A closely wound coil of 100 turns and of crosssection \(1 \mathrm{~cm}^2\) has coefficient of self inductance 1 mH . The magnetic induction at the centre of the core of a coil when a current of 2 A flows in it, will be (in \(\mathrm{Wb} / \mathrm{m}^2\) )

Given that, \(\mathrm{N}=100, \quad \mathrm{~A}=1 \mathrm{~cm}^2=1 \times 10^{-4} \mathrm{~m}^2\), \(\mathrm{L}=1 \mathrm{mH}=1 \times 10^{-3} \mathrm{H}\) and \(\mathrm{I}=2 \mathrm{~A}\)
Now, self-inductance of a coil is given by,
\(\mathrm{L}=\mu_0 \frac{\mathrm{~N}^2 \mathrm{~A}}{1}...(i)\)
Where
\(\mu_0=\) permeability of free space \(=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\) and \(l\) is the length of the coil
Also, magnetic induction B in the core, \(\mathrm{B}=\mu_0 \frac{\mathrm{NI}}{l}\)
From equation (i),
\(\begin{aligned}
& l=\mu_0 \frac{\mathrm{~N}^2 \mathrm{~A}}{\mathrm{~L}}=\left(4 \pi \times 10^{-7}\right) \frac{(100)^2 \times 1 \times 10^{-4}}{1 \times 10^{-3}} \\
& \quad=4 \pi \times 10^{-7} \times 10^4 \times 10^{-4} \times 10^3 \\
& l=4 \pi \times 10^{-3} \mathrm{~m}
\end{aligned}\)
Substituting this in equation (ii),
\(\begin{aligned}
B & =\left(4 \pi \times 10^{-7}\right) \frac{100 \times 2}{4 \pi \times 10^{-3}} \\
& =10^{-7} \times 200 \times 10^3=0.2 \mathrm{~Wb} / \mathrm{m}^2
\end{aligned}\)