Question

A current carrying loop is in the shape of an equilateral triangle of side length a. Its mass is M and it is in vertical plane. There exists a uniform horizontal magnetic field B in the region shown. The loop is in equilibrium for $y = \frac{\sqrt{3}}{4}a$. Neglect emf induced in the loop.

• A. The current in the loop is $\frac{2Mg}{aB}$
• B. If the loop is displaced slightly in its plane perpendicular to its side $AB$ and released it the loop performs SHM.
• C. If the loop is displaced slightly in its plane perpendicular to its side $AB$ and released its period of oscillation is $\pi \sqrt{\frac{\sqrt{3}a}{2g}}$
• D. If the current in the loop is changed to $\frac{\sqrt{3}Mg}{2aB}$ equilibrium shifts to $y = a$.

Answer 1

Answer: (A), (B), (C)

The correct options are the first three.

Answer 2

1. Only the part of the loop in the field (which is the chord joining the two intersection points with the boundary) produces net force. Calculating the intersections for an equilateral triangle with vertex C at (0,0) shows that at equilibrium (y = (√3/4)a) the chord length is a/2. Hence moment–balance gives I B (a/2)= M g ⇒ I= (2M g)/(aB).

2. A small displacement (rotation about AB) changes the chord length linearly so that the extra magnetic force is ∝ δ(displacement). This yields SHM about the equilibrium.

3. A careful linearization gives the period T = π √((√3a)/(2g)).

4. A change in current to I = (√3Mg)/(2aB) would mathematically require y = a for balance, but since the top vertex lies at y = √3a/2, such an equilibrium lies outside the physical loop.