(a)  Draw a schematic diagram of an ac generator. Explain its 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 a loop of write 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\cdot 0\times 10^{-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.

 

 

 
 
 
 
 

Answers (1)
S safeer

working of Ac generator:-
Systematic Diagram of Ac generator:

whenever a coil is placed in a uniform magnetic field and rotated, the flux linked with the coil changes and an emf induces in the coil 
The ends of the coil are connected to an external circuit by means of slip rings and brushes. when, the flux linked with the coil of area(A), placed in a uniform magnetic field (B)'
then, the flux
\phi _{B}= BA \cos \theta
or \phi _{B}= BA \cos wt---(i)
we know from faraday's law,
the emf induced in the coil is given as :-
\varepsilon = -N\frac{d\phi _B}{dt}  where, N= no. of turns.
\varepsilon = -NBA\frac{d}{dt}\cos wt
\varepsilon = NBA w\sinwt \ or \ \varepsilon = \varepsilon _{0}\sin wt
(b) (i) we have given the values:
r= 10 \: cm
N= 20\: turns
w= 50\: rads^{-1}
B= 3\cdot 0\times 10^{-2}T
we know, the induced emf
\varepsilon_{0} = NBAw
\varepsilon_{0} = 20\times 3\times 10^{-2}\times \pi \left ( 10\times 10^{-2} \right )^{2}50

\varepsilon_{0} =0\cdot 942 volt
\varepsilon_{av}= 0
The average value of induced emf=0
(ii) To calculate the maximum current 
we have
I_o= \frac{e_{0}}{R}= \frac{0\cdot 942}{10} where\: e_{0} \: is absolute permittivity, R is resistance.
I_0= 0\cdot 094A
To determine the average power loss, we have
P= \frac{1}{2}\varepsilon _{0}\times I_{0}
P= \frac{1}{2}\times 0\cdot 942\times 0\cdot 094
P= 0\cdot 045 watt

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