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(D) The mean radius $r$ of the toroid is given by the average of the inner and outer radii:$r = \frac{r_1 + r_2}{2} = \frac{24 \ cm + 25 \ cm}{2} = 24

Vedclass · Accepted Answer

$48$

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(D) The mean radius $r$ of the toroid is given by the average of the inner and outer radii:$r = \frac{r_1 + r_2}{2} = \frac{24 \ cm + 25 \ cm}{2} = 24.5 \ cm = 24.5 	imes 10^{-2} \ m$Given number of turns $N = 4900$ and current $I = 12 \ A$.The magnetic field $B$ inside the core of a toroid is given by the formula:$B = \frac{\mu_0 N I}{2 \pi r}$Substituting the values:$B = \frac{(4 \pi 	imes 10^{-7} \ T \cdot m/A) 	imes 4900 	imes 12}{2 \pi 	imes 24.5 	imes 10^{-2} \ m}$$B = \frac{2 	imes 10^{-7} 	imes 4900 	imes 12}{24.5 	imes 10^{-2}}$$B = \frac{2 	imes 4900 	imes 12}{24.5} 	imes 10^{-5}$$B = \frac{117600}{24.5} 	imes 10^{-5} = 4800 	imes 10^{-5} \ T = 48 	imes 10^{-3} \ T = 48 \ mT$.

$A$ toroid has a non-ferromagnetic core of inner radius $24 \ cm$ and outer radius $25 \ cm$,around which $4900$ 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: (in $mT$)

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