CBSE Class 12 Physics 2020 Outside Delhi Set 3 Paper

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Question : 13
Total: 13
Draw the labelled diagram of an AC generator. Briefly explain its working and obtain the expression for the emf produced in the coil.
Solution:  
AC generator:
Construction:
A coil of n turns rotates (called armature) between the magnetic poles by the means of an external agent (steam, running water, etc). The ends of the coils are connected to an external circuit through the means of carbon brushes and slip rings.

Principle of working:
AC generator works on the principle of Faraday's Laws of electromagnetic induction. According to it, when there is change of magnetic flux through a conducting loop, an emf is induced across the ends of the loop.
e=
dφB
dt

Working:
The strong magnetic field is created between the poles. The coil ABCD is rotated by external means in this field. As the coil rotates, the angle between the magnetic field and the coil changes which induces an alternating emf. The ends of the coil is connected to an external circuit by the means of carbon brushes (B1&B2) and slip rings R1&R2). When the external circuit is closed, an alternating current flows through the coil.
Expression of emf produced:
B= The magnetic field produced by the magnet
The coil be placed such that at t=0, the angle between the surface area vector of the coil and the magnetic field be
π
2
.
ω= angular speed of the coil
Angle between surface area vector of the coil and the magnetic field after time t is given by :
θ=θ0+ωt
Magnetic flux through the coil is then :
ΦB=BdA
Or, ΦB=BAcosθ
Or, ΦB=BAcos(θ0+ωt)
By Faraday's law of electromagnetic induction, magnitude of induced emf through a coil of n-turns is given by:
e=n
dϕB
dt

Or, e=n
d
dt
[BAcos(θ0+ωt)]

|e|=nBAωsin(θ0+ωt)
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