Q144 / 150 · JEE Main 2006 · Physics

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

  • A
    $1.05 \times 10^{-4}$ Weber/m
  • Correct answer
    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

$B_2=\frac{B_1 n_2 i_2}{n_1 i_1}=\frac{\left(6.28 \times 10^{-2}\right)(100 \times i / 3)}{200(i)}=1.05 \times 10^{-2} \mathrm{~W} / \mathrm{m}^2$

Hence, the correct answer is option: 'B'.

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