Two coils A and B have 180 and 360 turns resp | vedclass

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(B) The formula for self-inductance is $L = \frac{N \phi}{I}$, where $N$ is the number of turns, $\phi$ is the flux per turn, and $I$ is the current.F

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$2/5$

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(B) The formula for self-inductance is $L = \frac{N \phi}{I}$, where $N$ is the number of turns, $\phi$ is the flux per turn, and $I$ is the current.For coil $A$: $L_A = \frac{N_A \phi_A}{I} = \frac{180 	imes 0.8 	imes 10^{-3} 	ext{ Wb}}{1 	ext{ A}} = 144 	imes 10^{-3} 	ext{ H}$.For coil $B$: $L_B = \frac{N_B \phi_B}{I} = \frac{360 	imes 1 	imes 10^{-3} 	ext{ Wb}}{1 	ext{ A}} = 360 	imes 10^{-3} 	ext{ H}$.The ratio of self-inductances is $\frac{L_A}{L_B} = \frac{144 	imes 10^{-3}}{360 	imes 10^{-3}} = \frac{144}{360}$.Simplifying the fraction: $\frac{144}{360} = \frac{12}{30} = \frac{2}{5}$.

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