A certain volume of copper is drawn into a wire of radius a and is wrapped in shape of a helix having radius $r (> > a)$ . The winding are as close as possible without overlapping. Self-inductance of the inductor so obtained is $L_{1}$ . Another wire of radius $2 a$ is drawn using same volume of copper and wound in the fashion as described above. This time the inductance is $L_{2}$ . Find $\frac{L_{1}}{L_{2}}$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Length of wire $l = \frac{v o l u m e (V)}{π a^{2}}$
Number of turns $N = \frac{l}{2 π r}$
Length of the helix $b = 2 a . N = \frac{a l}{π r}$
Number of turns per meter length $n = \frac{1}{2 a}$
$∴$ Self-inductance
$\begin{array}{l} L = π μ_{0} n^{2} r^{2} b \\ = π μ_{0} . (\frac{1}{2 a}) . r^{2} . \frac{a l}{π r} \\ = \frac{1}{4} μ_{0} r \frac{l}{a} = \frac{1}{4} μ_{0} r \frac{V}{a (π a^{2})} \\ = \frac{μ_{0} r}{4 π} \frac{V}{a^{3}} \\ \frac{L_{1}}{L_{2}} = {(\frac{2 a}{a})}^{3} \end{array}$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET