Two coaxial solenoids are made by winding thi | vedclass

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Question

(A) The formula for the mutual inductance $M$ of two coaxial solenoids is given by $M = \frac{\mu_0 N_1 N_2 A}{\ell}$.Given values are:$A = 10 \ cm^2

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$2.4\pi 	imes 10^{-4} \ H$

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(A) The formula for the mutual inductance $M$ of two coaxial solenoids is given by $M = \frac{\mu_0 N_1 N_2 A}{\ell}$.Given values are:$A = 10 \ cm^2 = 10 	imes 10^{-4} \ m^2 = 10^{-3} \ m^2$$\ell = 20 \ cm = 0.2 \ m$$N_1 = 300$$N_2 = 400$$\mu_0 = 4\pi 	imes 10^{-7} \ T \ m \ A^{-1}$Substituting these values into the formula:$M = \frac{(4\pi 	imes 10^{-7}) 	imes 300 	imes 400 	imes 10^{-3}}{0.2}$$M = \frac{4\pi 	imes 10^{-7} 	imes 120000 	imes 10^{-3}}{0.2}$$M = \frac{4\pi 	imes 10^{-7} 	imes 120}{0.2}$$M = 4\pi 	imes 10^{-7} 	imes 600$$M = 2400\pi 	imes 10^{-7} \ H = 2.4\pi 	imes 10^{-4} \ H$.

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