The current in the windings of a toroid is 2. | vedclass

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(D) The magnetic field inside a toroid filled with a medium is given by the formula:$B = \frac{\mu_0 \mu_r N i}{L}$where $L = 2 \pi r$ is the mean cir

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$400$

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(D) The magnetic field inside a toroid filled with a medium is given by the formula:$B = \frac{\mu_0 \mu_r N i}{L}$where $L = 2 \pi r$ is the mean circumferential length.Given values:$i = 2.0\,A$$N = 400\,turns$$L = 40\,cm = 0.4\,m$$B = 1.0\,T$$\mu_0 = 4 \pi 	imes 10^{-7}\,T \cdot m/A$Rearranging the formula to solve for relative permeability $\mu_r$:$\mu_r = \frac{B \cdot L}{\mu_0 \cdot N \cdot i}$Substituting the values:$\mu_r = \frac{1.0 	imes 0.4}{4 \pi 	imes 10^{-7} 	imes 400 	imes 2.0}$$\mu_r = \frac{0.4}{3200 \pi 	imes 10^{-7}}$$\mu_r = \frac{0.4}{3.2 \pi 	imes 10^{-4}} = \frac{4000}{3.2 \pi} \approx \frac{4000}{10.05} \approx 398$Rounding to the nearest given option, we get $\mu_r \approx 400$.

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

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