A loop PQR formed by three identical uniform conducting rods each of length ‘a’ is suspended from one of its vertices (P) so that it can rotate about horizontal fixed smooth axis CD. Initially plane of loop is in vertical plane. A constant current ‘i’ is flowing in the loop. Total mass of the loop is ‘m’. At t = 0, a uniform magnetic field of strength B directed vertically upwards is switched on. Acceleration due to gravity is ‘g’.

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The equilibrium position (angle of the plane of the loop with vertical) is given by $\tan^{- 1} (\frac{3 B a i}{4 m g}) .$

The equilibrium position (angle of the plane of the loop with vertical) is given by $\tan^{- 1} (\frac{3 B a i}{2 m g}) .$

Minimum value of B so that the plane of the loop becomes horizontal (even for an instant) during its subsequent motion is $\frac{4 m g}{3 i a} .$

Minimum value of B so that the plane of the loop becomes horizontal (even for an instant) during its subsequent motion is $\frac{2 m g}{3 i a} .$

answer is B, D.

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Detailed Solution

Applying energy conservation, initially, kinetic energy = 0
Gravitational P.E = 0 (say) and magnetic P.E = 0
Where, $μ$ = Magnetic moment of the loop = $i (\frac{\sqrt{3} a^{2}}{4}) .$
Finally when the loop becomes horizontal, Kinetic energy = 0
Gravitational $P . E = m g (\frac{a}{\sqrt{3}})$ (because mg acts on the centre of mass)

Magnetic P.E = $- μ B \Rightarrow 0 + 0 + 0 = 0 + \frac{m g a}{\sqrt{3}} - μ B \Rightarrow B = \frac{m g a}{\sqrt{3} μ} = \frac{4 m g}{3 i a} .$

For equilibrium position. Torque due to magnetic force should be balanced by the torque due to gravitational force (mg). $τ_{B} = μ B \sin (90 - 0); = \frac{m g \sin 0 a}{\sqrt{3}} .$