A solenoid has 1000 turns per metre and a current of 2 A flows in its…

The correct answer is Option (3) → 2000 A/m

The magnetic intensity is defined by \( \mathbf{H}=\mathbf{B}/\mu_0-\mathbf{M} \). It is the part of the field set up by external factors such as the current in the solenoid. With no core, \( \mathbf{M}=0 \) and the solenoid field is \( B_0=\mu_0 nI \), so \( H=nI \).

    \( \begin{aligned} H &= nI \\ &= 1000\,\mathrm{m^{-1}}\times2\,\mathrm{A} \\ &= 2000\,\mathrm{A/m} \end{aligned} \)

NCERT (2026-27):
Magnetisation and Magnetic Intensity – “… Consider a long solenoid of n turns per unit length and carrying a current I. The magnetic field in the interior of the solenoid was shown to be given by

    \( \mathbf{B}_0=\mu_0\,nI \qquad (5.8) \)

… It is convenient to introduce another vector field H, called the magnetic intensity, which is defined by

    \( \displaystyle \mathbf{H}=\frac{\mathbf{B}}{\mu_0}-\mathbf{M} \)\( \qquad (5.11) \)

where H has the same dimensions as M and is measured in units of A m^–1. … 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.”