A long solenoid carries a steady current. Its interior is filled with an iron core. The core adds a magnetic field \(B_m=0.628\ \mathrm{T}\) to the field of the solenoid. The magnetisation of the core is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\) and \(\pi=3.14\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(5\times10^{5}\ \mathrm{A\,m^{-1}}\)
A magnetised core adds its own field \( B_m \) to the field of the solenoid. This added field depends only on the magnetisation \( M \) of the core, through \( B_m=\mu_0M \).
\( B_m=0.628\ \mathrm{T} \)
\( \begin{aligned} M &= \frac{B_m}{\mu_0} \\ &= \frac{0.628}{4\times3.14\times10^{-7}} \\ &= 5\times10^{5}\ \mathrm{A\,m^{-1}} \end{aligned} \)
NCERT (2026-27):
Magnetisation and Magnetic Intensity – “… where Bm is the field contributed by the material core. It turns out that this additional field Bm is proportional to the magnetisation M of the material and is expressed as”
\( \mathbf{B}_m=\mu_0\mathbf{M} \)\( \qquad (5.10) \)