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  • (a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular

(a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed '\omega' in a magnetic field \vec{B}, directed perpendicular to the axis of rotation.
(b) An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5\times 10^{-4}T and the angle of dip is 30^{\circ}\cdot

 

 

 
 
 
 
 

Answers (1)

a)
 
The working principle of the AC generator is electromagnetic induction.
When a coil is rotated in a uniform magnetic field the magnetic flux linked with the coil is changed continuously which induces an emf in the coil. the current produced in the coil is taken at through carbon brushes.
The expression for induced emf
Let.
 N   = number of turns in the coil
A = Area of each face of turns
\vec{B} = magnetic field strength
\theta = Angle between \vec{B}  and \vec{A}
w = Angular speed of coil
Then the magnetic flux linked with the coil at any instant t is
\phi = NBA \cos \theta = NBA \cos wt
According to faraday's law, the induced emf is given by
\varepsilon = \frac{-d\phi }{dt}= \frac{-d}{dl}\left ( NBA \cos wt \right )
\varepsilon= NBA \omega \sin \omega t
where V_{m} = maximum generated emf = NBA\omega
b) Given,
velocity

V= 900 \, km/hour= 250 \, m/s
using span l = 20 m
Earth magnetic field BH = 5\times 10^{-4}T
angle of \, \delta = 30^{\circ}
Potential difference developed between the end of wings  = BlV
the vertical component of earth magnetic field BV = BH \tan \delta
= 5\times 10^{4}\left ( \tan 30^{\circ} \right )T
The potential difference,
                                       = \frac{5\times 10^{4}\times 20\times 250}{\sqrt{3}}= 1\cdot 44V
Therefore, the potential difference developed = 1.44 V

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Safeer PP

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