Question

Calculate the number of turns in a toroid when the charges of $8\text{C}$ flows in $1\text{minute}$ through it, produce a magnetic field of $2.5\times {10}^{-4}\text{T}$.
[Radius of the toroid is $20\text{cm}$]

• A. $1275$
• B. $1875$
• C. $1675$
• D. $2075$

Answer 1

The correct option is B $1875$

Given:
$B=2.5\times {10}^{-4}\text{T};r=20\times {10}^{-2}\text{m}$
$q=8\text{C};t=1\text{min}=60\text{s}$

So, current will be

$i=\frac{q}{t}=\frac{8}{60}=\frac{2}{15}\text{A}$

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

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

$2.5\times {10}^{-4}=\frac{4\pi \times {10}^{-7}\times N\times 2}{2\pi \times 20\times {10}^{-2}\times 15}=\frac{{10}^{-5}\times N}{75}$

$N=\frac{2.5\times {10}^{-4}\times 75}{{10}^{-5}}$

$\therefore N=1875\times 10=1875$

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