(a) Draw a schematic diagram of an AC generator. Explain is working and obtain the expression for the instantaneous value of the emf in terms of the magnetic field B, number of turns N of the coil of area A rotating with angular frequency `omega`. Show how an alternating emf is generated by loop of wire rotating in a magnetic field.

(b) A circular coil of radius 10 cm and 20 turns is rotated about its vertical diameter with angular speed of 50 rad `s^(-1).` in a uniform horizontal magnetic field of `3.0xx10^(-2)T`.

(i) Calculate the maximum and average emf induced in the coil.

(ii) If the coil forms a closed loop of resistance `10 Omega`, calculate the maximum current in the coil and the average power loss due to Joule heating.

(b) A circular coil of radius 10 cm and 20 turns is rotated about its vertical diameter with angular speed of 50 rad `s^(-1).` in a uniform horizontal magnetic field of `3.0xx10^(-2)T`.

(i) Calculate the maximum and average emf induced in the coil.

(ii) If the coil forms a closed loop of resistance `10 Omega`, calculate the maximum current in the coil and the average power loss due to Joule heating.

(b) `r=(10)/(100)=(1)/(10)m`

N=20

`omega=50 rad//sec`

`B=3:0xx10^(-2)T`

(i) `epsilon_(0)=NBA omega=Nbpir^(2)omega`

`=20xx3xx10^(-2)xx3.14xx(1)/(10)xx(1)/(10)xx50`

=0.942 volt

Average emf over one complete cycle=0

for half cycle `=(2)/(pi)epsilon_(0)=(2)/(3.14)xx0.942=0.6 volt.`

(ii) `R=10 Omega`

Maximum current `I_(0)=(epsilon_(0))/(sqrt(2))`

`=(0.942)/(sqrt(2))=0.66 AmP`

Average power `P_(av)=I_(rms)^(2)R=(I_(0)^(2))/(2)R`

`=(0.66xx0.66)/(2)xx10=2.718 W.`