A current carrying loop is in the shape of an equilateral triangle of side length a. It's mass is M and it is in vertical plane. There exists a uniform horizontal magnetic field $\vec{B}$ of magnitude $B_{0}$ in the region (up to dotted line) as shown in the diagram. (neglect emf induced in the loop)

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If the loop is in equilibrium for $y = \frac{\sqrt{3}}{4} a$ , the corresponding current in the loop is $\frac{2 M g}{a B_{0}}$

If the loop is in equilibrium for $y = \frac{\sqrt{3}}{4} a$ , the corresponding current in the loop is $\frac{M g}{a B_{0}}$

If the loop is displaced slightly from it's equilibrium position $(y = \frac{\sqrt{3}}{4} a)$ in it's plane perpendicular to side AB and released, the time period of it's oscillation is $2 π \sqrt{\frac{\sqrt{3} a}{4 g}}$

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Detailed Solution

Effective length of loop DABE is line joining DE.

$\begin{array}{l} ℓ_{e} = 2 ytan 30^{\circ} \\ = \frac{2 y}{\sqrt{3}} \\ I f y = \frac{\sqrt{3} a}{4} t h e n ℓ_{e} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3} a}{4} = \frac{a}{2} \\ At equilibrium i ℓ_{e} B_{0} = Mg \\ \Rightarrow i = \frac{2 Mg}{{aB}_{0}} \\ When it is displaced slightly : \\ dF = {iB}_{0} d ℓ = - \frac{2 Mg}{{aB}_{0}} \times B_{0} \times \frac{2}{\sqrt{3}} dy = - (\frac{4 Mg}{\sqrt{3} a}) dy \\ M a = - (\frac{4 Mg}{\sqrt{3} a}) dy \\ c o m p a r i n g w i t h a n g u l a r S H M a = - ω^{2} y \\ ω^{2} = \frac{4 g}{\sqrt{3} a} \Rightarrow \frac{2 π}{T} = \sqrt{\frac{4 g}{\sqrt{3} a}} \\ ∴ T = 2 π \sqrt{\frac{\sqrt{3} a}{4 g}} \end{array}$