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Three solenoid coils of same dimension, same number of turns and same number of layers of winding are taken. Coil 1 with inductance $\mathrm{L}_{1}$ was wound using a Mn wire of resistance $11 \Omega / \mathrm{m}$; Coil 2 with inductance $\mathrm{L}_{2}$ was wound using the similar wire but the direction of winding was reversed in each layer; Coil 3 with inductance $\mathrm{L}_{3}$ was wound using a superconducting wire. The self inductance of the coils $\mathrm{L}_{1}, \mathrm{~L}_{2}, \mathrm{~L}_{3}$ are
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The correct answer is:
$\mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L}_{3}$
Self-inductance (L) of a solenoid $=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ where $\mu_{o}=$ absolute permittivily of space/ $\mathrm{N}=$ number of turns in the coil
$A=$ area of cross-section of the solenoid
$l=$ length of solenoid
This expression shows that for all three solenoids, the self-inductances will be equal.
$A=$ area of cross-section of the solenoid
$l=$ length of solenoid
This expression shows that for all three solenoids, the self-inductances will be equal.
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