A parallel plate capacitor is being charged by a current \(i\). A circular loop of radius \(r\) is drawn around the connecting wire, just outside the plates. A pot-shaped surface has this loop as its rim, and its bottom lies between the plates. Ampere’s circuital law \(\oint \vec{B}\cdot d\vec{l}=\mu_0 i\), without the displacement current, is applied to this surface. The magnetic field at a point on the loop comes out as
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → zero
Ampere’s circuital law in this form counts only the conduction current that passes through the chosen surface. For every surface with this loop as its rim, the left side is \( B(2\pi r) \).
Option (1) \( \mu_0 i/2\pi r \) — Incorrect. This value comes from a flat disc that the wire passes through. The pot-shaped surface does not meet the wire, so the current \( i \) does not cross it.
Option (2) \( \mu_0 i/2r \) — Incorrect. This is the field at the centre of a circular coil that carries the current \( i \). Here the loop is only a path drawn in space, and no current flows along it.
Option (3) \( \mu_0 i/4\pi r \) — Incorrect. Here the constant \( \mu_0/4\pi \) of the Biot-Savart law is mixed with the result for a straight wire. Also, no current crosses the pot-shaped surface at all.
Option (4) zero — Correct. The bottom of the pot lies in the gap between the plates, and no charge flows across the gap. So no current crosses the surface, and \( B(2\pi r)=0 \). This result goes against the flat-disc result, and Maxwell removed the contradiction with the displacement current.
NCERT (2026-27):
Displacement Current – “On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side … has not changed but the right hand side is zero and not μ0i, since no current passes through the surface … So we have a contradiction; calculated one way, there is a magnetic field at a point P; calculated another way, the magnetic field at P is zero.”