Case: Read the passage and answer the question.
1000 turns per metre are wound on a long solenoid with a core of ferromagnetic material. On passing a current of 2 A in the coil, a magnetic field of 10 T is produced in the core.
The magnetisation of the core (in \( \mathrm{A\,m^{-1}} \)) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 7.96\times10^{6} \)
Inside the core the total field is \( B=\mu_0(H+M) \). The magnetic intensity \( H \) is the part set up by the current in the solenoid, \( H=nI \). The magnetisation \( M \) is the part due to the core material.
\( \begin{aligned} H &= nI \\ &= 1000\,\mathrm{m^{-1}}\times2\,\mathrm{A} \\ &= 2\times10^{3}\,\mathrm{A\,m^{-1}} \end{aligned} \)
Taking \( \mu_0=4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \):
\( \begin{aligned} \frac{B}{\mu_0} &= \frac{10\,\mathrm{T}}{4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}}} \\ &\approx 7.96\times10^{6}\,\mathrm{A\,m^{-1}} \end{aligned} \)
\( \begin{aligned} M &= \frac{B}{\mu_0}-H \\ &= 7.96\times10^{6}\,\mathrm{A\,m^{-1}}-0.002\times10^{6}\,\mathrm{A\,m^{-1}} \\ &\approx 7.96\times10^{6}\,\mathrm{A\,m^{-1}} \end{aligned} \)
Therefore, in a ferromagnetic core almost the whole field comes from the magnetisation of the material. \( H \) is very small next to it.
NCERT (2026-27):
Magnetisation and Magnetic Intensity – “… Thus, the total magnetic field B is written as
\( \mathbf{B}=\mu_0(\mathbf{H}+\mathbf{M}) \)\( \qquad (5.12) \)
We repeat our defining procedure. We have partitioned the contribution to the total magnetic field inside the sample into two parts: one, due to external factors such as the current in the solenoid. This is represented by H. The other is due to the specific nature of the magnetic material, namely M.”