The self-inductance of a solenoid of length 3 | vedclass

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(B) Given:Length of solenoid,$\ell = 31.4 \ cm = 0.314 \ m$Area of cross-section,$A = 10^{-3} \ m^2$Total number of turns,$N = 500$Permeability of fre

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$1 	imes 10^{-3} \ H$

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(B) Given:Length of solenoid,$\ell = 31.4 \ cm = 0.314 \ m$Area of cross-section,$A = 10^{-3} \ m^2$Total number of turns,$N = 500$Permeability of free space,$\mu_0 = 4 \pi 	imes 10^{-7} \ T \cdot m/A$The formula for self-inductance of a solenoid is given by:$L = \frac{\mu_0 N^2 A}{\ell}$Substituting the values:$L = \frac{(4 	imes 3.14 	imes 10^{-7}) 	imes (500)^2 	imes 10^{-3}}{0.314}$$L = \frac{4 	imes 3.14 	imes 10^{-7} 	imes 250000 	imes 10^{-3}}{0.314}$$L = \frac{4 	imes 3.14 	imes 10^{-7} 	imes 2.5 	imes 10^5 	imes 10^{-3}}{0.314}$$L = \frac{4 	imes 3.14 	imes 2.5 	imes 10^{-5}}{0.314}$$L = 4 	imes 10 	imes 2.5 	imes 10^{-5} = 10 	imes 10^{-4} = 10^{-3} \ H$

The self-inductance of a solenoid of length $31.4 \ cm$,area of cross-section $10^{-3} \ m^2$ having a total number of turns $500$ will be nearly $\left[\mu_0 = 4 \pi \times 10^{-7} \ SI \ unit\right]$.

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