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.

Q: 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.

**Concept** Maximum torque on an N-turn coil is $\tau_{\max} = NIAB$. 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): $N(4s) = L \rightarrow s = L/4N$, so $A_{sq} = s^2 = L^2/16N^2$ - Circle (radius r): $N(2\pi r) = L \rightarrow r = L/2\pi N$, so $A_c = \pi r^2 = L^2/4\pi N^2$ **Ratio** $\tau_{sq} / \tau_c = A_{sq} / A_c = (L^2/16N^2) \div (L^2/4\pi N^2)$ = **$\pi/4 \approx 0.785$** The circular coil experiences the larger torque because a circle encloses the maximum area for a given perimeter.

Q: Why is the torque on the circular coil greater?

Torque depends on the enclosed area ($\tau_{\max} = NIAB$). 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 $\pi/4$.

Q: Does the number of turns N affect the final ratio?

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

Moving Charges and Magnetism

Torque on a current-carrying coil

Official Answer

Concept

Maximum torque on an N-turn coil is $\tau_{\max} = NIAB$ . 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): $N(4s) = L \rightarrow s = L/4N$ , so $A_{sq} = s^2 = L^2/16N^2$
Circle (radius r): $N(2\pi r) = L \rightarrow r = L/2\pi N$ , so $A_c = \pi r^2 = L^2/4\pi N^2$

Ratio

$\tau_{sq} / \tau_c = A_{sq} / A_c = (L^2/16N^2) \div (L^2/4\pi N^2)$ = $\pi/4 \approx 0.785$

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

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

Marking Scheme

11 mark: writing $\tau_{\max} = NIAB$ and reducing the ratio to $A_{sq}/A_c$ .
20.5 mark: correct areas $A_{sq} = L^2/16N^2$ and $A_c = L^2/4\pi N^2$ .
30.5 mark: final ratio $\pi/4$ (≈ 0.785). Accept 0.79 or equivalent.

Hint

Maximum torque = $NIAB$ ; 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 $NIAB$ , 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 $\tau_{square} : \tau_{circle} = \pi : 4$ .

Analysis & Explanation

Concept

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

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 $A_{sq}$ and $A_c$ .

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 $\pi/4 < 1$ 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.

Common Mistakes

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

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.

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Frequently Asked Questions

Why is the torque on the circular coil greater?

Torque depends on the enclosed area ( $\tau_{\max} = NIAB$ ). 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 $\pi/4$ .

Does the number of turns N affect the final ratio?

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