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 $\vec{B}$ is in plane of the coil. If due to a current i in the triangle a torque $ \tau $ acts on it, the side l of the triangle is

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

Answer 1

The Correct Option is B

To solve the given problem, let's start by analyzing the torque acting on a current-carrying coil in a magnetic field.

When a coil with current \( i \) is placed in a magnetic field \( \vec{B} \), and the plane of the coil is perpendicular to the direction of the magnetic field, the torque \( \tau \) on the coil is given by the formula:

\( \tau = n \cdot i \cdot A \cdot B \cdot \sin(\theta) \)

where:

• \( n \) = number of turns (for a single loop, \( n=1 \))
• \( i \) = current in the loop
• \( A \) = Area of the loop
• \( B \) = Magnetic field strength
• \( \theta \) = angle between the normal to the plane of the coil and the magnetic field

Given that the magnetic field \( \vec{B} \) is in the plane of the coil, the angle \( \theta = 90^\circ \), hence \( \sin(\theta) = 1 \).

For an equilateral triangle with side \( l \), the area \( A \) is given by:

\( A = \frac{\sqrt{3}}{4} l^2 \)

Substitute the area \( A \) in the torque formula:

\( \tau = i \cdot \left(\frac{\sqrt{3}}{4} l^2\right) \cdot B \)

Rearranging to find \( l^2 \), we have:

\( l^2 = \frac{4 \tau}{\sqrt{3} i B} \)

Solving for \( l \), we take the square root:

\( l = \sqrt{\frac{4 \tau}{\sqrt{3} i B}} \)

Simplifying further, we have:

\( l = 2 \left(\frac{\tau}{\sqrt{3} i B}\right)^{1/2} \)

Thus, the option \(2 \left(\frac{\tau}{\sqrt{3} i B}\right)^{1/2}\) is the correct answer, which matches the given correct answer.