No, as every material has some resistance. Yes, we can coil a wire to have resistance with inductance.

Induced current,
\(I=\frac { induced \ e.m.f/ }{ resistance } =\frac { e }{ R } \\ =\frac { d\Phi /dt }{ R } =\frac { -1 }{ R } \frac { d }{ dt } (BA \ cos{ 0 }^{ \circ })\\ I=-\frac { A }{ R } \frac { d }{ dt } ({ B }_{ 0 }sin\omega t)=-\frac { { AB }_{ 0 } }{ R } cos\omega t(\omega )\\ =-\frac { { A\omega B }_{ 0 } }{ R } cos\omega t\)

Yes, induced current will be there. This is because, at poles, earth's magnetic field is totally vertical. The orbiting satellite will intercept this field and emf will be induced.

Change in magnetic flux linked with the circuit.

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

This is because of eddy currents developed in copper sheet.

This is because induced e.m.f. opposes the current due to the actual source e.m.f.

No, because the pole intercepts neither H nor V.

The current leads the voltage in phaseHence, X_c > X_L
For resonance, we must have New value of X_c = XL
We, therefore, need to decrease X_c \((\frac { 1 }{ \omega c } )\) This
requires an increase in the value of C. Hence, capacitor c_o should be connected in parallel across C the diagram of the modified circuit is as shown.

For resonance, we then have
\(\frac { 1 }{ \omega (c+co) } =\omega L\)
\({ C }_{ o }=[\frac { 1 }{ { \omega }^{ 2 }L } -C\)

Self-inductance of the solenoid is the flux associated with the solenoid when the unit amount of a current passes through it.
Consider a solenoid of self-inductance L, length e, area of cross-section A.
Let alternating voltage
ε = ε₀ sin ωt applied across it, and back emf induced.

∴ e = -L \(\frac { dI }{ dt } \)
⇒ dL = \(\frac { e }{ L } \) dt
Small  amount of work done i.e
 dW = eI dt
Total work done 
 W = \(\int { el\ dt } \)
= \(\int _{ 0 }^{ 1 }{ LI\ dI\ =\frac { 1 }{ 2 } } \)
This work done will be stored in the form of magnetic energy in inductor
 U = \(\frac { 1 }{ 2 } \) LI²