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A solenoid of length 11.41 cm has 1079 turns of wire, and a radius of 2.07 cm.A. What is the self-inductance of the solenoid? HB. If a magnetizable core were inserted, such that, for a given current, the magnetic field were 5 times as strong, what would be the new self-inductance of the solenoid? H

Question

A solenoid of length 11.41 cm has 1079 turns of wire, and a radius of 2.07 cm.A. What is the self-inductance of the solenoid? HB. If a magnetizable core were inserted, such that, for a given current, the magnetic field were 5 times as strong, what would be the new self-inductance of the solenoid? H

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Solution

A. The self-inductance (L) of a solenoid can be calculated using the formula:

L = μ₀ * N² * A / l

where: μ₀ = permeability of free space = 4π x 10^-7 H/m N = number of turns = 1079 A = cross-sectional area = πr² = π * (2.07 cm)^2 = 0.0135 m² (converted from cm² to m²) l = length = 11.41 cm = 0.1141 m (converted from cm to m)

Substituting these values into the formula gives:

L = 4π x 10^-7 H/m * 1079² * 0.0135 m² / 0.1141 m = 0.034 H

B. If a magnetizable core is inserted and the magnetic field becomes 5 times as strong, the self-inductance will also increase by a factor of 5. Therefore, the new self-inductance will be:

L' = 5 * L = 5 * 0.034 H = 0.17 H

This problem has been solved

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