Question

A toroid has a non-ferromagnetic core of inner radius $20.5cm$ and outer radius $21.5cm$, around which $4200$ turns of a wire are wound. If the current in the wire is $10A$, the magnetic field inside the core of the toroid is $\left({\mu }_0=4\pi \times 10^{-7}H{m}^{-1}\right)$

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

Answer 1

Answer: (B)

Given, inner radius of non-ferromagnetic core, $r_i=20.5cm$
outer radius of non-ferromagnetic core, $r_0=21.5cm$
mean radius of toroid, $r=\frac{r_i+r_0}{2}=\frac{42}{2}=21cm$
number of turns of wire, $N=4200$
and current in wire, $I=10A$
${\mu }_0=4\pi \times 10^{-7}H{m}^{-1}$
$\therefore$ 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, $B={\mu }_{0}NI$
$=4\pi \times 10^{-7}\times \frac{n}{2\pi r}\times 10$
$=2\times 10^{-7}\times \frac{n}{r}\times 10$
$(\because$ Outer radius of core $r=21.5cm)$
$B=2\times 10^{-7}\times \frac{4200}{21}\times 10$
$=40\times 10^{-3}=40\times 10^{-3}T$
$B=40mT$