Question

A toroid made up of a circular core has inner circumference $30\pi \text{cm}$ and $2400$ turns. The relative permeability of the material is $1000$. What is the cross-sectional area of the core when a current of $0.75\text{A}$ passes through the winding, a magnetic field of strength $2\text{T}$ is produced at its mean radius ?

• A. $225\pi {\text{cm}}^2$
• B. $225{\text{cm}}^2$
• C. $3\pi {\text{cm}}^2$
• D. $9\pi {\text{cm}}^2$

Answer 1

The correct option is D $9\pi {\text{cm}}^2$

Given:
$N=2400;{\mu }_r=1000;I=0.75\text{A};B=2\text{T}$

Inner circumference, $2\pi r_1=30\pi \text{cm}$

$\therefore r_1=15\text{cm}$

We know that, magnetic field inside a toroid is given by,

$B=\frac{{\mu }_0{\mu }_{r}NI}{2\pi r}$

$\Rightarrow r=\frac{{\mu }_0{\mu }_{r}NI}{2\pi B}$

$r=\frac{4\pi \times {10}^{-7}\times 1000\times 2400\times 0.75}{2\pi \times 2}$

$\therefore r=18\times {10}^{-2}\text{m}=18\text{cm}$

So, cross-sectional area of core will be

$A=\pi (r-r_1{)}^2=\pi (18-15{)}^2$

$\therefore A=9\pi {\text{cm}}^2$

Therefore, option $\left(D\right)$ is correct.