A toroid has an average diameter of 2.5,m,400 | vedclass

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(B) The total magnetic field $B$ inside a toroid is given by $B = \mu_{0}(H + I)$,where $H$ is the magnetic field intensity and $I$ is the intensity o

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$\frac{10^{8}}{4 \pi}$

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(B) The total magnetic field $B$ inside a toroid is given by $B = \mu_{0}(H + I)$,where $H$ is the magnetic field intensity and $I$ is the intensity of magnetization.The magnetic field intensity $H$ for a toroid is given by $H = \frac{N i}{2 \pi r}$,where $N = 400$,$i = 2\,A$,and the radius $r = \frac{2.5}{2} = 1.25\,m$.$H = \frac{400 	imes 2}{2 \pi 	imes 1.25} = \frac{800}{2.5 \pi} = \frac{320}{\pi} \approx 101.86\,A/m$.Given $B = 10\,T$ and $\mu_{0} = 4 \pi 	imes 10^{-7}\,T\cdot m/A$,we have:$10 = 4 \pi 	imes 10^{-7} (H + I)$$H + I = \frac{10}{4 \pi 	imes 10^{-7}} = \frac{10^{8}}{4 \pi}$Since $H$ is very small compared to $I$ (as $I \approx \frac{10^{8}}{4 \pi} \approx 7.96 	imes 10^{6}$),the intensity of magnetization $I \approx \frac{10^{8}}{4 \pi}\,A/m$.

$A$ toroid has an average diameter of $2.5\,m$,$400$ turns,and a current of $2\,A$. If the total magnetic field inside the toroid is $10\,T$,what is the intensity of magnetization (in $A/m$)?

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