Two coaxial solenoids are made by winding thi | vedclass

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(D) 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:$N_1 = 300$$N_2 = 4

Vedclass · Accepted Answer

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

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(D) 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:$N_1 = 300$$N_2 = 400$$A = 10 \, cm^2 = 10 	imes 10^{-4} \, m^2 = 10^{-3} \, m^2$$\ell = 20 \, cm = 0.2 \, m = 2 	imes 10^{-1} \, m$$\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m \cdot 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$$M = 2.4\pi 	imes 10^{-4} \, H$.

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