Question

A toroid has an iron core with an internal magnetic field of $10\pi mT$, when the current in the winding of $1500$ turns per meter is $10A.$ Determine the field due to magnetisation $\left({\mu }_0=4\pi \times 10^{-7}H{m}^{-1}\right)$

• A. $(4\pi )mT$
• B. $(10\pi )mT$
• C. $\left(\frac{8}{\pi }\right)mT$
• D. $\left(\frac{\pi }{4}\right)mT$

Answer 1

Answer: (A)

Given, internal magnetic field, $B_{\text{net }}=10\pi mT$
current, flowing through winding, $I=10A$ and
number of turns per unit length, $n=1500$
Since, magnetic field of toroid,
$B_{\text{toroid }}={\mu }_{0}nI$
$=4\pi \times 10^{-7}\times 1500\times 10=6\pi mT$
Hence, the field due to magnetisation,
$B_m=B_{\text{net }}-B_{\text{toroid }}=10\pi -6\pi$
$\Rightarrow B_m=4\pi mT$