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Q)

A torroid has a core outer radius $26\;cm$ around which 3500 turns of a wire are wound . If the current in the wire is $11\;A$ , What is the magnetic field, inside the core of the torriod .

$\begin{array}{1 1} 3.02 \times 10^{-3}T \\ 3.02 \times 10^{-2}T \\ 3.02 \times 10^{-4}T \\ 3.02 \times 10^{-1}T \end{array} $

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A)
Solution :
$r_1=25\;cm=0.25\;m$
$r_2=26\;cm=0.26\;m$
$N=3500$
$I= 11A$
Mean length of the torriod $l= 2 \pi \bigg( \large\frac{r_1+r_2}{2}\bigg)$
$\qquad= \pi (r_1 +r_2)$
$\qquad= 0.51 \pi$
$\B= \large\frac{\mu_0 N I}{l}$
$\quad= \large\frac{(4 \pi \times 10^{-7}) \times 3500 \times 11}{0.51 \;\pi}$
$\quad= 3.02 \times 10^{-2}\;T$
Answer : $ 3.02 \times 10^{-2}T$
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