Question 192 marksVery Short AnswerSection B

A wire of length L is bent round into (i) a square coil having N turns and (ii) a circular coil having N turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.

Moving Charges and Magnetism
Torque on a current-carrying coil

Official answer

Concept


Maximum torque on an N-turn coil is τmax⁡=NIAB\tau_{\max} = NIAB. Since N, I and B are the same for both coils, the ratio of torques equals the ratio of their areas.


Areas from the same wire length L


  • Square (side s): N(4s)=L→s=L/4NN(4s) = L \rightarrow s = L/4N, so Asq=s2=L2/16N2A_{sq} = s^2 = L^2/16N^2
  • Circle (radius r): N(2πr)=L→r=L/2πNN(2\pi r) = L \rightarrow r = L/2\pi N, so Ac=πr2=L2/4πN2A_c = \pi r^2 = L^2/4\pi N^2

Ratio


τsq/τc=Asq/Ac=(L2/16N2)÷(L2/4πN2)\tau_{sq} / \tau_c = A_{sq} / A_c = (L^2/16N^2) \div (L^2/4\pi N^2) = π/4≈0.785\pi/4 \approx 0.785


The circular coil experiences the larger torque because a circle encloses the maximum area for a given perimeter.

Answer keywords:maximum torqueNIABmagnetic momentsquare coilcircular coilenclosed arearatio pi/4same wire length

Marking scheme

  1. 1.1 mark: writing τmax⁡=NIAB\tau_{\max} = NIAB and reducing the ratio to Asq/AcA_{sq}/A_c.
  2. 2.0.5 mark: correct areas Asq=L2/16N2A_{sq} = L^2/16N^2 and Ac=L2/4πN2A_c = L^2/4\pi N^2.
  3. 3.0.5 mark: final ratio π/4\pi/4 (≈ 0.785). Accept 0.79 or equivalent.

Hint

Maximum torque = NIABNIAB; with N, I, B equal, only the enclosed area differs — express side and radius in terms of L and N.

Quick oral answer

Maximum torque is NIABNIAB, and since N, I and B are identical, the torques are in the ratio of the coils' areas; the circle encloses more area for the same wire, giving τsquare:τcircle=π:4\tau_{square} : \tau_{circle} = \pi : 4.

Common mistakes

  1. 1.Ignoring the N turns while relating perimeter to L (using 4s=L4s = L instead of N⋅4s=LN \cdot 4s = L); although N cancels, the setup shown must be correct for full marks.
  2. 2.Computing the area ratio upside-down and reporting 4/π4/\pi instead of π/4\pi/4.
  3. 3.Forgetting that maximum torque needs sin⁡θ=1\sin\theta = 1 and inserting an unnecessary orientation factor.

Analysis and explanation

Concept


A current loop in a magnetic field is a magnetic dipole of moment m=NIAm = NIA. The torque it feels is τ=mBsin⁡θ=NIABsin⁡θ\tau = mB \sin\theta = NIAB \sin\theta, which is maximum (τmax⁡=NIAB\tau_{\max} = NIAB) when the plane of the coil is parallel to the field (θ=90∘\theta = 90^\circ).


Why area is the deciding factor


Both coils are made from the same wire length L and carry the same current with the same number of turns N in the same field B. Everything cancels except the enclosed area of one turn, so the problem reduces to comparing AsqA_{sq} and AcA_c.


Exam trap


  • Students often forget that N turns use up wire as N × (perimeter of one turn), so the side/radius already contains a factor of N — but N cancels in the final ratio.
  • The result π/4<1\pi/4 < 1 confirms the classic isoperimetric fact: for a fixed perimeter, a circle encloses more area than a square.

Real-world link


This is exactly why moving-coil galvanometers and motors favour compact geometries — maximising enclosed area per unit conductor maximises the deflecting torque.

Interesting facts

The result reflects the isoperimetric inequality: among all plane figures of a given perimeter, the circle encloses the greatest area — proved rigorously only in the 19th century by Weierstrass.

Real galvanometer coils are wound rectangular, not circular, because a rectangular coil is easier to suspend and sits neatly in the radial field gap, trading a little area for mechanical convenience.

Question details

Q19 of 39 questions

Board
CBSE
Class
Class 12
Subject
Physics
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Moving Charges and Magnetism
Topic
Torque on a current-carrying coil
Bloom's level
apply
Syllabus topic
Unit III: Moving Charges and Magnetism — Torque on a current loop in a magnetic field (magnetic dipole)
Author
CBSE
Reviewed by
Padhantu Academic Team

Spotted a mistake or something unclear?

Tell us and we will fix it.

Frequently asked questions

Why is the torque on the circular coil greater?

Torque depends on the enclosed area (τmax⁡=NIAB\tau_{\max} = NIAB). For a fixed wire length, a circle encloses more area than a square, so the circular coil has the larger magnetic moment and hence the larger maximum torque — the square-to-circle ratio being π/4\pi/4.

Does the number of turns N affect the final ratio?

No. N appears in both the length constraint (N·perimeter = L) and in τmax⁡=NIAB\tau_{\max} = NIAB, and it cancels completely. The ratio π/4\pi/4 is independent of N, I, B and L.