A long straight wire carries a current I. The magnetic intensity H is measured at a point P close to the wire. A long cylindrical iron rod is brought close to the wire so that the point P is at the centre of the rod. The value of H at P will
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → remain almost constant
The total field inside a material is \( \mathbf{B}=\mu_0(\mathbf{H}+\mathbf{M}) \). The magnetic intensity H is the part set up by external factors such as the current. The magnetisation M depends on the material placed in the field.
Option (1) increase many times — Incorrect. The iron is strongly magnetised. So it is B that rises many times through the large \( \mathbf{M} \). The material does not raise H.
Option (2) decrease many times — Incorrect. Iron is magnetised along the field and adds to it. It does not reduce the part of the field set up by the current.
Option (3) remain almost constant — Correct. The current in the wire and the distance of P from it fix H at P, \( H=I/2\pi r \). Neither changes when the rod is brought in. So H stays almost the same, while B inside the iron becomes much larger.
Option (4) become zero — Incorrect. The wire still carries the current \( I \), so H at a distance \( r \) from it is \( I/2\pi r \), which is not zero.
NCERT (2026-27):
Magnetisation and Magnetic Intensity – “… 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.”