The answer is A. Zero.

Here's why:

Self-induced emf in a coil is the emf induced in the coil due to its own changing magnetic field. This is given by Faraday's law of electromagnetic induction as ε = -L (dI/dt), where L is the inductance of the coil and dI/dt is the rate of change of current.

However, in this case, it is mentioned that a steady current is passing through the coil. A steady current means that the current is not changing with time, i.e., dI/dt = 0.

Therefore, the self-induced emf will be ε = -L * 0 = 0.

So, the magnitude of the self-induced emf in the coil will be zero.

Question

The answer is A. Zero.

Here's why:

Self-induced emf in a coil is the emf induced in the coil due to its own changing magnetic field. This is given by Faraday's law of electromagnetic induction as ε = -L (dI/dt), where L is the inductance of the coil and dI/dt is the rate of change of current.

However, in this case, it is mentioned that a steady current is passing through the coil. A steady current means that the current is not changing with time, i.e., dI/dt = 0.

Therefore, the self-induced emf will be ε = -L * 0 = 0.

So, the magnitude of the self-induced emf in the coil will be zero.

Knowee AI · Accepted Answer

The answer is A. Zero.

Here's why:

Self-induced emf in a coil is the emf induced in the coil due to its own changing magnetic field. This is given by Faraday's law of electromagnetic induction as ε = -L (dI/dt), where L is the inductance of the coil and dI/dt is the rate of change of current.

However, in this case, it is mentioned that a steady current is passing through the coil. A steady current means that the current is not changing with time, i.e., dI/dt = 0.

Therefore, the self-induced emf will be ε = -L * 0 = 0.

So, the magnitude of the self-induced emf in the coil will be zero.

A steady current is passing through a coil, magnitude of self- induced emf in it,will be:A. zero ⃝ B. maximum ⃝C.  = LIt ⃝ D.  =  ILt

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