What is the magnitude of the magnetic field due to an infinitely long…

The correct answer is Option (2) → \( \displaystyle\frac{\mu_0 I}{2\pi r} \)

By Ampere’s circuital law, the line integral of \( \vec{B} \) around a closed loop equals \( \mu_0 \) times the current enclosed. For a long straight wire the field lines are circles centred on the wire. So we choose a circular loop of radius \( r \). On this loop, \( B \) is tangential and has the same magnitude everywhere.

    \( B\times2\pi r=\mu_0 I \)

    \( \displaystyle \Rightarrow\; B=\frac{\mu_0 I}{2\pi r} \)

Option (1) \( \displaystyle\frac{\mu_0 I}{4\pi r} \) — Incorrect. The circular loop has length \( 2\pi r \). So \( \mu_0 I \) is divided by \( 2\pi r \), and a factor \( 4\pi \) has no place here.

Option (2) \( \displaystyle\frac{\mu_0 I}{2\pi r} \) — Correct. Ampere’s law over the circle of length \( 2\pi r \) gives this field. It is directly proportional to the current and inversely proportional to the distance from the wire.

Option (3) \( \displaystyle\frac{\mu_0 I}{2r} \) — Incorrect. This is the field at the centre of a circular loop of radius \( r \). For a straight wire the loop length \( 2\pi r \) brings in the factor \( \pi \).

Option (4) \( \displaystyle\frac{\mu_0 I}{2\pi r^{2}} \) — Incorrect. The field of a long straight wire falls as \( 1/r \). It does not follow an inverse-square law.

NCERT (2026-27):
Ampere’s Circuital Law – “When there is a system with a symmetry such as for a straight infinite current-carrying wire …, the Ampere’s law enables an easy evaluation of the magnetic field, much the same way Gauss’ law helps in determination of the electric field. … We find that the magnetic field at a distance r outside the wire is tangential and given by

    \( B\times2\pi r=\mu_0 I, \)

    \( B=\mu_0 I/(2\pi r) \)\( \qquad (4.14) \)”