Question

The current in the winding of a toroid is $2.0\text{A}$. There are $950$ turns and the mean circumferential length is $19\pi \text{cm}$. If the magnetic field inside the toroid is $2\text{T}$, then the magnetic susceptibility of core is

Answer 1

Given:
$I=2.0\text{A};N=950;B=2\text{T}$

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

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

$\Rightarrow \mu =\frac{B\left(2\pi r\right)}{NI}....\left(1\right)$

Here, mean circumferential length is, $2\pi r=19\pi \text{cm}$.

So, on substituting the values in $\left(1\right)$, we get

$\mu =\frac{4\times 19\pi \times {10}^{-2}}{950\times 2}$

The relative permeability will be

${\mu }_r=\frac{\mu }{{\mu }_0}=\frac{4\times 19\pi \times {10}^{-2}}{950\times 2}\times \frac{1}{4\pi \times {10}^{-7}}$

$\therefore {\mu }_r=1000$

Using formula, ${\mu }_r=1+\chi$

$\therefore \chi ={\mu }_r-1=1000-1=999$

Accepted answers: $999,999.0,999.00$