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 $10\, A.$ Determine the field due to magnetisation $\left(\mu_{0}=4 \pi \times 10^{-7} Hm ^{-1}\right)$

• 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, $I=10\, A$ and
number of turns per unit length, $n=1500$
Since, magnetic field of toroid,
$B_{\text {toroid }} =\mu_{0} n I$
$=4 \pi \times 10^{-7} \times 1500 \times 10=6 \pi\, mT$
Hence, the field due to magnetisation,
$B_{m}=B_{\text {net }}-B_{\text {toroid }}=10 \pi-6\, \pi$
$\Rightarrow B_{m} =4 \pi\, mT$