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(A) The mutual inductance $M$ of two co-axial solenoids is given by the formula: $M = \mu_0 n_1 n_2 A l$,where $n_1$ and $n_2$ are the number of turns

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$9$

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(A) The mutual inductance $M$ of two co-axial solenoids is given by the formula: $M = \mu_0 n_1 n_2 A l$,where $n_1$ and $n_2$ are the number of turns per unit length,$A$ is the cross-sectional area of the inner solenoid,and $l$ is the length of the solenoids.Given:Length $l = 60 \ cm = 0.6 \ m$Turns per unit length of outer solenoid $n_1 = 15 \ 	ext{turns/cm} = 1500 \ 	ext{turns/m}$Turns per unit length of inner solenoid $n_2 = 40 \ 	ext{turns/cm} = 4000 \ 	ext{turns/m}$Area of inner solenoid $A = 2 	imes 10^{-3} \ m^2$Permeability of free space $\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$Substituting the values:$M = (4\pi 	imes 10^{-7}) 	imes 1500 	imes 4000 	imes (2 	imes 10^{-3}) 	imes 0.6$$M = (4 	imes 3.14 	imes 10^{-7}) 	imes (6 	imes 10^6) 	imes (1.2 	imes 10^{-3})$$M = 12.56 	imes 10^{-7} 	imes 7.2 	imes 10^3$$M \approx 9.04 	imes 10^{-3} \ H = 9 \ mH$.

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