Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A long solenoid has 200 turns per $\mathrm{cm}$ and carries a current $\mathrm{i}$. The magnetic field at its centre is $6.28 \times 10^{-2}$ Weber $/ \mathrm{m}^2$. Another long solenoid has 100 turns per $\mathrm{cm}$ and it carries a current $\mathrm{i} / 3$. The value of the magnetic field at its centre is
PhysicsMagnetic Effects of CurrentJEE MainJEE Main 2006
Options:
  • A
    $1.05 \times 10^{-4}$ Weber/m
  • B
    $1.05 \times 10^{-2}$ Weber $/ \mathrm{m}^2$
  • C
    $1.05 \times 10^{-5} \mathrm{Weber} / \mathrm{m}^2$
  • D
    $1.05 \times 10^{-3} \mathrm{Weber} / \mathrm{m}^2$
Solution:
2361 Upvotes Verified Answer
The correct answer is:
$1.05 \times 10^{-2}$ Weber $/ \mathrm{m}^2$
$B_2=\frac{B_1 n_2 i_2}{n_1 i_1}=\frac{\left(6.28 \times 10^{-2}\right)(100 \times i / 3)}{200(i)}=1.05 \times 10^{-2} \mathrm{~W} / \mathrm{m}^2$

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.