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)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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 $2 π \sqrt{\frac{\sqrt{3} a}{2 g}}$

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

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Effective length of loop DABE is line joining DE.

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