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(A) Given:Inner radius $r_1 = 24 \ cm = 0.24 \ m$Outer radius $r_2 = 26 \ cm = 0.26 \ m$Number of turns $N = 2000$Current $I = 12 \ A$The mean radius

Vedclass · Accepted Answer

$1.92 	imes 10^{-2} \ T$

Vedclass · Answer

(A) Given:Inner radius $r_1 = 24 \ cm = 0.24 \ m$Outer radius $r_2 = 26 \ cm = 0.26 \ m$Number of turns $N = 2000$Current $I = 12 \ A$The mean radius $r$ of the toroid is given by:$r = \frac{r_1 + r_2}{2} = \frac{24 + 26}{2} = 25 \ cm = 0.25 \ m$The magnetic field $B$ inside the core of a toroid is given by the formula:$B = \mu_0 n I$where $n$ is the number of turns per unit length, $n = \frac{N}{2 \pi r}$.Substituting the values:$B = \mu_0 \left( \frac{N}{2 \pi r} ight) I$$B = (4 \pi 	imes 10^{-7}) 	imes \left( \frac{2000}{2 \pi 	imes 0.25} ight) 	imes 12$$B = (2 	imes 10^{-7}) 	imes \left( \frac{2000}{0.25} ight) 	imes 12$$B = (2 	imes 10^{-7}) 	imes 8000 	imes 12$$B = 192000 	imes 10^{-7} \ T$$B = 1.92 	imes 10^{-2} \ T$

$A$ toroid has a core (non-ferromagnetic) of inner radius $24 \ cm$ and outer radius $26 \ cm$ around which $2000$ turns of a wire are wound. If the current in the wire is $12 \ A$, the magnetic field inside the core of the toroid is:

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