Thank you for visiting Find the self inductance of a 100 turn solenoid with a cross sectional area of 10 cm 2 and a length of 62 8 cm. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
The self-inductance of a solenoid can be calculated using the formula[tex]L = (μâ‚€ * N² * A) / l[/tex], where L is the self-inductance, N is the number of turns, A is the cross-sectional area, and l is the length of the solenoid.
In this case, we are given that the solenoid has 100 turns, a cross-sectional area of 10 cm², and a length of 62.8 cm.
First, let's convert the given values into SI units. The cross-sectional area can be converted from cm² to m² by dividing by [tex]10000 (1 m² = 10000 cm²)[/tex]. So, the cross-sectional area is[tex]10 cm² / 10000 = 0.001 m².[/tex]
Next, let's convert the length from cm to m by dividing by 100 (1 m = 100 cm). So, the length is 62.8 cm / 100 = 0.628 m.
Now, we can substitute the values into the formula:
[tex]L = (μâ‚€ * N² * A) / l = (4π * 10^-7 T·m/A * 100² turns² * 0.001 m²) / 0.628 m[/tex]
Calculating this expression gives us the self-inductance of the solenoid.
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