The magnetic field inside a long solenoid is 0.4 T. The solenoid is now filled with a paramagnetic material of susceptibility \(3\times10^{-5}\). The field inside the solenoid increases by
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1.2\times10^{-5}\ \mathrm{T}\)
The core gets magnetised and adds its own field \( B_m=\mu_0M \) to the field \( B_0 \) of the solenoid. For such a material \( M=\chi H \), and the magnetic intensity inside the solenoid is \( H=B_0/\mu_0 \).
Field of the solenoid, \( B_0=0.4\ \mathrm{T} \)
Susceptibility, \( \chi=3\times10^{-5} \)
\( \begin{aligned} B_m &= \mu_0M \\ &= \mu_0\chi H \\ &= \chi B_0 \\ &= 3\times10^{-5}\times0.4 \\ &= 1.2\times10^{-5}\ \mathrm{T} \end{aligned} \)
So the field rises only slightly, as expected for a paramagnetic core.
NCERT (2026-27):
Magnetisation and Magnetic Intensity – “If the interior of the solenoid is filled with a material with non-zero magnetisation, the field inside the solenoid will be greater than B0. The net B field in the interior of the solenoid may be expressed as”
\( \mathbf{B}=\mathbf{B}_0+\mathbf{B}_m \)\( \qquad (5.9) \)
“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”
\( \mathbf{B}_m=\mu_0\mathbf{M} \)\( \qquad (5.10) \)
Summary – “5. Consider a material placed in an external magnetic field B0. The magnetic intensity is defined as, H = B0/μ0 … 6. For a linear material M = χ H. …”