Problem 35
Question
A coil of wire of certain radius has 100 turns and a self-inductance of \(15 \mathrm{mH}\). The self-inductance of a second similar coil of 500 turns will be (a) \(75 \mathrm{mH}\) (b) \(375 \mathrm{mH}\) (c) \(15 \mathrm{mH}\) (d) None of these
Step-by-Step Solution
Verified Answer
The self-inductance of the second coil is 375 mH, option (b).
1Step 1: Understanding Self-Inductance
Self-inductance, denoted by \( L \), of a coil is a measure of how much voltage is induced in the coil when the current through it changes. It depends on the square of the number of turns in the coil.
2Step 2: Using the Formula for Self-Inductance
For a coil, the self-inductance \( L \) is proportional to the square of the number of turns \( N \) and can be expressed by the relationship \( L \propto N^2 \). If we have an initial coil with \( N_1 = 100 \) turns and \( L_1 = 15 \mathrm{mH} \), and a second similar coil with \( N_2 = 500 \) turns, we can use the proportional relationship to determine \( L_2 \).
3Step 3: Setting Up Proportional Equation
The relationship between the self-inductances of the two coils can be set up as \( \frac{L_2}{L_1} = \left( \frac{N_2}{N_1} \right)^2 \). Here, \( N_1 = 100 \) and \( N_2 = 500 \). Substituting the known values, the equation becomes \( \frac{L_2}{15} = \left( \frac{500}{100} \right)^2 \).
4Step 4: Calculating the Ratio
Calculate \( \left( \frac{500}{100} \right)^2 = \left( 5 \right)^2 = 25 \). So, the equation becomes \( \frac{L_2}{15} = 25 \).
5Step 5: Solving for L2
To find \( L_2 \), multiply both sides of the equation by 15: \( L_2 = 25 \times 15 \).
6Step 6: Final Calculation
Calculate \( 25 \times 15 = 375 \). Therefore, the self-inductance \( L_2 = 375 \mathrm{mH} \).
Key Concepts
Number of Turns in a CoilProportional RelationshipsElectromagnetic Induction
Number of Turns in a Coil
The number of turns in a coil, often denoted as \( N \), plays a crucial role in determining the coil's electrical properties. Each turn of the coil contributes to its ability to produce magnetic fields and store energy as the coil is energized. A higher number of turns results in a stronger magnetic field, enhancing the coil's inductive properties.
- More turns lead to increased self-inductance, making the coil more effective at storing magnetic energy.
- Coils with more turns will typically have higher resistance due to the longer length of wire required.
Proportional Relationships
Proportional relationships are an essential part of understanding how different quantities interact in physics. When one quantity changes, another can change in a predictable manner according to a set rule.
The concept of proportionality is key when calculating self-inductance in coils. The self-inductance \( L \) is proportional to the square of the number of turns \( N \): \( L \propto N^2 \). This means:
The concept of proportionality is key when calculating self-inductance in coils. The self-inductance \( L \) is proportional to the square of the number of turns \( N \): \( L \propto N^2 \). This means:
- If you double the number of turns, the self-inductance will increase by a factor of four (since \((2N)^2 = 4N^2\)).
- If the number of turns is increased five times (as in our 100 to 500 turns scenario), the self-inductance will increase by 25 times the original (since \((5N)^2 = 25N^2\)).
Electromagnetic Induction
Electromagnetic induction is the process by which a changing magnetic field within a coil induces a voltage (or electromotive force). This principle underpins the operation of many electrical devices, such as transformers and inductors, and is also the basis for self-inductance.
- It occurs when the magnetic field around a conductor varies, either by changing the current passing through the conductor or by moving the conductor within the magnetic field.
- Induced voltage depends on the rate of change of the magnetic field and the number of turns in the coil.
Other exercises in this chapter
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