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A magnetic field of flux density 1.0Wbm2 acts normal to a 80 turn coil of 0.01m2area. 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

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,
ϕ=B×A
where B is the magnetic field
A is the area
According to Faraday’s law,
emf=dϕdt

Complete step-by-step solution
The magnetic flux,
ϕ=B×A
where, B is the magnetic field
A is the area
Then the magnetic flux in 80 turns is given by,
ϕ=0.01×80ϕ=0.8T
Differentiating ϕ with respect to time t we get,
dϕdt=0.8t
Given that t=0.1second
dϕdt=0.80.1dϕdt=8V
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=dϕdt
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.