Question

If current in a 130 mH coil changes steadily from 20 mA to 28 mA in 140 ms. Find the magnitude and direction of the induced emf.​

Answer 1

Answer:

see the attachment please

Explanation:

Hope the answer will help you

Answer 2

The magnitude of the induced emf is |ε| = 2.57 * 10^-5 V, and its direction is opposing the change in current.

To Find:

• Magnitude and direction of the induced emf in the coil.

Given:

• Inductance of a coil: 130 mH
• Initial current in the coil: 20 mA
• Final current in the coil: 28 mA
• Time for the change in current: 140 ms

Solution:

The induced emf in a coil can be calculated using Faraday's law of electromagnetic induction:

ε = -dΦ/dt

where ε is the induced emf, Φ is the magnetic flux, and t is time.

In this case, the magnetic flux is proportional to the current in the coil and can be represented by Φ = L * I, where L is the inductance (130 mH) and I is the current in the coil.

So, the induced emf can be calculated as:

ε = -d(L * I)/dt

The change in current from 20 mA to 28 mA in 140 ms gives us a change in magnetic flux of:

dΦ = (28 - 20) * 10^-3 * 130 * 10^-3 = 0.36 * 10^-3 Weber

And the induced emf is:

ε = -dΦ/dt = -0.36 * 10^-3 / 140 * 10^-3 = -2.57 * 10^-5 V

The negative sign indicates that the induced emf opposes the change in current, which is in accordance with Lenz's law.

So, the magnitude of the induced emf is |ε| = 2.57 * 10^-5 V, and its direction is opposing the change in current.

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