A toroidal solenoid has 550 turns, cross-sectionalarea 6.00cm2, and – [Free] B16
Part AA toroidal solenoid has 550 turns, cross-sectionalarea 6.00cm2, and mean radius 4.60 cm .Calculate the coil’s self-inductance.Express your answer in henries.Previous AnswersRequest AnswerIncorrect; Try Again; 5 attempts remainingAl Study ToolsLooking for some guidance? Let’s work through a few relatedpractice questions before you go back to the real thing.This won’t impact your score, so stop at anytime and ask forclarification whenever you need it.Ready to give it a try?Part BIf the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms ,calculate the self-induced emf in the coil.Express your answer in volts.
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Answer
Self-Inductance and Self-Induced EMF in a Toroidal Solenoid
🔹 Part A: Self-Inductance Calculation
For a toroidal solenoid, the self-inductance is calculated using the formula:
Given values:
- Permeability of free space, μ₀ = 4π × 10⁻⁷ H/m
- Number of turns, N = 550
- Cross-sectional area, A = 6 cm² = 6 × 10⁻⁴ m²
- Mean radius, r = 4.60 cm = 4.6 × 10⁻² m
Substituting the values into the formula:
Calculating further:
🔹 Part B: Self-Induced EMF Calculation
The self-induced emf in the solenoid is given by:
Given:
- Initial current, I₁ = 5.00 A
- Final current, I₂ = 2.00 A
- Time interval, Δt = 3.00 ms = 3.00 × 10⁻³ s
Calculating rate of change of current:
Taking absolute value:
Now calculating emf:
✅ Final Answers:
- Self-Inductance: L = 7.89 × 10⁻⁴ H
- Self-Induced EMF: ε = 0.789 V