If a conductor cuts 2 webers of flux in 2 seconds and has 50 turns, then emf induced in the conductor is:

Explanation:

Understanding Electromagnetic Induction:

Electromagnetic induction is the process of generating an electromotive force (emf) across an electrical conductor in a changing magnetic field. This phenomenon was first discovered by Michael Faraday in 1831 and is described by Faraday's Law of Electromagnetic Induction.

Faraday's Law of Electromagnetic Induction:

Faraday's Law states that the induced emf in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit. Mathematically, it can be expressed as:

emf = -N × (dΦ/dt)

Where:

• emf is the electromotive force in volts (V).
• N is the number of turns in the coil.
• dΦ/dt is the rate of change of magnetic flux (Φ) in webers per second (Wb/s).

The negative sign indicates the direction of the induced emf and current, as given by Lenz's Law, which states that the induced emf will always work to oppose the change in magnetic flux that produced it.

Given Problem:

In the given problem, we have the following information:

• The magnetic flux (Φ) cut by the conductor is 2 webers (Wb).
• The time (t) taken to cut the flux is 2 seconds (s).
• The number of turns (N) in the conductor is 50.

We need to calculate the induced emf in the conductor using the provided data.

Step-by-Step Solution:

1. Calculate the rate of change of magnetic flux (dΦ/dt):

Since the magnetic flux (Φ) changes from 0 to 2 webers in 2 seconds, the rate of change of magnetic flux can be calculated as:

dΦ/dt = Φ / t = 2 Wb / 2 s = 1 Wb/s

2. Use Faraday's Law to calculate the induced emf:

Substitute the values of N and dΦ/dt into Faraday's Law equation:

emf = -N × (dΦ/dt)

Since the rate of change of flux is 1 Wb/s and the number of turns is 50:

emf = -50 × 1 = -50 V

The negative sign indicates the direction of the induced emf as per Lenz's Law. However, for the magnitude of the emf, we consider only the absolute value:

emf = 50 V

Therefore, the induced emf in the conductor is 50 volts.

Correct Option Analysis:

The correct option is:

Option 2: 50 V

This option correctly represents the induced emf calculated using Faraday's Law of Electromagnetic Induction. The given data and the step-by-step solution lead us to the conclusion that the induced emf is indeed 50 volts.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 200 V

This option is incorrect. If we were to calculate the induced emf as 200 volts, it would imply an error in the calculation or misinterpretation of the given data. The correct calculation using Faraday's Law shows that the induced emf is 50 volts.

Option 3: 100 V

This option is also incorrect. To arrive at an induced emf of 100 volts, either the number of turns or the rate of change of magnetic flux would need to be different from the given values. The accurate calculation shows an induced emf of 50 volts.

Option 4: 250 V

This option is incorrect as well. An induced emf of 250 volts does not align with the given data and the application of Faraday's Law. The correct induced emf is 50 volts.

Understanding the principles of electromagnetic induction and the application of Faraday's Law is essential for solving problems related to induced emf. In this case, accurately applying the formula and substituting the correct values leads us to the correct conclusion that the induced emf is 50 volts.