The calculated mutual inductance is 8.544 x 10⁻⁵ H.
Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.
Length of the solenoid= 5.0 cm
Area of cross-section=1.0 cm²
no of spaced turns=300 turns
turns of insulated wire=180 turns
Mutual inductance (M) = μ₀μr N1N2 A/ L
=(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5
=79.12 x 10⁻¹⁰ x 54000 / 5
=8.544 x 10⁻⁵ H
hence, the mutual inductance is 8.544 x 10⁻⁵ H.
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