Question

A coil in the shape of an equilateral triangle of side \(l\) is suspended between the pole pieces of a permanent magnet such that \(\mathbf{B}\) is in the plane of the coil. If due to current \(i\) in the triangle, a torque \(\tau\) acts on it. The side \(l\) of the triangle is

• A. \(\frac{2}{\sqrt{3}}\left(\frac{\tau}{Bi}\right)\)
• B. \(2\left(\frac{\tau}{\sqrt{3}Bi}\right)^{1/2}\)
• C. \(\frac{2}{\sqrt{3}}\left(\frac{\tau}{Bi}\right)^{1/2}\)
• D. \(\frac{1}{\sqrt{3}}\frac{\tau}{Bi}\)

Hint:

Torque is maximum when plane of coil is parallel to magnetic field (\(\theta=90^\circ\)).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Torque on a current loop: \(\tau = NiAB\sin\theta\). Here, \(\theta = 90^\circ \Rightarrow \sin\theta = 1\).
Step 2: Detailed Explanation:
Area of equilateral triangle: \(A = \frac{\sqrt{3}}{4}l^2\). Number of turns: \(N = 1\). Torque expression: \[ \tau = i \times \frac{\sqrt{3}}{4}l^2 \times B \] Solve for \(l^2\): \(l^2 = \frac{4\tau}{\sqrt{3}Bi}\). Taking square root: \[ l = 2\sqrt{\frac{\tau}{\sqrt{3}Bi}} = 2\left(\frac{\tau}{\sqrt{3}Bi}\right)^{1/2} \]
Step 3: Final Answer:
\[ \boxed{2\left(\frac{\tau}{\sqrt{3}Bi}\right)^{1/2}} \]