Problem 68 - Master Resource Book in JEE Main Physics

Question

If the self-inductance of 500 turn coil is \(125 \mathrm{mH}\), then the self- inductance of similar coil of 800 turns is [Kerala CET 2008] (a) \(48.8 \mathrm{mH}\) (b) \(200 \mathrm{mH}\) (c) \(187.5 \mathrm{mH}\) (d) \(320 \mathrm{mH}\)

Step-by-Step Solution

Answer

The self-inductance of the coil with 800 turns is 320 mH.

Step 1: Understand the Formula

The self-inductance of a coil is directly proportional to the square of the number of turns of the coil. Mathematically, this can be expressed as \( L \propto N^2 \), where \( L \) is the inductance and \( N \) is the number of turns.

Step 2: Set up the Proportional Relationship

Given that \( L_1 = 125 \text{ mH} \) for a coil with \( N_1 = 500 \) turns, and \( N_2 = 800 \) turns, we need to find \( L_2 \). The relationship can be expressed as: \[ \frac{L_2}{L_1} = \left(\frac{N_2}{N_1}\right)^2 \].

Step 3: Insert Given Values into the Formula

Insert the given values into the proportionality relationship: \[ \frac{L_2}{125} = \left( \frac{800}{500} \right)^2 \].

Step 4: Calculate the Proportion

First, simplify the fraction \( \frac{800}{500} = \frac{8}{5} \). Then, calculate \( \left(\frac{8}{5}\right)^2 = \left( \frac{64}{25} \right) = 2.56 \).

Step 5: Solve for L2

Use the proportionality to solve for \( L_2 \): \[ L_2 = 125 \times 2.56 = 320 \text{ mH} \].

Key Concepts

Number of Turns in a Coil

The number of turns in a coil is a crucial factor in determining the self-inductance of the coil. This is because the inductance is directly affected by how many loops the wire makes in the coil. In simple terms, more turns mean more magnetic interactions. As electric current passes through the coil, it generates a magnetic field. The more turns there are, the greater the magnetic field and thus the higher the inductance.
Consider a coil with several loops of wire. Each loop adds to the total magnetic field generated. If a coil has 500 turns and generates a certain level of inductance, increasing the turns to 800 can significantly increase the inductance.
- More turns = stronger magnetic field - Stronger magnetic field = higher self-inductance
Remember, though, it's not just about the number of turns. The shape and material of the coil and whether it has a core can also influence the overall inductance.

Inductance Proportionality

Inductance proportionality helps us understand how the inductance of a coil changes with different numbers of turns. The relationship is directly proportional to the square of the number of turns. This means if the number of turns in a coil doubles, the inductance doesn't just double, it increases by fourfold.
In mathematical terms, this is expressed as: \[ L \propto N^2 \] where- \(L\) = Inductance- \(N\) = Number of turns
This proportion demonstrates why coils with more turns tend to have much higher inductance values. A change in the number of turns exponentially affects the magnetic field and energy stored within the coil.

Calculation of Inductance

Calculating inductance involves applying the proportional relationship between the number of turns and inductance. For practical calculations, we can use the formula: \[ \frac{L_2}{L_1} = \left(\frac{N_2}{N_1}\right)^2 \] Consider a scenario where you have two coils, one with 500 turns and an inductance of 125 mH, and you want to calculate the inductance of another coil with 800 turns. Here's how:

• Find the ratio of the number of turns: \( \frac{800}{500} = \frac{8}{5} \)
• Square this ratio: \( \left( \frac{8}{5} \right)^2 = \frac{64}{25} = 2.56 \)
• Multiply by the original inductance: \( L_2 = 125 \times 2.56 = 320 \text{ mH} \)

This step-by-step breakdown simplifies how we discover the increased inductance based on turns. It demonstrates the power of mathematical relationships in understanding real-world applications like coils in electric circuits.