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The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is
  • a)
    1 henry
  • b)
    4 henry
  • c)
    10–3 henry
  • d)
    4 × 10–3 henry
Correct answer is option 'D'. Can you explain this answer?
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To find the self-inductance of the coil, we can use Faraday's law of electromagnetic induction, which states that the induced emf (ε) in a coil is equal to the rate of change of magnetic flux (Φ) through the coil.

Mathematically, ε = -L * (dI/dt), where ε is the induced emf, L is the self-inductance of the coil, and (dI/dt) is the rate of change of current.

Given that the current changes from 1 ampere to zero in one millisecond, we can calculate the rate of change of current as (dI/dt) = (0 - 1 A) / (0.001 s) = -1000 A/s.

We are also given that the induced emf is 4 volts, so we can plug these values into the equation to solve for the self-inductance:

4 V = -L * (-1000 A/s)
4 V = 1000 L
L = 4 V / 1000 A/s
L = 0.004 H

Therefore, the self-inductance of the coil is 0.004 henry, which is approximately equal to 4 millihenries. Thus, the correct option is (b) 4 henry.
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The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil isa)1 henryb)4 henryc)10–3 henryd)4 × 10–3 henryCorrect answer is option 'D'. Can you explain this answer?
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