A circular loop of radius \(R\) carries a current \(I\). The magnetic field at its centre is \(B_0\). Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) Field at the centre when the current and the radius are both doubled | (I) \(B_0/8\) |
| (B) Field at the centre when the radius is halved and the current is \(I\) | (II) \(B_0/2\) |
| (C) Field at a point on the axis at a distance \(\sqrt{3}R\) from the centre | (III) \(B_0\) |
| (D) Field at the centre of a semicircular arc of radius \(R\) carrying current \(I\) (the straight leads lie along its diameter) | (IV) \(2B_0\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The field at the centre of a loop is \( B_0=\mu_0I/2R \). It is directly proportional to the current and inversely proportional to the radius. On the axis, the field becomes weaker as we move away from the centre.
(A) Field at the centre when the current and the radius are both doubled — This matches (III) \( B_0 \). Doubling \( I \) doubles the field and doubling \( R \) halves it, so the field stays \( B_0 \).
(B) Field at the centre when the radius is halved and the current is \( I \) — This matches (IV) \( 2B_0 \). With \( R/2 \) in place of \( R \), the field is \( \mu_0I/R \), which is \( 2B_0 \).
(C) Field at a point on the axis at a distance \( \sqrt{3}R \) from the centre — This matches (I) \( B_0/8 \). Here \( x^{2}+R^{2}=4R^{2} \) and \( (4R^{2})^{3/2}=8R^{3} \). So the axial field is \( \mu_0IR^{2}/16R^{3}=\mu_0I/16R \), which is \( B_0/8 \).
(D) Field at the centre of a semicircular arc of radius \( R \) carrying current \( I \) — This matches (II) \( B_0/2 \). The straight leads give no field at the centre, and the arc gives half the field of a full loop.
NCERT (2026-27):
Magnetic Field on the Axis of a Circular Current Loop – “… Thus, the magnetic field at P due to entire circular loop is”
\( B=\frac{\mu_0 I R^{2}}{2(x^{2}+R^{2})^{3/2}} \)\( \qquad (4.11) \)
Magnetic Field on the Axis of a Circular Current Loop – “As a special case of the above result, we may obtain the field at the centre of the loop. Here x = 0, and we obtain,”
\( B_0=\frac{\mu_0 I}{2R} \)\( \qquad (4.12) \)
Example 4.5 – “(b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. …”