Question

A toroid having a square cross-section of side $5\text{cm}$ and inner radius of $15\text{cm}$, has $500$ turns and carries a current of $0.8\text{A}$. What is the magnetic field just inside the outer radius of the toroid ?

Answer 1

Given:

$r_1=15\text{cm};N=500;I=0.8\text{A}$

Side of cross-section, $a=5\text{cm}$

The outer radius of the toroid will be,

$r_2=r_1+a=15+5=20\text{cm}$

The magnetic field just inside the outer radius of the toroid will be,

$B=\frac{{\mu }_{0}NI}{2\pi r_2}$

$B=\frac{4\pi \times {10}^{-7}\times 500\times 0.8}{2\pi \times 20\times {10}^{-2}}$

$\therefore B=4\times {10}^{-4}\text{T}$

Hence, option $\left(C\right)$ is the correct answer.