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Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1: 2 . The ratio of self-inductance of two solenoids is
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1: 2
From $L=\frac{\mu_{0} N^{2} A}{1} \alpha \frac{N^{2}}{1}$
we get, $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{(1 / 2)^{2}}{1 / 2}=\frac{1}{2}$
we get, $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{(1 / 2)^{2}}{1 / 2}=\frac{1}{2}$
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