Search any question & find its solution
Question:
Answered & Verified by Expert
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $\mathrm{A}=10 \mathrm{~cm}^{2}$ and length $=20 \mathrm{~cm} .$ If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}\right)$
Options:
Solution:
1996 Upvotes
Verified Answer
The correct answer is:
$2.4 \pi \times 10^{-4} \mathrm{H}$
$\begin{aligned} M &=\frac{\mu_{0} \mathrm{~N}_{1} \mathrm{~N}_{2} \mathrm{~A}}{\ell} \\ &=\frac{4 \pi \times 10^{-7} \times 300 \times 400 \times 100 \times 10^{-4}}{0.2} \\ &=2.4 \pi \times 10^{-4} \mathrm{H} \end{aligned}$
#
#
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.