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Q)

A circular coil of wire consisting of 100 times of radius $8.0\;cm$ carries a current of $0.40\;A$ What is the magnitude of the magnetic field B at the center of the coil ?

$\begin{array}{1 1} 3.1 \times 10^{-4}\;T \\ 3.1 \times 10^{-3}\;T \\ 3.1 \times 10^{4}\;T \\ 3.1 \times 10^{3}\;T\end{array} $

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A)
Solution :
given $n=100$
$r= 8\;cm$
$\quad= 8 \times 10^{-2}\;m$
$I=0.40\;A$
$B= \large\frac{\mu_0}{4 \pi} \frac{2 \pi I_n}{r}$
$\quad= \large\frac{10^{-7} \times 2 \times 3.14 \times 0.4 \times 100}{8 \times 10^{-7}}$
$\qquad= 3.1 \times 10^{-4}\;T$
Answer : $ 3.1 \times 10^{-4}\;T$
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