The current in the windings of a toroid is $2.0\,A$. There are $400\,turns$ and the mean circumferential length is $40\,cm$. If the inside magnetic field is $1.0\,T,$ the relative permeability is near to
Medium
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Toroid
$B=\frac{\mu_{0} N i}{2 \pi r}$ in free space
$B=\frac{\mu_{0} \mu_{r} N i}{2 \pi r}$ in a medium
Given: $i=2 A$
$N=400$ turns
circumferential length $=2 \pi r=40 \mathrm{cm}=0.4 \mathrm{m}$
$B=I T$
substituting the values in formula;
$I=\frac{\mu_{r} \mu_{0}(400)(2)}{0.4} .$ also $\mu_{0}=4 \pi \times 10^{-7}$
hence $\mu_{r}=\frac{0.4}{4 \pi \times 10^{-7} \times 400 \times 2}=400$
hence $\mu_{r}=400$
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