Question

An emf of 2 V is induced in a coil when the current in it is changed from 0 A to 10 A in 0.40 s. Find the coefficient of self-inductance of the coil.

(a) 0.1 H
(b) 0.2 H
(c) 0.3 H
(d) 0.4 H

Answer 1

Using Faraday's law of induction, the coefficient of self-inductance (L) is calculated as 0.08 H. However, this result is not within the provided options; the closest is 0.1 H, which may indicate a typo in the options.

Step-by-step explanation:

To find the coefficient of self-inductance of the coil, we can use the formula for induced electromotive force (emf) given by Faraday's law of induction, which is:

E = -L ΔI/Δt

Where:

• E is the induced emf
• L is the coefficient of self-inductance
• ΔI is the change in current
• Δt is the change in time

According to the question:

• E = 2 V
• ΔI = 10 A - 0 A = 10 A
• Δt = 0.40 s

Plugging these values into the formula, we get:

2 V = -L (10 A) / 0.40 s

Solving for L, the self-inductance of the coil:

L = -2 V × 0.40 s / 10 A = 0.08 H

However, since the induced emf (E) is in the direction to oppose the change in current, we take the magnitude of L which gives us:

L = 0.08 H

This is not one of the options provided, but this result indicates that the closest option would be (b) 0.1 H, which may suggest that there is a possible typo in the provided options.