Number of turns, N = 500
Area of cross section, A = 30 x 30 x 10^-4 
= 900 x 10^-4 m²
Magnetic field, B = 0.4 T
Angle of inclination θ = 90⁰ - 30⁰= 60⁰
∴ Magnetic flux Φ = NAB cos θ
∴ Φ = 500 x 900 x 10^-4 x 0.4 x cos60⁰
= 45 x 10⁴ x 10^-4 x 4 x 10^-1 x \(\frac{1}{2}\)
Φ= 9 x 10^-1 = 9.0 Wb
∴ Magnetic flux Φ = 9.0 Wb

Resonant frequency is the frequency of the applied alternating source at which the current in the circuit reaches its maximum value.

When an electric current passing through a coil changes with time, an emf is induced in the neighboring coil. This phenomenon is known as mutual induction.

Whenever the magnetic flux linked with a closed coil changes, an emf is induced and hence an electric current flows in the circuit. This current is called an induced current and the emf giving rise to such current is called an induced emf. This phenomenon is known as electromagnetic induction.

Yes! An emf will be induced in the wire because it moves perpendicular to the horizontal component of Earth’s magnetic field and hence it cuts the magnetic lines of Earth's magnetic field.

B = 4.1 x 10^–5 T; θ = 90^o – 47^o = 43° ;
A = 3m²
We know that   \(\Phi_{B}=B A \cos \theta\) 
\(\Phi_{\mathrm{B}}\) = 4.1 x 10^–5 x 3 x cos 43^o  
= 4.1 x 10^–5 x 3 x 0.7314
= 89.96 \(\mu \mathrm{Wb}\).

V_p = 12 V; V_s = 240 V
I_s = 50 mA; N_p = 100 turns
\(\frac { { V }_{ s } }{ { V }_{ P } } =\frac { { N }_{ s } }{ { N }_{ p } } =\frac { { I }_{ P } }{ { I }_{ S } } =K\)
Transformation ratio, K = \(\frac{240}{12}=20\)
The number of turns in the secondary
N_S = N_P x K = 100 x 20 = 2000
Primary current,
I_P = K x I_s = 20 x 50 mA = 1 A

l = 0.5 m; h = 7.2 m; u = 0 m s^–1;
g = 10 m s^–2; B_H = 4.04 x 10^–5 T
The final velocity of the rod is
V² = u² = + 2g h = 0 + (2 x 10 x 7.2) =144
v = 12 ms^-1
The magnitude of the induced emf when the rod is about to touch the ground is
ε = B_H lv = 4.04 × 10^–5 × 0.5 × 12
= 242.4 µV

\(L=\frac{20}{\pi^2} \mathrm{H}, \mathrm{f}=50 \mathrm{~Hz}  \)
\(f=\frac{1}{2 \pi \sqrt{L C}}  \)
\(50=\frac{1}{2 \pi \sqrt{\frac{20}{\pi^2} \times C}}  \)
\(50=\frac{1}{2 \times \sqrt{20 C}}  \)
\(\therefore(50)^2=\frac{1}{4 \times 20 C}  \)
\(\therefore C=\frac{1}{2500 \times 4 \times 20}=5 \times 10^{-6}=5 \mu \mathrm{F}\)

\(Q_{midpoint}=\frac{Q}{\sqrt{1^2+1^2}}=\frac{Q}{\sqrt2}\)

L = 20 x 10^-3H. C = 50 x 10^-6 F, R= 40Ω
enf V = 10 sin 340 t
\(\therefore V_0=10 \mathrm{~V}, \omega=340  \)
\(X_1=1 \omega^{\prime}=20 \times 10^3 \times 340  \)
\(=6800 \times 10^{-1}=6.8 \Omega  \)
\(X_C=\frac{1}{C .}  \)
\(=\frac{1}{50 \times 10^{-\alpha} \times 340}=\frac{10^{\circ}}{17000}=\frac{10^{\prime}}{17}=58.823 \Omega   \)
\(Z=\sqrt{R^2+\left(X_6-X_1\right)^2}  \)
\(=\sqrt{(40)^2+(58.82-6.8)^2}  \)
\(=\sqrt{(40)^2+(52.02)^2}  \)
\(=65.62 \Omega\)
The peak current in the circuit is,
\(I_0=\frac{V_0}{Z}=\frac{10}{65.62}  \)
\(\cos 0=\frac{R}{Z}=\frac{40}{65.62} \)
\(\text{Power loss in A.C. circuit }=V_{r m} 1_{r \rightarrow \infty} \cos \phi \)
\(=\frac{1}{2} V_{\mathrm{o}} I_{\mathrm{c}} \cos \phi  \)
\(=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}\)
\(\frac{2000}{4305.98}\)
= 0.46 W

\(A_1  =4 \times 10^{-4} \mathrm{~m}^2  \)
\(N_1  =10 \text { turns }  \)
\(A_2  =4 \times 10^{-4} \mathrm{~m}^2  \)
\(N_2  =15 \times 10^{-}=1500 \mathrm{turn} / \mathrm{m}  \)
\(\phi  =B_2 A_2=\left(\mu_0 \mathrm{n}_2 I_2\right) \mathrm{A}_1 \)
\(Where, \mathrm{n}_2=\frac{\mathrm{N}_2}{l}=1500 \mathrm{turn} / \mathrm{m}\)
The mutual Inductance is,
\(M  =\frac{N_1 o_{12}}{I_2}=\mu_0 n_2 N_1 A_1  \)
\(=4 \pi \times 10^{-7} \times 1500 \times 10 \times 4 \times 10^{-4}  \)
\(=7.54 \times 10^{-6} \mathrm{H}=7.54 \mu \mathrm{H}\)

\(\phi_B =10 t^2-50 t+250  \)
\(e  =\frac{-d \phi_B}{d t}  \)
\(=\frac{-d}{d t}\left(10 t^2-50 t+250\right) \)
=-(20 t - 50) 
=-20 t + 50
When, t = 3 s, e =-20(3) + 50 = -60 + 50
e = -10V