Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

The torque \(\tau\) acting on a current-carrying coil of \(N\) turns, area \(A\), and current \(I\) in a uniform magnetic field \(B\) is given by:

$$\tau = N I A B \sin\theta$$

For maximum torque, \(\sin\theta = 1\), hence \(\tau_{max} = N I A B\). Since \(N, I, B\) are constant for both coils, the ratio of torques depends on the ratio of their areas: \(\frac{\tau_{square}}{\tau_{circle}} = \frac{A_{square}}{A_{circle}}\).

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

Let the total length of the wire be \(L\). For a coil with \(N\) turns, the length of wire used for one turn is \(l = \frac{L}{N}\).

• For the square coil:

  Perimeter \(4a = \frac{L}{N} \implies a = \frac{L}{4N}\)

  Area \(A_{square} = a^2 = \left(\frac{L}{4N}\right)^2 = \frac{L^2}{16N^2}\)

• For the circular coil:

  Circumference \(2\pi r = \frac{L}{N} \implies r = \frac{L}{2\pi N}\)

  Area \(A_{circle} = \pi r^2 = \pi \left(\frac{L}{2\pi N}\right)^2 = \frac{L^2}{4\pi N^2}\)

Calculating the ratio:

$$\frac{\tau_{square}}{\tau_{circle}} = \frac{A_{square}}{A_{circle}} = \frac{\frac{L^2}{16N^2}}{\frac{L^2}{4\pi N^2}} = \frac{4\pi}{16} = \frac{\pi}{4}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The ratio of the maximum torque acting on the square coil to that on the circular coil is \(\pi : 4\).

💡 EL10 Concept Insight:

Think of it like a fence: for a fixed amount of wire, a circle is the most efficient shape to enclose the largest possible area. Because the square encloses less area than the circle, it experiences less "push" (torque) from the magnetic field.