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  • A coil has turn N = 8000 , average radius r = 10 cm rotating at f = 40 revolution per second in a  constant magnetic field of intensity <img alt="B_0= 1 \: \: gauss" src="http://entrancec

A coil has turn N = 8000 , average radius r = 10 cm rotating at f = 40 revolution per second in a  constant magnetic field of intensity B_0= 1 \: \: gauss. EMF(maximum ) induced (in volts) in coil is (give answer till 2 decimal places)

Option: 1

6.31


Option: 2

4.20


Option: 3

8.34


Option: 4

0


Answers (1)

best_answer

As we have learned

Area (A) change -

\varepsilon = -NB\cos \Theta \left ( \frac{dA}{dt} \right )

 \phi = NBA \cos \omega t= N\pi r^2 B_0\cos \omega t

 induced emf 

\varepsilon = \frac{-d\phi }{dt}= N \pi r^2 \omega \sin \omega t

NOw for \varepsilon(max )\: \: \sin \omega t = 1  Thus 

\varepsilon_{max }= N \pi r^2 B_0 w \Rightarrow N\pi r^2 \pi f

= 8000 \times 3014\times 10^{-2}\times 10^{-4} \times 2\times 3.14 \times 40 = 6.31 V

 

Posted by

Nehul

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