An insulated square frame ABCD(side a) is able to rotate abt 1 f its sides takn as + z axis.B=Bo(j).A small block of mass m ,charge q moveable along side CB is initially near C, when frame lies in x-z plane.Now frame is given ang. vel. w abt z-axis. Whole system lies in gravity free space. If after time t block reaches to corner B, find Bo in terms of t.

Torque due to weight of coil
τ→=(a2iˆ)×(−mgkˆ)=mg.a2(jˆ)τ→=(a2i^)×(-mgk^)=mg.a2(j^)
For the equilibrium of loop, torque on it must be along negativey−y-axis . Let the magnetic moment of loop beMkˆMk^. As the loop lies inxy−xy-plane, its magnetic moment vector ( from right−-hand thumb rule ) either points up to down. Torque due to magnetic force,
τ→B=μ→×B→=μkˆ×(3iˆ+4kˆ)B0=3μB0jˆτ→B=μ→×B→=μk^×(3i^+4k^)B0=3μB0j^
(b)(b)Force acting on armRS=I(lˆ×B→)RS=I(l^×B→)
=I[(−bjˆ)×(3iˆ+4kˆ)B0]=IB0b(3kˆ−4iˆ)=I[(-bj^)×(3i^+4k^)B0]=IB0b(3k^-4i^)
(c)(c)In equilibrium ,τ→gravity+τ→B=0τ→gravity+τ→B=0
Hence3(abI)B0=mga23(abI)B0=mga2orI=mg6B0b