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GUJCET · Physics · Electromagnetic Induction

The number of turns in the coil of an AC generator are 100 and its cross sectional area is \(2.5 m^2\). The coil is revolving in a uniform magnetic field of strength 0.3 T with the uniform angular velocity of \(60 rad s ^{-1}\). The value of maximum induced emf is _________ kV.

  1. A 1.25
  2. B 4.5
  3. C 6.75
  4. D 2.25
Verified Solution

Answer & Solution

Correct Answer

(B) 4.5

Step-by-step Solution

Detailed explanation

B
\(\varepsilon=\) N B A \(\omega \sin \omega t\)
For \(\varepsilon_{\max } \sin \omega t=1\)
\(\begin{aligned} \varepsilon_{\max } & = NBA \omega \\ \varepsilon_{\max } & =500 \times 0.3 \times 2.5 \times 60 \\ \varepsilon_{\max } & =4500 V \\ \varepsilon_{\max } & =4.5 kV \end{aligned}\)