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Question: The flux linked with a coil at any instant \( 't' \) is given by \( \phi = 10{t^2} - 50t + 250 \) . ...

The flux linked with a coil at any instant ′t′'t' is given by ϕ=10t2−50t+250\phi = 10{t^2} - 50t + 250 . The induced emf at t=3st = 3s is
(A)−190V(A) - 190V
(B)−10V(B) - 10V
(C)10V(C)10V
(D)190V(D)190V

Explanation

Solution

To solve this question, Faraday’s second law will be used. Faraday’s second law states that the induced voltage in a circuit is proportional to the rate of change over time of the magnetic flux through that circuit. This law was given by the English scientist Michael Faraday.

Complete step by step solution:
Faraday gave two laws which are as follows:
(1) Faraday’s first law: Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced. If the conductor circuit is closed, a current is induced, which is called induced current.
(2) Faraday’s second law: Faraday’s second law of electromagnetic induction states that the magnitude of induced emf is directly proportional to the time rate of change in magnetic flux linked with the circuit.
ε∝dϕdx\varepsilon \propto \dfrac{{d\phi }}{{dx}}
ε=−dϕdx.......(1)\varepsilon = - \dfrac{{d\phi }}{{dx}}.......(1)
Where, ϕ\phi is the flux of the magnetic field through the area
Another important law which is used to find the direction is the Lenz law. Lenz’s law states that the polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
The negative sign in equation (1) represents this effect. Thus, the negative sign indicates that the direction of the induced emf and change in the direction of magnetic fields is opposite.
Now, according to the Faraday’s second law,
ε=−dϕdx\varepsilon = - \dfrac{{d\phi }}{{dx}}
On putting the value of flux ϕ\phi , we get,
ε=−d(10t2−50t+250)dx\varepsilon = - \dfrac{{d(10{t^2} - 50t + 250)}}{{dx}}
On applying differentiation,
ε=−(20t−50)V\varepsilon = - (20t - 50)V
On putting t=3st = 3s , we get,
ε=−[(20×3)−50]V\varepsilon = - [(20 \times 3) - 50]V ε=−10V\varepsilon = - 10V
ε=−[60−50]V\varepsilon = - [60 - 50]V
On opening the bracket,
ε=−10V\varepsilon = - 10V
So, the final answer is (B)−10V(B) - 10V .

Note:
If the flux increases with time, dϕdx\dfrac{{d\phi }}{{dx}} positive and ε\varepsilon is negative. The current which flows in the circuit is negative. It means the direction will be opposite to the arrow put on the loop. But if flux decreases with time, dϕdx\dfrac{{d\phi }}{{dx}} is negative and ε\varepsilon is positive. The current which flows in the circuit is positive. It means the direction of current is along the arrow.