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(A) The average radius of the toroid is given by $r = \frac{r_{1} + r_{2}}{2}$.The magnetic field $B$ inside a toroid is given by the formula $B = \mu

Vedclass · Accepted Answer

$\frac{\mu_{0} NI}{\pi(r_{1}+r_{2})}$

Vedclass · Answer

(A) The average radius of the toroid is given by $r = \frac{r_{1} + r_{2}}{2}$.The magnetic field $B$ inside a toroid is given by the formula $B = \mu_{0} n I$,where $n$ is the number of turns per unit length.The number of turns per unit length is $n = \frac{N}{2 \pi r}$.Substituting the value of $r$,we get $n = \frac{N}{2 \pi \left(\frac{r_{1} + r_{2}}{2}\right)} = \frac{N}{\pi(r_{1} + r_{2})}$.Therefore,the magnetic field is $B = \mu_{0} I \left(\frac{N}{\pi(r_{1} + r_{2})}\right) = \frac{\mu_{0} NI}{\pi(r_{1} + r_{2})}$.

$A$ toroid has a non-ferromagnetic 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 ($\mu_{0} =$ permeability of free space).

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