Question

A wire of length $314\text{cm}$ is used to make a toroid of cross- sectional radius $2\text{cm}$. Find the magnetic field at the inner core of the toroid having average radius of $50\text{cm}$ and current flowing through it is $2\text{A}$.
[wires are closely wounded]

• A. $2\times {10}^{-5}\text{T}$
• B. $0.5\times {10}^{-5}\text{T}$
• C. $1\times {10}^{-4}\text{T}$
• D. $0.25\times {10}^{-3}\text{T}$

Answer 1

The correct option is A $2\times {10}^{-5}\text{T}$

Given:
$r=2\text{cm};R=50\text{cm};i=2\text{A}$
$L=314\text{cm}$

Length of one turn, $l=2\pi r$

Let $N$ be the total number of turns in the toroid.

So, Total length of wire, $L=N\times l$

$\therefore L=N\times 2\pi r$

$314=N\times 2\pi \left(2\right)=N\times 2\left(3.14\right)\left(2\right)$

$\therefore N=25\text{turns}$

Now magnetic field produced by toroid inside the core of toroid is given by

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

$B=\frac{4\pi \times {10}^{-7}\times 25\times 2}{2\pi \times 50\times {10}^{-2}}$

$\therefore B=2\times {10}^{-5}\text{T}$

Hence, option $\left(a\right)$ is correct.