Question

A coil of wire of radius $r$ has $600$ turns and self-inductance of $108mH$ . The self-inductance of a coil with the same radius and $500$ turns is

• A. $80mH$
• B. $75mH$
• C. $108mH$
• D. $90mH$

Answer 1

Answer: (B)

Given $L_1=108mH,N_1=600$ 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
$\frac{L_1}{L_2}={\left(\frac{N_1}{N_2}\right)}^2$ $\left(\because a_1=a_2\right)$
$\frac{108}{L_2}={\left(\frac{600}{500}\right)}^2$
$L_2=\frac{108\times 25}{36}$
$L_2=3\times 25=75mH$