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  • Current in the windings of the toroid is 2A. There are 400 turns & the mean circumferential length is 40cm. If the inside magnetic field is

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B= \mu in=\frac{\mu_r \mu_o iN}{2\pi r}

Putting the Values:

\therefore 1= \mu in=\frac{\mu_r* 4 \pi*10^{(-7)}*400*2}{0.4}

Then \Rightarrow \mu _r =400

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