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(B) The inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$,which implies $L \propto \frac{N^2}{l}$.When a coil is divided into $n = 6$ e

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$\frac{L}{36}$

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(B) The inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$,which implies $L \propto \frac{N^2}{l}$.When a coil is divided into $n = 6$ equal parts,the number of turns in each part becomes $N' = \frac{N}{6}$ and the length of each part becomes $l' = \frac{l}{6}$.The inductance of each part $L'$ is given by $L' = L \left( \frac{N'}{N} \right)^2 \left( \frac{l}{l'} \right) = L \left( \frac{1}{6} \right)^2 \left( \frac{l}{l/6} \right) = L \left( \frac{1}{36} \right) (6) = \frac{L}{6}$.When these $6$ parts are connected in parallel,the equivalent inductance $L_e$ is given by $\frac{1}{L_e} = \sum \frac{1}{L'} = \frac{1}{L'} + \frac{1}{L'} + \dots + \frac{1}{L'} = \frac{6}{L'}$.Substituting $L' = \frac{L}{6}$,we get $\frac{1}{L_e} = \frac{6}{L/6} = \frac{36}{L}$.Therefore,$L_e = \frac{L}{36}$.

$A$ coil of inductance $L$ is divided into $6$ equal parts. All these parts are connected in parallel. The resultant inductance of this combination is

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