A steady current is passing through a coil, magnitude of self- induced emf in it, will be:
Correct answer: A. Zero
- A. Zero
- B. Maximum
- C. ϵ = L(∆ I/ ∆t)
- D. ϵ =- L(∆ I/ ∆t)
Explanation
The magnitude of self-induced electromotive force (emf) in a coil, also known as back emf, is given by Faraday's law of electromagnetic induction. The law states that the induced emf is equal to the negative rate of change of magnetic flux through the coil.However, if a steady current is passing through the coil, the magnetic field produced by the coil is constant. This means that the magnetic flux through the coil is not changing with time. Therefore, the rate of change of magnetic flux is zero.In mathematical terms, this can be expressed as: emf=− ΔΦ/ Δtwhere:emf is the induced electromotive force,Φ is the magnetic flux, andt is time.Since ΔΦ/ Δt=0 for a steady current (because the magnetic flux Φ is not changing with time), the self-induced emf will be zero.So, the magnitude of self-induced emf in a coil with a steady current passing through it will be zero.
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About Faraday's Law
Faraday's law states that an induced electromotive force equals the rate of change of magnetic flux linkage through a circuit. Questions cover changing field, area, angle or motion, the negative sign associated with Lenz's law, and the distinction between induced emf and the current that may flow in a closed circuit.
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