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A magnetic field of flux density $1.0Wb{{m}^{-2}}$ acts normal to a 80 turn coil of $0.01{{m}^{2}}$area. The e.m.f induced in it, if this coil is removed from the field in 0.1 seconds is:
(A) 8V
(B) 4V
(C ) 10V
(D) 6V

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Last updated date: 18th Jun 2024
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Answer
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Hint: The magnetic flux depends on two factors. That is the magnetic field and the area of the loop. Thus calculate the magnetic flux for 80 turns which is the product of the magnetic field and area. Then find the time derivative of flux. This is obtained by differentiating magnetic flux density with respect to time which is equal to its emf. According to Faraday’s law rate of change of magnetic flux is equal to the induced emf.
Formula used:
The magnetic flux,
$\phi =B\times A$
where B is the magnetic field
A is the area
According to Faraday’s law,
$emf=-\dfrac{d\phi }{dt}$

Complete step-by-step solution
The magnetic flux,
$\phi =B\times A$
where, B is the magnetic field
A is the area
Then the magnetic flux in 80 turns is given by,
$\begin{align}
  & \phi =0.01\times 80 \\
 & \therefore \phi =0.8T \\
\end{align}$
Differentiating $\phi $ with respect to time t we get,
$\dfrac{d\phi }{dt}=\dfrac{0.8}{t}$
Given that t=0.1second
$\begin{align}
  & \Rightarrow \dfrac{d\phi }{dt}=\dfrac{0.8}{0.1} \\
 & \therefore \dfrac{d\phi }{dt}=8V \\
 & \\
\end{align}$
According to Faraday’s law, the rate of change of magnetic flux is equal to the induced emf. The induced emf is equal to the negative rate of change in magnetic flux. The negative sign shows that the induced emf is opposite to the change in magnetic flux.
We know that,
$emf=-\dfrac{d\phi }{dt}$
Thus the induced emf is 8V. Hence, option(A) is correct.

Note: According to Faraday’s law rate of change of magnetic flux is equal to the induced emf. The magnetic flux depends on two factors. That is the magnetic field and the area of the loop. The induced emf is equal to the negative rate of change in magnetic flux. The negative sign shows that the induced emf is opposite to the change in magnetic flux.