Question

A toraid of non-ferromagnetic has core if inner radius24 $cm$ and outer radius 25 $cm$ , it has 3000 turns and carries a current of 11 $A$ , then the magnetic field at a point in the internal cavity of toroid

• A. Zero
• B. $3\times {10}^{-3}T$
• C. Infinity
• D. $2.69\times {10}^{-2}T$

Answer 1

The correct option is D $2.69\times {10}^{-2}T$

Given: $r_1=24cm=0.24m$

$r_2=25cm=0.25m$

$N=3000turns$

$I=11A$

To find: magnetic field in internal cavity,$B=?$

Solution: As we know that,

magnetic field in internal cavity is,

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

here, r is the mean radius.

i.e. $r=(r_1+r_2)/2$

$==>r=(0.24+0.25)/2$

$==>r=0.245m$

Put this in $\left(1\right)$, we get

$B=\frac{2\ast {10}^{-7}\ast 3000\ast 11}{0.245}$

$B=269.38\ast {10}^{-4}$

$B=2.69\ast {10}^{-2}T$