Question

A straight wire of length 90 cm carrying a current of 3 A is bent in the form of an equilateral triangular loop and is placed in a uniform magnetic field of $8 \times 10^{-4}$ T such that the plane of the loop makes an angle of $30^{\circ}$ with the direction of the magnetic field. The torque acting on the triangular loop is

• A. $81\sqrt{3} \times 10^{-6}$ Nm
• B. $27 \times 10^{-6}$ Nm
• C. $27\sqrt{3} \times 10^{-6}$ Nm
• D. $81 \times 10^{-6}$ Nm

Hint:

Angle $\theta$ in $\tau = MB \sin \theta$ is between normal and field. Angle $\alpha$ in $\tau = MB \cos \alpha$ is between plane and field. Here $\alpha = 30^{\circ}$, so torque depends on $\cos(30) = \sin(60)$.

Answer 1

The Correct Option is D