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 vartically upwards is swiched on.Acceleration due to gravity is `g` then Find the minimum valeu of `B` so that the plane of the loop becomes horizontal (even for an instant)during its subsequent motion.

Explanation

Applying Energy conservation, initially kinetic energy =0 gravitational P.E.=0 (say) & Magnetic P.E.=`muB`
where, `mu`=magnetic moment of the loop =i.`((sqrt3a^(2))/4)`
Finally when the loop becomes horizontal, Kinetic energy=0 gravitational P.E.=`mg(a/sqrt3)` (because mg acts on the centre of mass) magnetic P.E.=0
`rArr 0+0+muB=0+((mga)/sqrt3)+0 rArr B=((mga)/(sqrt3 mu))=(4mg)/(3ia)`
using values, `B=400 mT`