Question

The current in the winding of a toroid is $2.0\text{A}$. There are $400$ turns and the mean circumferential length is $40\text{cm}$. If the magnetic field inside the toroid is $0.5\text{T}$, the permeability is nearly:

• A. $5\times {10}^{-4}\text{Wb/A-m}$
• B. $25\times {10}^{-4}\text{Wb/A-m}$
• C. $2.5\times {10}^{-4}\text{Wb/A-m}$
• D. $1.25\times {10}^{-4}\text{Wb/A-m}$

Answer 1

The correct option is C $2.5\times {10}^{-4}\text{Wb/A-m}$

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

$B=\frac{{\mu }_0{\mu }_{r}NI}{2\pi r}$

Here, $\mu ={\mu }_o\times {\mu }_r$

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

$\Rightarrow \mu =\frac{B2\pi r}{NI}$

Here, mean circumferential length = $2\pi r$

$\therefore \mu =\frac{0.5\times 40\times {10}^{-2}}{400\times 2}$

$\therefore \mu =2.5\times {10}^{-4}\text{Wb/A-m}$

Hence, the correct answer is option $\left(C\right)$.