Yes, this would happen when phase difference between alternating voltage and alternating current is \({ 90 }^{ \circ }.\) It can happen when the circuit contains pure L or pure C.

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.

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

Induced e.m.f., \(e=V=\frac { work }{ charge } =\frac { M{ L }^{ 2 }{ T }^{ -2 } }{ AT } \)
\(=[{ M }^{ 1 }{ L }^{ 2 }{ T }^{ -3 }{ A }^{ -1 }]\)
Rate of change of magnetic flux\(=\frac { d\Phi }{ dt } =\frac { BA }{ t } \)
 \(=\frac { FA }{ q\upsilon t } =\frac { { (MLT }^{ -2 })({ L }^{ 2 }) }{ (AT)({ L }T^{ -1 })(T) } \ (\because F=Bq\upsilon )\\ \ \ =[{ M }^{ 1 }{ L }^{ 2 }T^{ -3 }A^{ -1 }]\)
Both have the same units/dimensions.

(i) Team spirit and curiosity.
(ii) When the wire in N-S direction is dropped feely, none of the components of earth's magnetic field is intercepted. So, no induced emf is produced. When the wire is dropped freely in E-W direction horizontal component of earth's magnetic field is intercepted. So emf is induced in the coil.

Step down transformer:

Principle: When the current flowing through the primary coil changes, an emf is induced in the secondary coil due to the change in magnetic flux linked with it i.e., it works on the principle of mutual induction. For step down transformer,
N_s < N_p, hence \({\epsilon}_{s}<{\epsilon}_{p}\).
\({ { {\epsilon}_{s} }\over{ {\epsilon}_{p} } }={{{N}_{s}}\over{{N}_{p}}}\)
For an ideal transformer, P_in = P_out
\({\epsilon }_{p }{ I}_{p }={ \epsilon}_{s }{I }_{s }\Rightarrow{{{I}_{p}}\over{{I}_{s}}}={{{\epsilon}_{s}}\over{{\epsilon}_{p}}}={{{N}_{s}}\over{{N}_{p}}}\)
P_in = P_out = 550 W = \({\epsilon}_{p}{I}_{p}\) = 550
\(220\times{I}_{p}=550\Rightarrow{I}_{p}={{550}\over{220}}={{5}\over{2}}=2.5 \ A\)

Here, B₁ = 2\(\times\)10^-2 Wb m^-2, \(\theta ={ 0 }^{ \circ }\),
A = 1000 cm² = 1000\(\times\)10^-4m² = 10^-1m², N = 50,
B₂ = 0, dt = \(\frac { 1 }{ 10 } \)s, e = ?
\(e=\frac { -d\phi }{ dt } =-\frac { N\left( { B }_{ 2 }-{ B }_{ 1 } \right) A }{ dt } =-\frac { 50\left( 0-2\times { 10 }^{ -2 } \right) { 10 }^{ -1 } }{ 1/10 } =1.0 \ V\)

\(Here, \ R=100\Omega , \ { E }_{ v }=220V, \ v=50Hz\)
\((a) \ { I }_{ v }=?, \ { I }_{ v }=\frac { { E }_{ v } }{ R } =\frac { 220 }{ 100 } =2.2A\)
\((b) \ Net \ power \ consumed \ over \ a \ full \ cycle,\)
\(P={ E }_{ v }{ I }_{ v }=220\times 2.2=484W\)