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Two coaxial solenoids are wound with thin insulated wire on the same pipe of cross-sectional area A = 10 cm² and length 20 cm. One solenoid has 300 turns and the other has 400 turns. Find their mutual inductance. (mu0 = 4*pi*10⁻⁷ T m/A)

1. 2.4*pi*10⁻⁵ H
2. 4.8*pi*10⁻⁴ H
3. 4.8*pi*10⁻⁵ H
4. 2.4*pi*10⁻⁴ H

Correct answer: 2.4*pi*10⁻⁵ H

Solution

The mutual inductance of two coaxial solenoids sharing the same cross-section is M = mu0*N1*N2*A/l. Substituting: M = (4*pi*10⁻⁷)*(300)*(400)*(10*10⁻⁴)/(0.20). Compute the numerator coefficient: 4*pi*10⁻⁷ * 1.2*10⁵ * 1.0*10⁻³ = 4*pi*10⁻⁷ * 1.2*10² = 4.8*pi*10⁻⁵; dividing by 0.20 gives 2.4*pi*10⁻⁵ H.

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