Given, N_pB = 2N_pA NsB = 2N_sA, E_pB = 2E_pA
As we know, \(\frac{N_{s}}{N_{p}}=\frac{E_{s}}{E_{p}}\)
For transformer B, \(\frac{N_{s B}}{N_{p B}}=\frac{2 N_{s A}}{2 N_{p A}}=\frac{E_{s B}}{E_{p B}}\)
\(\Rightarrow \quad \frac{E_{s B}}{E_{p B}}=\frac{E_{s B}}{2 E_{p A}}=\frac{E_{s A}}{E_{p A}}\)
\(\Rightarrow \quad E_{s B}=2 E_{s A}\)
\(\therefore\)  Secondary voltage on transformer B is equal to the twice of secondary voltage on transformer A.

Substance, which at room temperature, retain their ferromagnetic property for a long period of time are called permanent magnets. Iron, cobalt, steel and Nickel.

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.

The path of the charged particle will be helix. As, the charge moves linearly in the direction of the magnetic field with velocity v \(\cos { \theta } \) and also describe the circular path due to velocity v \(\sin { \theta } \)
Time taken by the charge to complete one circular rotation,
\(T=\frac { 2\pi r }{ { v }_{ \bot } } \)   ......(i)
\(\Rightarrow\)  \(\\ f=q{ v }_{ \bot }B\)    
and \(\\ \frac { m{ { v }^{ 2 } }_{ \bot } }{ r } =q{ v }_{ \bot }B\)
\(\Rightarrow \ \frac{v_{\perp} m}{q B}=r\)     .....(ii)
From Eqs. (i) and (ii), we get
\(\Rightarrow T=\frac { 2\pi { v }_{ \bot }m }{ qB.{ v }_{ \bot } } =\frac { 2\pi m }{ Bq }\)
Distance moved by the particle along the magnetic field in one rotation (pitch of the helix path)
 \(={ v }_{ || }\times T \ (\because { v }_{ || }={ v }_{ parallel })\)
\(\\ =v\cos { \theta \times \frac { 2\pi m }{ Bq } } \)
\(\\ p=\frac { 2\pi mv\cos { \theta } }{ qB } \\ \\\)

Here,  \(V=220 \ V, \ R=11\Omega , \ E=198V\)
Current at full speed \(I=\frac { V-E }{ R } \)
\(=\frac { 220-198 }{ 11 } =\frac { 22 }{ 11 } =2A\)
When motor is just switched on, \(E=0\)
\(\therefore d I=\frac { V }{ R } =\frac { 220 }{ 11 } =20 \ A\)

In LCR circuit, resonance occurs, when \({ X }_{ L }={ X }_{ C }or{ V }_{ L }={ V }_{ C }.\) The resonance frequency is \({ v }_{ r }=\frac { 1 }{ 2\pi \sqrt { LC } } .\)

e = 3.12 V
\({ Here,\upsilon =450m/s,H=4 \times { 10 }^{ -4 }T, }\)
\( \delta ={ 30 }^{ \circ },l=30m\)
A jet plane flying horizontally intercepts vertical component (V) of earth's magnetic field.
\(\therefore B=V=H \ tan\delta =4 \times { 10 }^{ -4 }tan{ 30 }^{ \circ }\)
\(=\frac { 4 \times { 10 }^{ -4 } }{ \sqrt { 3 } } T\)
\( e=B/\upsilon\)
\(e=\frac { 4 }{ \sqrt { 3 } } \times { 10 }^{ -4 } \times 30 \times 450 \ V=3.12 \ volt\)

(a) (i) Induced current along abcda (Using FLH Rule).
(ii) Induced current along same direction i.e. abcda 
(iii) Induced current zero (the loop is normal to the field lines.)
(b) Induced e.m.f. greatest (loop is parallel to field).

e = Blv sin θ

With resistor, current is same both for 100 Hz and 100 kHz. With inductor, the current is 11.9 A and 11.9 mA respectively

Impedance of series LCR circuit is given by \(Z=\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } } \)
or For Z to be minimum, \({ X }_{ L }={ X }_{ e }(or\omega =\frac { 1 }{ \sqrt { LC } } )\)
For wattless current to flow, circuit should not have any ohmic resitance R = 0
Alternatively : Power = \({ V }_{ rms }{ I }_{ rms }cos\phi \)
\(\phi ={ 90 }^{ 0 }=\frac { \pi }{ 2 } \)
Power = 0

Actually there is no permanent strong magnet inside the earth responsible for earth's magnetism, Dr. Gilbert considered a fictitious strong magnet inside the earth, whose magnetic axis is inclined roughly 11.3^o west of axis of rotation of earth, which is along geographic north and geographic south.

Through pure inductor or pure capacitor, phase diff.between alternating voltage and alternating current is \({ 90 }^{ \circ }.\) Therefore, power factor, \(cos\phi =cos{ 90 }^{ \circ }=0.\)

Hence the current is said to be wattless.

Wb/m² is the SI unit of magnetic field induction B, which is a vector quantity.

\(X_L=\omega L=2 \pi \nu L=2 \pi \times 50 \times 0.25\)
\(=25 \pi \mathrm{ohm}\)

Large induced e.m.f. at break causes the sparking.

\(R =200 \Omega \)
\(C =\frac{50}{\pi} \times 10^{-6} \mathrm{~F} \)
\(E_0 =100 \mathrm{~V} \)
\( f =50 \mathrm{~Hz} \)
i) Capacitive reactance, Xc \(=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)
\( =\frac{1}{2 \times 3.14 \times 50 \times \frac{50}{\pi} \times 10^{-6}} \)
\(=\frac{10^6}{2 \times 2500}=\frac{10^6}{5000}=\frac{1000000}{5000}=200 \Omega \)
\(Z =\sqrt{X_C^2+R^2}=\sqrt{(200)^2+(200)^2} \)
\(=\sqrt{40000+40000}=\sqrt{80000}=2 \sqrt{2} \times 100 =200 \sqrt{2} \Omega \)
ii) Phase angle, \(\begin{aligned} \phi & =\tan ^{-1}\left(\frac{X_c}{R}\right)=\tan ^{-1}\left(\frac{200}{200}\right) =\tan ^{-1}(\mathrm{l})=45^{\circ} \end{aligned}\)
iii) \( V_R=I R=\frac{E_0 / \sqrt{2}}{Z} \times R=\frac{100}{\sqrt{2} \times 200 \sqrt{2}} \times 200=50 \mathrm{~V}\)

(i) Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence, an emf will be induced in the neighboring coil or circuit. This phenomenon is called 'mutual induction'. According to question, let the current in big coil of radius r₂ be I₁ so, magnetic field at point O due to this coil will be \(\frac { { \mu }_{ 0 }{ I }_{ I } }{ { 2r }_{ 2 } } .\)

Change in magnetic flux in the coil of radius r₁ is
\(\phi =BA=\frac { { \mu }_{ 0 }{ I }_{ 1 } }{ { 2r }_{ 2 } } \times { \pi r }_{ 1 }^{ 2 }\)
Mutual inductance,
\(M=\frac { \phi }{ { I }_{ 1 } } =\frac { { \mu }_{ 0 }{ I }_{ 1 }{ \pi r }_{ 1 }^{ 2 } }{ { 2r }_{ 2 }\times { I }_{ 1 } } =\frac { { \mu }_{ 0 }{ I }_{ 1 }{ \pi r }_{ 1 }^{ 2 } }{ { 2r }_{ 2 } } \)
This is the required expression.
(ii) According to the question, if the coil rotates with an angular velocity of 00 and N turns through an angle \(\theta \) in time t, thus \(\theta =\omega t\)
∴ \(\phi =BAcos\theta =BAcos\omega t\)
As the coil rotates, the magnetic flux linked with it changes. An induced emf Js set up in the coil which is given by
\(e=\frac { -d\phi }{ dt } =\frac { -d }{ dt } (BAcos\omega t)\)
\(=BA\omega sin\omega t\)
For N number of turns, \(e=NBA\omega sin\omega t\)
For maximum value of emf \(\omega t\)  must be equals to 90°.
So, maximum emf induced is =\(NBA\omega \)
i.e.\(e=NBA2\pi f\)   \(\left[ \because \omega =e=2\pi f \right] \)

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.

Here, n = 12, l = 0.2 m.
\(v=60rpm=\frac { 60 }{ 60 } =1rps\)
\(B=0.4\times { 10 }^{ -4 }T, \ e=?\)
\(v=r\omega =r(2\pi v)=2\pi vl\)
\(=2\times 3.14\times 1\times 0.2=1.256 \ { ms }^{ -1 }\)
As the spoke rotates about the axle, therefore linear velocity of each spoke at the axle v₁= 0. And linear velocity of each spoke at the rim, v₂ = 1.256 m/s.
Average velocity of spoke = \(=\frac { v_{ 1 }+{ v }_{ 2 } }{ 2 } =\frac { 0+1.256 }{ 2 } =0.628\ m/s\)
As well spokes are in parallel, therefore, e.m.f. induced between two ends of every spoke = e = Bvl = 0.4\(\times\)10^-4 \(\times\) 0.628\(\times\)0.2 = 5\(\times\)10^-6 V