Question

A toroid has a core of inner radius $20\text{cm}$ and outer radius $22\text{cm}$ around which $4200$ turns of a wire are wound. If the current in the wire is $10\text{A}$, what is the magnetic field inside the core ?

• A. $3\times {10}^{-2}\text{T}$
• B. $1\times {10}^{-2}\text{T}$
• C. $4\times {10}^{-2}\text{T}$
• D. $2\times {10}^{-2}\text{T}$

Answer 1

The correct option is C $4\times {10}^{-2}\text{T}$

Given:
$N=4200;i=10\text{A}$
inner radius, $r_1=20\text{cm}$
outer radius, $r_2=22\text{cm}$

$\therefore$ mean radius, $r=\frac{r_1+r_2}{2}$

$r=\frac{20+22}{2}=21\text{cm}=0.21\text{m}$

Magnetic field inside the core of toroid is given by

$B=\frac{{\mu }_{0}Ni}{2\pi r}$

$B=\frac{4\pi \times {10}^{-7}\times 4200\times 10}{2\pi \times 0.21}$

$\therefore B=4\times {10}^{-2}\text{T}$

Therefore, option $\left(c\right)$ is the right choice.