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(B) To find the magnetic field inside the toroid,we use Ampere's Circuital Law.Consider an Amperian loop of radius $r$ inside the toroid,where $r = \f

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$\frac{\mu_0 N I}{\pi(r_1+r_2)}$

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(B) To find the magnetic field inside the toroid,we use Ampere's Circuital Law.Consider an Amperian loop of radius $r$ inside the toroid,where $r = \frac{r_1 + r_2}{2}$.According to Ampere's Law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{	ext{enclosed}}$.Since the magnetic field $B$ is uniform along the circular path and parallel to the length element $d\vec{l}$,the angle between them is $0^\circ$.Thus,$B \oint dl = \mu_0 N I$.The circumference of the loop is $2 \pi r = 2 \pi \left( \frac{r_1 + r_2}{2} ight) = \pi(r_1 + r_2)$.Substituting this into the equation: $B \cdot \pi(r_1 + r_2) = \mu_0 N I$.Therefore,the magnetic field is $B = \frac{\mu_0 N I}{\pi(r_1 + r_2)}$.

$A$ toroid has a core of inner radius $r_1$ and outer radius $r_2$,around which $N$ turns of wire are wound. If the current in the wire is $I$,then the magnetic field inside the toroid is $(\mu_0 = \text{permeability of free space})$

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