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An emf of 5 V is produced by self - inductance, when the current changes at a steady rate from 3 A to 2 A in 1 ms. The value of self - inductance is
- a)500 H
- b)250 H
- c)5 mH
- d)zero
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An emf of 5 V is produced by self - inductance, when the current chang...
Concept:
- The self-inductance of a coil is defined as the ratio of the magnetic flux and the current flowing through it.
- In a conducting coil, magnetic flux is the product of the magnetic field due to the current flowing in the coil and the area of the coil.
- According to Faraday's law, in a conducting wire, the change in electric current changes the magnetic flux associated with it, and the change in the magnetic flux induces an electric current.
- According to Lenz law, the direction of the induced current is such that the initial change in the electric current is opposed.
- As soon as the change in electric current stops, the induced current also stops flowing.
- The formula to calculate the electromotive force due to induction is,
Where E is the electromotive force or the potential, t is the time elapsed, L is the self-inductance of the coil and ΔI is the change in the electric current of the coil.
- The S.I unit of self-inductance is Henry(H).
Calculation:
Given: E = 5 V, t = 1 ms = 0.001 s, ΔI = (2A - 3A) = - 1A
According to Faraday's Law,
Given: E = 5 V, t = 1 ms = 0.001 s, ΔI = (2A - 3A) = - 1A
According to Faraday's Law,
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An emf of 5 V is produced by self - inductance, when the current chang...
Given:
- EMF (electromotive force) = 5 V
- Change in current (ΔI) = 3 A - 2 A = 1 A
- Time taken for the change (Δt) = 1 ms = 0.001 s
Formula:
The induced EMF (ε) in a coil is given by the formula:
ε = -L * (ΔI/Δt)
Where:
ε = induced EMF
L = self-inductance of the coil
ΔI = change in current
Δt = time taken for the change
Calculation:
Given that ε = 5 V, ΔI = 1 A, and Δt = 0.001 s, we can rearrange the formula to solve for L:
L = -ε * (Δt/ΔI)
Substituting the given values:
L = -5 V * (0.001 s / 1 A)
L = -5 * 0.001 V·s/A
L = -0.005 V·s/A
Since self-inductance is always positive, the magnitude of L is considered:
L = 0.005 V·s/A = 5 mH (millihenries)
Conclusion:
The value of self-inductance is 5 mH (option C).
- EMF (electromotive force) = 5 V
- Change in current (ΔI) = 3 A - 2 A = 1 A
- Time taken for the change (Δt) = 1 ms = 0.001 s
Formula:
The induced EMF (ε) in a coil is given by the formula:
ε = -L * (ΔI/Δt)
Where:
ε = induced EMF
L = self-inductance of the coil
ΔI = change in current
Δt = time taken for the change
Calculation:
Given that ε = 5 V, ΔI = 1 A, and Δt = 0.001 s, we can rearrange the formula to solve for L:
L = -ε * (Δt/ΔI)
Substituting the given values:
L = -5 V * (0.001 s / 1 A)
L = -5 * 0.001 V·s/A
L = -0.005 V·s/A
Since self-inductance is always positive, the magnitude of L is considered:
L = 0.005 V·s/A = 5 mH (millihenries)
Conclusion:
The value of self-inductance is 5 mH (option C).
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An emf of 5 V is produced by self - inductance, when the current changes at a steady rate from 3 A to 2 A in 1 ms. The value of self - inductance isa)500 Hb)250 Hc)5 mHd)zeroCorrect answer is option 'C'. Can you explain this answer?
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