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Question: A field of \(\dfrac{{5 \times {{10}^4}}}{\pi }\) ampere-turns/meter acts at right angle to a coil of...

A field of 5×104π\dfrac{{5 \times {{10}^4}}}{\pi } ampere-turns/meter acts at right angle to a coil of 5050 turns of area 10−2m2{10^{ - 2}}{m^2}. The coil is removed from the field in 0.1sec⁡ond0.1\sec ond . Then the induced e.m.f in the coil is
A) 0.1V0.1V
B) 80KV80KV
C) 7.96V7.96V
D) None of the above

Explanation

Solution

The process of inducing a voltage when the magnetic flux is changed is known as electromagnetic Induction. According to Faraday law of electromagnetic induction, how an electric current produces a magnetic field and how a changing magnetic field generates an electric field.

Complete step by step answer:
Given the strength of magnetic field is H=5×104πH = \dfrac{{5 \times {{10}^4}}}{\pi }
Magnetic flux density is calculated by B=μoHB = {\mu _o}H
μo=4π×10−7{\mu _o} = 4\pi \times {10^{ - 7}}
Substituting the values, and solving
⇒B=4π×10−7×5×104π\Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 5 \times {{10}^4}}}{\pi }
⇒B=2×10−2Wm−2\Rightarrow B = 2 \times {10^{ - 2}}W{m^{ - 2}}
The amount of magnetic field going through any particular area is known as magnetic flux. Since the area vector is parallel to the strength of the magnetic field therefore the angle between them is 0.
Flux for a coil with one turn is given by
Φ=B.A\Phi = B.A
⇒Φ=BAcos⁡0\Rightarrow \Phi = BA\cos 0
⇒Φ=BA\Rightarrow \Phi = BA
But for N turns in the coil, the flux is given as
Φ=N.B.A\Phi = N.B.A
NN is the number of turns in the coil
Where BB is the magnetic field and
AA is the area vector.
Induced e.m.f is the rate of change of magnetic flux and is given by
E=−dΦdtE = \dfrac{{ - d\Phi }}{{dt}}
E=−d(N.B.A)dtE = \dfrac{{ - d(N.B.A)}}{{dt}}
⇒E=−NAdBdt\Rightarrow E = - NA\dfrac{{dB}}{{dt}}
Given that the flux is removed after 0.1sec⁡0.1\sec , therefore the induced e.m.f will become zero. But the magnetic flux density is 2×10−22 \times {10^{ - 2}}
⇒E=−50×10−2×(0−2×10−2)0.1\Rightarrow E = \dfrac{{ - 50 \times {{10}^{ - 2}} \times (0 - 2 \times {{10}^{ - 2}})}}{{0.1}}
⇒E=0.1V\Rightarrow E = 0.1V

∴\therefore The induced e.m.f in the coil is 0.1V0.1V. Therefore, option (A) is the right answer.

Note:
It is to be noted that if the number of turns of the coil increases then the amount of induced e.m.f will also change. The negative sign tells the direction of the induced e.m.f. produced. The faster the magnetic field changes in the circuit the greater will be the voltage produced in the circuit. The direction of change in the magnetic field determines the direction of the current.