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.

Answer

`(4mg)/(3ia)`

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(1/sqrt3)` (because `mg` acts on the centre of mass) magnetic `P.E.=0`
`rArr 0+0+muB=0+(mga)/(sqrt3)+0 rArr B=(mga)/(sqrt3mu)=(4mg)/(3ia)`