Physics Answer the following in Detail

a) Consider a rectangular loop of conducting wire ‘PQRS’ partly placed in uniform
magnetic field of induction ‘B’ as shown in figure.

b) Let ‘l’ be the length of the side PS and ‘x’ be the length of the loop within the field.
∴ A = lx = area of the loop, which lies inside the field.
c) The magnetic flux (φ) through the area A at certain time ‘t’ is Φ = BA = Blx
d) The loop is pulled out of the magnetic field of induction ‘B’ to the right with a uniform
velocity ‘v’.
e) The rate of change of magnetic flux is given by, \(\frac{d \phi}{d t}=\frac{d}{d t}(B l x)\)
\(\therefore \quad \frac{ d \phi}{ dt }= B l\left(\frac{ dx }{ dt }\right)\)
But, \(\left(\frac{d x}{d t}\right)=v\)
\(\therefore \quad \frac{ d \phi}{ dt }= B / v\)...(i)
f) Due to change in magnetic flux, induced current is set up in the coil. The direction of
this current is clockwise according to Lenz’s law. Due to this, the sides of the coil
experience the forces, F1, F2 and F as shown in figure. The directions of these forces is
given by Fleming’s left hand rule.
g) The magnitude of force ‘F’ acting on the side PS is given by, F = BIl.
h) The force \(\vec{F}_1\)  and \(\vec{F}_2\) are equal in magnitude and opposite in direction, therefore they cancel out. The only unbalanced force which opposes the motion of the coil is \(\vec{F}\) Hence, work must be done against this force in order to pull the coil.
i) The work done in time ‘dt’ during the small displacement ‘dx’ is given by, dW = −Fdx
− ve sign shows that F and ‘dx’ are opposite to each other.
∴ dW = − (BIl) dx ….(ii)
j) Mechanical power is given by,
\(P =\frac{ dW }{ dt }= BI l\left(\frac{ dx }{ dt }\right)\)
\(\therefore \quad P = BI / v \quad\left[\because \frac{ dx }{ dt }= v \right]\)
k) This external work provides the energy needed to maintain the induced current I
through the loop (coil).
l. If ‘e’ is the e.m.f induced then,
electric power \(=\frac{d W}{d t}=e I\)
∴ dW = eIdt .…(iii)
m) From equations (ii) and (iii), we have, eIdt = − BIl dx
\(\therefore e=-B l \frac{d x}{d t}\)
\(\therefore e =- B / v \ldots\) (iv)
n. From equation (i) and (iv), we have,
\(e=-\frac{d \phi}{d t}\)
o) Graph of variation of induced e.m.f. against phase (θ = ωt) over one cycle

Magnetic flux Φ = NABcos ωt
N → No.s of turns
A → Area of a coil
B → Magnetic field {external}
ω → Angular speed of the coil
\(\therefore \max \phi_0=N A B, \omega t \rightarrow 0\)
\(\therefore \phi=\phi_0 \cos \omega t\)
induce \(\operatorname{EMF}( e )\)
\(e =-\frac{d \phi}{d t}=+\omega N A B \sin \omega t\)
\(e_0=\omega N A B, \omega t \rightarrow 90\)
\(\therefore e=e_0 \sin \omega t\)
So there is phase difference of \(\frac{\pi}{2}\) with \(\phi\) and e
 

A transformer is a device used for changing the voltage of alternating current from low value to high value or vice versa.
It works on the principle of mutual induction. Here, a changing alternating current in primary coil produces a changing magnetic field, which, in turn, induces a changing alternating current in the secondary coil.

An LCR circuit is a circuit that contains inductance (L), capacitance (C), and resistance (R) components connected in series. It's a fundamental circuit used in electronics and electrical engineering for various applications like filtering, oscillators, and signal processing.
Impedance in an LCR Circuit
The total impedance (Z) in a series LCR circuit is the total opposition to the flow of alternating current. It's a complex quantity and is calculated as the vector sum of the individual impedances of the inductor \(\left(Z_L\right)\),capacitor \(\left(Z_C\right)\),and resistor \(\left(Z_R\right)\).
The impedance expressions for each component in a series LCR circuit are:
Inductive impedance: \(Z_L=j \omega L\)
Capacitive impedance:\(Z_C=1 /(j \omega C)\)
Resistive impedance: \(Z_R=R\)
The total impedance in a series LCR circuit is given by the formula:\(Z=\sqrt{\left(Z_L=Z_C+Z+R\right)^2}\)
Conditions for Series Resonance
Series resonance in an LCR circuit occurs when the inductive reactance \(\left(X_L\right)\) equals the capacitive reactance \(\left(X_C\right)\) and the circuit's impedance reaches its minimum value. This happens when \(X_L=X_C\).
Conditions are:
\(X_L=X_C\)
At resonance, net reactance \(\left(X_L-X_C\right)=0\).
The total impedance of the circuit is equal to the resistance \((Z=R)\).
Derivation of Resonant Frequency
At resonance, \(X_L=X_C\)
\(X_L=\omega L\)
\(X_C=1 /(\omega C)\)
∴ After substitution,
\(\omega L=1 /(\omega C)\)
\(\omega=1 /(\sqrt{L C})\)
Hence, the expression for the resonant frequency in a series LCR circuit is
\(\omega_0=1 /(\sqrt{L C})\)