Question

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current 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) \(1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\)
(b) \(1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\)
(c) \(1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\)
(d) \(1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\).

Answer 1

The correct answer Is:
(b) : \(\ln\) first case, \(B_{1}=\mu_{0} \mu_{1} l_{1}\)
In secoond case, \(B_{2}=\mu_{9} r_{2} l_{2}\)
$$
\begin{array}{ll}
\therefore \quad & \frac{B_{2}}{B_{1}}=\frac{\pi_{1}}{n_{1}} \times \frac{I_{2}}{I_{1}}=\frac{100}{200} \times \frac{1 / 3}{t}=\frac{1}{6} \\
\therefore \quad & B_{2}-\frac{B_{1}}{6}-\frac{628 \times 10^{-2}}{6} \\
& =1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2} .
\end{array}
$$

Answer 2

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Answer 3

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