In a coil consisting of 200 turns there is a magnetic flux of 25mVb. Determine the self-induction
| March 1, 2021 | education
In a coil consisting of 200 turns there is a magnetic flux of 25mVb. Determine the self-induction emf in the coil when it is opened if the current in it disappears in 0.02s.
n = 200.
Ф0 = 25 mVb = 0.025 Vb.
Ф = 0 Wb.
t = 0.02 s.
EMF -?
For the phenomenon of self-induction, Faraday’s law is valid, according to which the electromotive force of self-induction EMF is directly proportional to the rate of change of the magnetic flux: EMF = n * ΔF / t, where n is the number of turns in the coil.
The change in the magnetic flux ΔФ is expressed by the formula: ΔФ = Ф0 – F.
Since the magnetic flux decreases to 0, Ф = 0 Wb, then ΔФ = Ф0.
EMF = n * F / t.
EMF = 200 * 0.025 Wb / 0.02 s = 250 V.
Answer: when the coil is turned off, self-induction EMF = 250 V.
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