It works on the principle of electromagnetic induction, i.e. when a coil continuously rotates in a magnetic field, the magnetic flux associated with it keeps on changing; thus an emf is induced in it.
(b) When the coil rotates in a magnetic field, its effective area i.e. A cos \(\theta \) , (i.e. area normal to the magnetic field) keeps on changing. Hence magnetic flux  \(\phi \) = NBAcos S, keeps on changing.
(c) Let the coil be rotating with angular velocity 'm',
at any instant 't' when the normal to the plane of the coil makes an angle \(\theta \) with the magnetic field. Hence magnetic flux, 12   \(\phi \) = NBA cos mt, Therefore induced emf (e)
\(\epsilon =-\frac { d\phi }{ dt } \)
\(\Rightarrow \epsilon =NBA\omega sin\quad \omega t\)
Induced emf will be maximum when mt = 90° 12 Hence, max = NBA\(\omega \)
Direction of induced emf can be determined using Flemming's Right-hand rule. Alternatively: Statement of the above rule.

(a)

(b) Principle of a transformer: When alternating current flows through the primary coil, an emf is induced in the neighbouring (secondary) coil
Let \(\frac { d\phi }{ dt } \) be the rate of change of flux through each turn of the primary and the secondary coil
\(\frac { { \varepsilon }_{ 1 } }{ { \varepsilon }_{ 2 } } =-{ N }_{ 1 }\frac { d\phi }{ dt } /-{ N }_{ 2 }\frac { d\phi }{ dt } =\frac { { N }_{ 1 } }{ { N }_{ 2 } } \\ \frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { { N }_{ 1 } }{ { N }_{ 2 } } \)
But for an ideal transformer,
\(\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { I_{ 2 } }{ I_{ 1 } } \)
From equation (1) and (2)
\(\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { N_{ 1 } }{ N_{ 2 } } =\frac { I_{ 2 } }{ I_{ 1 } } \)
(c) Main assumptions
(i) The primary resistance and current are small
(ii) The flux linked with the primary and secondary coils is same / there is no leakage of flux from the core.
(iii) Secondary current is small.
(d) Reason due to which energy loses may occur Flux leakage/Resistance of the coils/Eddy currents/Hysteresis.

Given,  
\(L=80mH=80\times { 10 }^{ -3 }H, \ R=0, \ v=50Hz\)
\(C=60\mu F=60\times { 10 }^{ -6 }F,\)
\(\omega =2\pi v=100\pi \ rad/s \)
\({ V }_{ rms }=230 \ V,\)
and \(\\ { V }_{ 0 }=\sqrt { 2{ V }_{ rms } } =\sqrt { 2 } \times 230V\)
(i) I₀ = ? and I_rms = ?
\(\begin{aligned} \Rightarrow I_0 & =\frac{V_0}{Z}=\frac{V_0}{\left|\omega L-\frac{1}{\omega C}\right|} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{230 \sqrt{2}}{\left|100 \pi \times 80 \times 10^{-3}-\frac{1}{100 \pi \times 60 \times 10^{-6}}\right|} \\ \end{aligned}\)
\(\begin{aligned} =\frac{230 \sqrt{2}}{\left|8 \pi-\frac{1000}{6 \pi}\right|}=-11.63 \mathrm{~A} \\ \end{aligned}\)
\(\begin{aligned} I_{\mathrm{rms}} & =\frac{I_0}{\sqrt{2}}=\frac{-11.63}{\sqrt{2}}=8.23 \mathrm{~A} \end{aligned}\)
(ii) For L, V_L = I_rms \(\omega \)L = 8.23 \(\times\) 100\(\pi\)\(\times\)80 \(\times\)10^-3
= 206.84 V
For C, \(V_C=I_{\mathrm{rms}} \frac{1}{\omega C}=8.23 \times \frac{1}{100 \pi \times 60 \times 10^{-6}}\)
= 436.84 V
Since, voltage across L and C are 180° out of phase, therefore they are subtracted.
Thus, applied rms voltage = 436.84 - 206.84
= 230.0 V
(iii) Average power transferred per cycle by source to inductor is always zero because of phase difference of \(\pi\)/2 between voltage and current through inductor.
(iv) Average power transferred per cycle by source to inductor is always zero because of phase difference of \(\pi\)/2 between voltage and current through inductor
(v) \(\therefore\) Total average power absorbed by the circuit is also zero.

\(Here,\quad V=80V,\quad I=10A,\)
\(R=\frac { V }{ I } =\frac { 80 }{ 10 } =8\Omega \)
\({ E }_{ v }=100V,\quad v=50Hz,\quad { I }_{ v }=I=10A,\quad L=?\)
If Z is impedance of lamp and choke coil,
\(then \ Z=\frac { { E }_{ v } }{ { I }_{ v } } =\frac { 100 }{ 10 } =10\Omega \)
\(As \ { R }^{ 2 }+{ X }_{ L }^{ 2 }={ Z }^{ 2 }\)
\(\\ \therefore \ \ { X }_{ L }^{ 2 }={ Z }^{ 2 }-{ R }^{ 2 }={ 10 }^{ 2 }-{ 8 }^{ 2 }=36, \ { X }_{ L }=6\Omega \)
\(Now \ \ \ { X }_{ L }=\omega l=2\pi vL\)
\(\\ \therefore \ \ L=\frac { { X }_{ L } }{ 2\pi v } =\frac { 6\times 7 }{ 2\times 22\times 50 } =1.9\times { 10 }^{ -2 }H\)

0.066 H, \(12\Omega \)

Let (I_rms)_a = rms current in circuit (a)
(I_rms)_b  = rms current in circuit (b)
\({ \left( I_{ rms } \right) }_{ a }=\frac { { V }_{ rms } }{ Z } =\frac { V }{ R } \)
\({ \left( I_{ rms } \right) }_{ b }=\frac { { V }_{ rms } }{ Z } b\)
\(=\frac { V }{ \sqrt { { R }^{ 2 }+\left( { X }_{ L }-{ X }_{ C } \right) ^{ 2 } } } \)
When \(({ I }_{ rms })_{ a } =({ I }_{ rms })_{ b }\)
\(R=\sqrt { { R }^{ 2 }+\left( { X }_{ L }-{ X }_{ C } \right) ^{ 2 } } \)
\(\Rightarrow \ { X }_{ L }={ X }_{ C }\) in resonance condition
(ii) As,  \( Z\ge R\)
\(\Rightarrow \frac { ({ I }_{ rms })_{ a } }{ ({ I }_{ rms })_{ a } } =\frac { \sqrt { R=\sqrt { { R }^{ 2 }+\left( { X }_{ L }-{ X }_{ C } \right) ^{ 2 } } } }{ R } \)
\(=\frac { Z }{ R } \ \ge 1\)
\(({ I }_{ rms })_{ a }\ge ({ I }_{ rms })_{ b }\)
No, the rms current in circuit (b) cannot be larger than that in (a)

Here, A =  \(20cm\times 30cm\)
= \(6\times { 10 }^{ -2 }{ m }^{ -2 }\)
B = 0.3T
Let \(\theta \) be the angle made by field B with the normal to the plane of the coil.
(i) Here, \(\theta \) = 90\(^{0}\)- 90\(^{0}\) = 0\(^{0}\)
So, flux, \(\phi \)= BA cos\(\theta \)
\(\phi =0.3\times 6\times { 10 }^{ -2 }\times cos{ 0 }^{ \circ }\)
\(=1.8\times { 10 }^{ -2 }Wb\)  (1)
(ii) Here, \(\theta \) = 90\(^{0}\)- 30\(^{0}\)
= 60\(^{0}\)
\(\phi =0.3\times 6\times { 10 }^{ -2 }\times cos{ 60 }^{ \circ }\\ \phi =0.9\times { 10 }^{ -2 }Wb\)
(iii) Here, \(\theta \) = 90\(^{0}\)
\(\phi =0.3\times 6\times { 10 }^{ -2 }\times cos{ 90 }^{ \circ }\\ \phi =0\)

In an inductor, the current lags the voltage by 90⁰ . If the source voltage is sinusoidal, then the current is also sinusoidal but shifted in phase. The instantaneous power defined as the product of the instantaneous voltage and current can also be seen to be sinusoidal in time. However, in contrast to the resistive load, the instantaneous power in the inductor goes negative for part of the cycle of the source driving it.
As, \(V(t)={ V }_{ m }\sin { \omega t } \)
 \(I(t)=-{ I }_{ m }cos\omega t\)
instantaneous power P_i = V(t).I(t)
\(={ V }_{ m }sinn\omega t\times (-{ I }_{ m }cos\omega t)\)
 \(=-\frac { { V }_{ m }{ I }_{ m } }{ 2 } \times 2sin\omega tcos\omega t\)
 \(=-\frac { { V }_{ m }{ I }_{ m } }{ 2 } [sin\omega t+sin\ 0]\)
 \(=-\frac { { V }_{ m }{ I }_{ m } }{ 2 } sin2\omega t\)
The variation of P_i with \(\omega t\) is as given in the figure.

The instantaneous power altertes positive and negative at twice the ferquency of source supplying it.

Since the North pole of the bar magnet is approachng the loop, therefore the induced current wall flow in such a way that when loop viewed from left side. it will behave like a north pole and when viewed from right side, it will behave like a South pole. So, the flow of induced current in the loop will be clockwise. Hence, A acquires positive polarity and B acquires negative polarity.

\(dt=10s,{ \phi }_{ 1 }=8 \times { 10 }^{ -4 }Wb, \ e=?,{ \phi }_{ 2 }=0\)
\(e=\frac { -d\phi }{ dt } =\frac { -({ \phi }_{ 2 }-{ \phi }_{ 1 }) }{ dt } =\frac { -(0-8 \times { 10 }^{ -4 }) }{ 10 }\)
\(=8 \times { 10 }^{ -5 }V\)

In LR circuit, the impedance is given by
\({ Z }_{ L }=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } \quad ...(i)\)
In LCR circuit, the impedance is given by
\(Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } \quad ...(ii)\)
From equations, (i) and (ii), we find that
\(Z<{ Z }_{ L }\)
\(since \ I=\frac { E }{ Z } \)
As Z decreases on connecting a  capacitor in series with LR circuit, hence the current I in the circuit increases.

The back emf in the coil opposes the change in the current as per Lenz's law.

Current through the bulb
\({ I }_{ \upsilon }=\frac { { E }_{ \upsilon } }{ Z } =\frac { { E }_{ \upsilon } }{ \sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } } ,where\ { X }_{ L }=\omega L\)
On introducing an iron rod into the inductor its inductance L increases. \({ X }_{ L }={ \omega }L\) increases. Therefore, impedance Z increases. Hence brightness of bulb decreases.

\(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi \omega v C}\) therefore, when v is tripled

For transfer of power \(\left( =V\times I \right) \) at higher voltage (220 V instead of 110 V), current carried by wires is just half. Therefore, such wires need not be very thick, saving lot of transmission material and reducing the cost of transmission. This is one advantage of 220 V supply. But to design a device of particular wattage, \(P=\frac { { V }^{ 2 } }{ R } \) , as \({ V }^{ 2 }\) is 4 times, R must be four times. If not, the dissipation of power in the form of heat will be larger on 220 V supply. This is one disadvantage of this supply.

(i) On increasing distance between the two coils, magnetic flux passing from one coil to the other decreases. Therefore, mutual inductance decreases.
(ii) When an iron sheet is placed between the two coils, mutual inductance increases because \(M\propto \mu \) (the permeability).

\(e=-L \frac{d l}{d t}\)
Where symbols have usual meaning.

No, because the pole intercepts neither H nor V.