Correct Answer - A::B::C::D
Let us consider the current in the clockwise direction in loop PQRS. Force on wire QR.
`vec(F_(QR))=I(vec(l)xxvec(B))=I[(a hat i)xx(3hat i_4 hat k)(B_(0)]`
`=I(B_0)[3a hat i xx hat i xx4 a hat i xxhat k)=I(B_0)[0+4a(- hat j)]=-4a(B_0)I(hat j)`
Force on wire PS
`vec(F_(PS))=I(vec(l)xx vec(B))=I[a(- hat i)xx(3 hat i + 4 hat k)B_(0)]=4 a B_(0)I(hat j)`
Thus we see that force on QR is equal and opposite to htat on PS and balance each other.
The force on RS is
`vec(F_(RS))=I(vec(l)xx vec(B))=I[b(- hat j)xx(3 hat i + 4 hat k)B_(0)]`
`=IbB_(0)[3 hat k -4 hat i]` ...(i) and The torque about PQ by this force is
`vec(tau_(RS))=vec(r)xxvec(F)=(hat i a)xx(3 hat k - 4hat i)IbB_(0)`
`I ab B_(0)(3 hat j)`...(ii)
The torque about PQ due to weight of hte wire PQRS is
`(tau)=mg(a/2)` ...(iii)
For the wire loop to be horizontal we have to equate (ii) and (iii) `3IabB_(0)=mg(a/2)`
`implies I+(mg)/(6 b b_(0)` ...(iv)
Therefore,
(a) The direction of current assumed is right. This is because torque due to mg and current are in opposite directions. Therefore, current is from p to Q.
(b) from (i), `vec(F_(RS))=IbB_(0)(3 hat k - 4hati)`
(c ) from (iv) `I=(mg)/(6aB_(0)`.