 
But electric flux passing through the surface
\(\phi =-3\times { 10 }^{ -14 }N-{ m }^{ 2 }/C\)
\(=-26.55\times { 10 }^{ -26 }C\)
\(q=-2.655\times { 10 }^{ -25 }C\)

(ii) Electric flux passing through the surface remains unchanged because it depends only on charge enclosed by the surface and is the independence of its size.

Here,  P = \(16 \ \Omega\), Q = \(12 \ \Omega\), R = \(9 \ \Omega\), S = \(4 \ \Omega\) 
Let \(9 \ \Omega\) be shunted with resistance x to balance the bridge. Then effective resistance of \(9 \ \Omega\) and x ohm in parallel,
\(R'=\frac{9x}{9 + x}\)
For balanced bridge,
\(\frac{P}{Q}=\frac{R'}{S}\)
\(\therefore \frac{16}{12}=\frac{9x/(9+x)}{4}\)
or  \(\frac{4}{3}=\frac{9x}{4(9+x)}\)
or  144 + 16 x = 27 x
or x = \(13.1 \ \Omega\)

\(Here,{ N }_{ 1 }=400, \ { N }_{ 2 }=700\)
\( I_{ 1 }=3.5A \ \phi =300 \times { 10 }^{ -6 }{ Tm }^{ 2 }\)
\({ \phi }_{ 2 }=900 \times { 10 }^{ -6 }{ Tm }^{ 2 }\)
\(M=\frac { { N }_{ 2 }{ \phi }_{ 2 } }{ I_{ 1 } } =\frac { 700 \times 90 \times { 10 }^{ -6 } }{ 3.5 } =1.8 \times { 10 }^{ -2 }H\)
\({ L }_{ 1 }=\frac { { N }_{ 1 }{ \phi }_{ 1 } }{ I_{ 1 } } =\frac { 400 \times 300 \times { 10 }^{ -6 } }{ 3.5 } =3.43 \times { 10 }^{ -2 }H\)
\({ e }_{ 2 }=M(\frac { dI_{ 1 } }{ dt } )=1.8 \times { 10 }^{ -2 } \times (0.5)\)
\(=9 \times { 10 }^{ -3 }V\)

Using Eq, the magnitude of B is
\(B=\frac { E }{ c } \) 
\(=\frac { 6.3V/m }{ 3\times { 10 }^{ 8 }m/s } =2.1\times { 10 }^{ -8 }T\)
To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.
Since, \((+\overrightarrow{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\overrightarrow{\mathbf{i}}, \mathbf{B}\) is along the z-direction.
Thus, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)