Question

An air cored toroid $10$ turns per $\text{cm}$ carries a current of $1\text{mA}.$ The intensity of magnetic field inside it is

• A. $4\pi \times {10}^{-6}\text{T}$
• B. $4\pi \times {10}^{-7}\text{T}$
• C. $4\pi \times {10}^{-8}\text{T}$
• D. $4\pi \times {10}^{-9}\text{T}$

Answer 1

The correct option is B $4\pi \times {10}^{-7}\text{T}$

Given:
$I=1\text{mA}=1\times {10}^{-3}\text{A}$

$n=10\frac{\text{turns}}{\text{cm}}=10\times 100\frac{\text{turns}}{\text{m}}={10}^3\frac{\text{turns}}{\text{m}}$

Because it is air cored, so

$\mu ={\mu }_0=4\pi \times {10}^{-7}{\text{Hm}}^{-1}$

The field inside the toroid is,

$B={\mu }_{0}nI$

$B=4\pi \times {10}^{-7}\times {10}^3\times {10}^{-3}$

$\therefore B=4\pi \times {10}^{-7}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.