Question

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

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

Answer 1

Answer: (B)

The 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 Eqs. (i) and (ii), we get
$\frac{L_1}{L_2}={\left(\frac{N_1}{N_2}\right)}^2\left(\because a_1=a_2\right)$
$\frac{108}{L_2}={\left(\frac{600}{500}\right)}^2$
$L_2=\frac{108\times 25}{36}$
$L_2=3\times 25=75mH$