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 $\mathrm{Wb} / \mathrm{m}^2$ )

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Aicharya · Accepted Answer

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), $\begin{aligned} B & =\left(4 \pi \times 10^{-7}\right) \frac{100 \times 2}{4 \pi \times 10^{-3}} \\ & =10^{-7} \times 200 \times 10^3=0.2 \mathrm{~Wb} / \mathrm{m}^2 \end{aligned}$

A closely wound coil of 100 turns and of crosssection $1 \mathrm{~cm}^2$ has coefficient of self inductance…

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