A coil wrapped around a toroid has an inner r | vedclass

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(A) Given: Inner radius $r_1 = 20 \,cm = 0.2 \,m$, Outer radius $r_2 = 25 \,cm = 0.25 \,m$, Number of turns $N = 800$, Current $I = 12 \,A$.The magnet

Vedclass · Accepted Answer

$9.6 \,mT, 7.68 \,mT$

Vedclass · Answer

(A) Given: Inner radius $r_1 = 20 \,cm = 0.2 \,m$, Outer radius $r_2 = 25 \,cm = 0.25 \,m$, Number of turns $N = 800$, Current $I = 12 \,A$.The magnetic field $B$ inside a toroid at a distance $r$ from the center is given by $B = \frac{\mu_0 N I}{2 \pi r}$.Since $B \propto \frac{1}{r}$, the magnetic field is maximum at the minimum radius $(r_1)$ and minimum at the maximum radius $(r_2)$.Maximum magnetic field: $B_{\max} = \frac{4 \pi 	imes 10^{-7} 	imes 800 	imes 12}{2 \pi 	imes 0.2} = \frac{2 	imes 10^{-7} 	imes 9600}{0.2} = 9.6 	imes 10^{-3} \,T = 9.6 \,mT$.Minimum magnetic field: $B_{\min} = \frac{4 \pi 	imes 10^{-7} 	imes 800 	imes 12}{2 \pi 	imes 0.25} = \frac{2 	imes 10^{-7} 	imes 9600}{0.25} = 7.68 	imes 10^{-3} \,T = 7.68 \,mT$.

$A$ coil wrapped around a toroid has an inner radius of $20 \,cm$ and an outer radius of $25 \,cm$. If the wire wrapping makes $800$ turns and carries a current of $12 \,A$, the maximum and minimum values of the magnetic field within the toroid are:

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