Question

The average radius of a toroid, made out of a non-magnetic material, is $0.1\text{m}$ and it has $500$ turns. If it carries a $0.5\text{A}$ current, the intensity of magnetic field along its circular axis will be-

• A. $\pi \times {10}^4\text{T}$
• B. $2\pi \times {10}^4\text{T}$
• C. $2.5\times {10}^{-4}\text{T}$
• D. $5\times {10}^{-4}\text{T}$

Answer 1

The correct option is D $5\times {10}^{-4}\text{T}$

Given, $N=500;R=0.1\text{m};I=0.5\text{A}$



The magnetic field inside a toroid is,

$B=\frac{{\mu }_{0}NI}{2\pi R}......\left(1\right)$

Putting this value in (1) we get,

$=\frac{4\pi \times {10}^{-7}\times 500\times 0.5}{2\pi \times 0.1}$

$=5\times {10}^{-4}\text{T}$

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