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

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(D) The magnetic field $B$ inside a toroid is given by the formula: $B = \mu_0 \mu_r n i$,where $n$ is the number of turns per unit length.Given:$i =

Vedclass · Accepted Answer

$400$

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(D) The magnetic field $B$ inside a toroid is given by the formula: $B = \mu_0 \mu_r n i$,where $n$ is the number of turns per unit length.Given:$i = 2.0\,A$$N = 400$$L = 40\,cm = 0.4\,m$$B = 1.0\,T$Number of turns per unit length $n = \frac{N}{L} = \frac{400}{0.4} = 1000\,turns/m$.Substituting the values in the formula:$1.0 = (4\pi 	imes 10^{-7}) 	imes \mu_r 	imes 1000 	imes 2.0$$1.0 = 8\pi 	imes 10^{-4} 	imes \mu_r$$\mu_r = \frac{1}{8\pi 	imes 10^{-4}} \approx \frac{1}{25.12 	imes 10^{-4}} \approx \frac{10000}{25.12} \approx 398$.Rounding to the nearest option,the relative permeability is $400$.

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 magnetic field inside the toroid is $1.0\,T$,the relative permeability is near to:

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