Magnetic Field Of A Toroid

Applying Ampere's circuital law to a circular loop of radius $r$ running along the axis of a toroid encloses all $N$ turns, each carrying current $I$, giving $B(2\pi r)={\mu }_{0}NI$, hence $B=\frac{{\mu }_{0}NI}{2\pi r}$. The field exists only within the windings and is essentially zero outside an ideal toroid. Substituting $N=3000$, $I=5\text{A}$ and $r=0.15\text{m}$ gives $B=\frac{(4\pi \times 10^{-7})(3000)(5)}{2\pi (0.15)}=\frac{2\times 10^{-7}\times 15000}{0.15}=2.0\times 10^{-2}\text{T}$. The option $1.0\times 10^{-2}\text{T}$ halves the enclosed turns. The option $4.0\times 10^{-2}\text{T}$ uses half the radius. The option $6.3\times 10^{-3}\text{T}$ confuses the toroid with a straight solenoid using turns per unit length carelessly. NCERT derives this exact expression for the toroid as a key Ampere's-law application. A check confirms the field of a couple of centitesla, larger than an open solenoid because the closed magnetic circuit confines the flux. The fact that the field is confined almost entirely inside the windings makes the toroid the geometry of choice for transformers and inductors where stray external fields must be minimised. Because the enclosed current depends only on the total number of turns crossing the Amperian loop, the result is remarkably robust, and the slight variation of field across the cross-section is often neglected when the mean radius greatly exceeds the bore radius.