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Q.
A coil of wire of radius $r$ has $600$ turns and self-inductance of $108\,mH$ . The self-inductance of a coil with the same radius and $500$ turns is
NTA AbhyasNTA Abhyas 2022
Solution:
Given $L_{1}=108 \, mH, \, N_{1}=600 \, $ turns, $\, N_{2}=500 \,$ turns and $L_{2}=?$
By self inductance of a plane coil
$L_{1}=\frac{\mu _{0} \pi N_{1}^{2} a_{1}}{2}$ ......(i)
$L_{2}=\frac{\mu _{0} \pi N_{2}^{2} a_{2}}{2}$ ......(ii)
From the equations (i) and (ii), we get
$\frac{L_{1}}{L_{2}}=\left(\frac{N_{1}}{N_{2}}\right)^{2} \, $ $\left(\because \, a_{1} = a_{2}\right)$
$\frac{108}{L_{2}}=\left(\frac{600}{500}\right)^{2}$
$L_{2}=\frac{108 \times 25}{36}$
$L_{2}=3\times 25=75 \, mH$