Question

A toroid has a non-ferromagnetic core of inner radius $20.5 \,cm$ and outer radius $21.5\, cm$, around which $4200$ turns of a wire are wound. If the current in the wire is $10\, A$, the magnetic field inside the core of the toroid is $\left(\mu_{0}=4 \pi \times 10^{-7}\, Hm ^{-1}\right)$

• A. $20\, mT$
• B. $40\, mT$
• C. $20 \,\pi \,mT$
• D. $40\, \pi\, mT$

Answer 1

Answer: (B)

Given, inner radius of non-ferromagnetic core, $r_{i}=20.5\, cm$
outer radius of non-ferromagnetic core, $r_{0}=21.5\, cm$
mean radius of toroid, $r=\frac{r_{i}+r_{0}}{2}=\frac{42}{2}=21 \, cm$
number of turns of wire, $N=4200$
and current in wire, $I=10\, A$
$\mu_{0}=4 \pi \times 10^{-7} \, Hm ^{-1}$
$\therefore $ Circumference of core $=$ length of the wire
$2 \pi r=l$
Then, number of turns per unit length,
$N=\frac{n}{l}=\frac{n}{2 \pi r}$
Now, magnetic field inside the core of the toroid is magnetic field, $B=\mu_{0} \, N I$
$= 4 \pi \times 10^{-7} \times \frac{n}{2 \pi r} \times 10 $
$= 2 \times 10^{-7} \times \frac{n}{r} \times 10 $
$(\because$ Outer radius of core $ r=21.5\, cm ) $
$ B= 2 \times 10^{-7} \times \frac{4200}{21} \times 10 $
$= 40 \times 10^{-3}=40 \times 10^{-3} \, T$
$ B= 40 \, mT $