Question 231: Electromagnetic Induction

Question

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

Accepted Answer

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

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Electromagnetic Induction

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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