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

Step 1: Calculate Loop Dimensions:
Total length $L = 90$ cm = 0.9 m. Equilateral triangle side $a = L/3 = 0.3$ m. Area of loop $A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (0.3)^2 = \frac{0.09\sqrt{3}}{4} \, m^2$.
Step 2: Magnetic Moment ($M$):
$M = NIA$ (Here $N=1$). $M = 1 \times 3 \times \frac{0.09\sqrt{3}}{4} = \frac{0.27\sqrt{3}}{4} \, Am^2$.
Step 3: Calculate Torque ($\tau$):
Formula: $\tau = MB \sin \theta$. Important: $\theta$ is the angle between the area vector (normal to loop) and the magnetic field. Given: Plane of loop makes $30^{\circ}$ with field. So, $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$. \[ \tau = \left( \frac{0.27\sqrt{3}}{4} \right) (8 \times 10^{-4}) \sin(60^{\circ}) \] \[ \tau = \left( \frac{0.27\sqrt{3}}{4} \right) (8 \times 10^{-4}) \left( \frac{\sqrt{3}}{2} \right) \] \[ \tau = \frac{0.27 \times (\sqrt{3} \times \sqrt{3}) \times 8 \times 10^{-4}}{8} \] The 8 in numerator cancels $4 \times 2$ in denominator. \[ \tau = 0.27 \times 3 \times 10^{-4} \] \[ \tau = 0.81 \times 10^{-4} \, Nm \] \[ \tau = 81 \times 10^{-6} \, Nm \]
Step 4: Final Answer:
The torque is $81 \times 10^{-6}$ Nm.