Question:

The magnetic flux linked with the coil varies with time as $\phi = 3t^2+4t+9$.The magnitude of the induced emf of $2s$ is

Updated On: Aug 1, 2022
  • 9 V
  • 16 V
  • 3 V
  • 4 V
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The Correct Option is B

Solution and Explanation

As emf, $e=\frac{d \phi}{\text { dt }}\left(\frac{d \phi}{d t}\right.$. Aate of charge of magnetic flux) $=\frac{d}{d t}\left(3 t^{2}+4 t+9\right) $ $=6 t+4+0 $ So, at $ t =2 s,$ $ e=6 \times 2+4=16\, V$
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Concepts Used:

Faradays Laws of Induction

There are two laws, given by Faraday which explain the phenomena of electromagnetic induction:

Faraday's First Law:

Whenever a conductor is placed in a varying magnetic field, an emf is induced. If the conductor circuit is closed, a current is induced, known as the induced current.

Faraday's Second Law:

The Emf induced inside a coil is equal to the rate of change of associated magnetic flux.

This law can be mathematically written as:

\(-N {\triangle \phi \over \triangle t}\)