Question

If a thin conducting wire of length \(12m\) carrying a current of \(2\sqrt3 A\) is bent into a regular hexagonal loop and placed in a uniform magnetic field of \(2T\), then the maximum torque acting on the loop is:

• A. \(36Nm\)
• B. \(48Nm\)
• C. \(72Nm\)
• D. \(24Nm\)

Hint:

Remember: \[ \tau=IAB\sin\theta. \] Maximum torque occurs when the magnetic field is parallel to the plane of the loop.

Answer 1

The Correct Option is C

Concept: Maximum torque on a current loop: \[ \tau=IAB. \] For maximum torque, \[ \sin\theta=1. \]

Step 1: Find side of hexagon.
Perimeter \[ =12m. \] For regular hexagon, \[ 6a=12. \] \[ a=2m. \]

Step 2: Area of regular hexagon.
\[ A = 6\left(\frac{\sqrt3}{4}a^2\right). \] \[ = 6\left(\frac{\sqrt3}{4}\times4\right). \] \[ = 6\sqrt3. \]

Step 3: Calculate torque.
\[ \tau = IAB. \] \[ = (2\sqrt3)(6\sqrt3)(2). \] \[ = 72Nm. \] Hence, \[ \boxed{72Nm}. \]