Question

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.

Hint:

For a current-carrying coil, \[ \tau_{\max}=NIAB. \] If \(N\), \(I\), and \(B\) are the same for different shapes of coils, then \[ \tau_{\max}\propto A. \] Also, among all plane figures with the same perimeter, the circle encloses the maximum area. Therefore, the circular coil always experiences a larger maximum torque than the square coil.

Answer 1

Concept: The torque experienced by a current-carrying coil placed in a uniform magnetic field is given by \[ \tau = N I A B \sin\theta, \] where
• \(N\) is the number of turns of the coil,
• \(I\) is the current flowing through the coil,
• \(A\) is the area enclosed by the coil,
• \(B\) is the magnetic field strength,
• \(\theta\) is the angle between the normal to the plane of the coil and the magnetic field. The maximum torque is obtained when \[ \sin\theta =1 \qquad \text{or} \qquad \theta=90^\circ. \] Hence, \[ \tau_{\max}=NIAB. \] Since both coils have the same number of turns, carry the same current and are placed in the same magnetic field, the ratio of their maximum torques depends only on the ratio of their areas. Thus, \[ \frac{\tau_s}{\tau_c} = \frac{A_s}{A_c}, \] where \(A_s\) and \(A_c\) denote the areas of the square and circular coils respectively.

Step 1: Find the area of the square coil.
The total length of wire is \(L\) and the coil has \(N\) turns. Therefore, the length of wire available for one turn of the square is \[ \text{Perimeter of one square} = \frac{L}{N}. \] If \(a\) is the side of the square, then \[ 4a=\frac{L}{N}. \] Hence, \[ a=\frac{L}{4N}. \] The area enclosed by one turn of the square coil is \[ A_s=a^2 = \left(\frac{L}{4N}\right)^2 = \frac{L^2}{16N^2}. \]

Step 2: Find the area of the circular coil.
The length of wire available for one turn of the circular coil is also \[ \frac{L}{N}. \] If \(r\) is the radius of the circular coil, then \[ 2\pi r=\frac{L}{N}. \] Therefore, \[ r=\frac{L}{2\pi N}. \] Hence, the area of one turn of the circular coil is \[ A_c=\pi r^2 = \pi\left(\frac{L}{2\pi N}\right)^2. \] \[ A_c = \pi\left(\frac{L^2}{4\pi^2N^2}\right) = \frac{L^2}{4\pi N^2}. \]

Step 3: Determine the ratio of maximum torques.
Since \[ \frac{\tau_s}{\tau_c} = \frac{A_s}{A_c}, \] we get \[ \frac{\tau_s}{\tau_c} = \frac{\dfrac{L^2}{16N^2}} {\dfrac{L^2}{4\pi N^2}}. \] Cancelling \(L^2\) and \(N^2\), \[ \frac{\tau_s}{\tau_c} = \frac{4\pi}{16} = \frac{\pi}{4}. \] Therefore, \[ \boxed{ \frac{\tau_{\text{square}}} {\tau_{\text{circular}}} = \frac{\pi}{4} } \] Hence, the required ratio of the maximum torque acting on the square coil to that on the circular coil is \[ \boxed{\frac{\pi}{4}:1}. \]