Ampere’s circuital lawA. relates the magnetic field to the current fo…

The correct answer is Option (1) → A, C and D only

Ampere’s circuital law states that the line integral of \( \vec{B} \) around any closed loop equals \( \mu_0 \) times the steady current passing through the surface bounded by the loop.

    \( \displaystyle \oint \vec{B}\cdot d\vec{l} = \mu_0 I \)

(A) relates the magnetic field to the current for a closed magnetic loop. — Correct. The law connects the circulation of \( \vec{B} \) around a closed (Amperian) loop with the current enclosed by it.

(B) holds for steady currents as well as the currents fluctuating with time. — Incorrect. In this form the law holds only for steady currents. For time-varying currents it gives contradictions (for example, near a charging capacitor), which Maxwell removed by adding the displacement current.

(C) can be conveniently applied to calculate the magnetic field produced by a solenoid. — Correct. A rectangular Amperian loop partly inside a long solenoid gives \( BL = \mu_0 nLI \), so \( B = \mu_0 nI \) in one line.

(D) holds valid for closed loop of any shape. — Correct. The law is true for every closed loop; a symmetric loop is chosen only to make the integral easy to evaluate.

NCERT (2026-27):
Ampere’s Circuital Law – “… Ampere’s circuital law is not new in content from Biot-Savart law. Both relate the magnetic field and the current, and both express the same physical consequences of a steady electrical current. … We also note that Ampere’s circuital law holds for steady currents which do not fluctuate with time.”

Ampere’s Circuital Law – “… It should be noted that while Ampere’s circuital law holds for any loop, it may not always facilitate an evaluation of the magnetic field in every case.”