For a toroid,N = 500,radius R = 40 text{ cm}, | vedclass

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(A) The magnetic field inside a toroid is given by $B = \frac{\mu_0 N i}{2 \pi R}$.The total magnetic flux linkage is $\phi = B A N = \frac{\mu_0 N^2

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) The magnetic field inside a toroid is given by $B = \frac{\mu_0 N i}{2 \pi R}$.The total magnetic flux linkage is $\phi = B A N = \frac{\mu_0 N^2 i A}{2 \pi R}$.Since $\phi = L i$,the self-inductance $L$ is given by $L = \frac{\mu_0 N^2 A}{2 \pi R}$.Given values: $N = 500$,$R = 0.4 	ext{ m}$,$A = 10 	imes 10^{-4} 	ext{ m}^2$,and $\mu_0 = 4 \pi 	imes 10^{-7} 	ext{ T m/A}$.Substituting these values:$L = \frac{(4 \pi 	imes 10^{-7}) 	imes (500)^2 	imes (10 	imes 10^{-4})}{2 \pi 	imes 0.4}$$L = \frac{2 	imes 10^{-7} 	imes 250000 	imes 10^{-3}}{0.4}$$L = \frac{2 	imes 10^{-7} 	imes 250 	imes 10^0}{0.4} = \frac{500 	imes 10^{-7}}{0.4} = 1250 	imes 10^{-7} 	ext{ H} = 1.25 	imes 10^{-4} 	ext{ H} = 125 \mu	ext{H}$.

For a toroid,$N = 500$,radius $R = 40 \text{ cm}$,and area of cross-section $A = 10 \text{ cm}^2$. Find the inductance in $\mu\text{H}$.

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