Which of the following is the expression of Ampere’s circuital law (I is the current enclosed by the closed loop)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \oint \vec{B}\cdot d\vec{l}=\mu_0 I \)
Option (1) \( \oint \vec{B}\cdot d\vec{l}=2\mu_0 I \) — Incorrect. The line integral equals \( \mu_0 I \), without any factor 2. For a straight wire it gives \( B\times2\pi r=\mu_0 I \), the known field \( \mu_0 I/2\pi r \).
Option (2) \( \oint \vec{B}\cdot d\vec{S}=\mu_0 I \) — Incorrect. \( \vec{B}\cdot d\vec{S} \) summed over a surface is a magnetic flux. Ampere’s law uses the line elements \( d\vec{l} \) along the boundary of the loop.
Option (3) \( \oint \vec{B}\cdot d\vec{S}=2\mu_0 I \) — Incorrect. It has both faults. It uses a surface element in place of the line element \( d\vec{l} \), and it has the factor 2.
Option (4) \( \oint \vec{B}\cdot d\vec{l}=\mu_0 I \) — Correct. The line integral of \( \vec{B} \) round the closed loop equals \( \mu_0 \) times the total current \( I \) passing through the surface bounded by the loop.
NCERT (2026-27):
Ampere’s Circuital Law – “… Ampere’s law states that this integral is equal to μ0 times the total current passing through the surface, i.e.,
\( \oint \mathbf{B}\cdot d\mathbf{l}=\mu_0 I \qquad [4.13(a)] \)
where I is the total current through the surface. The integral is taken over the closed loop coinciding with the boundary C of the surface.”