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