A rectangular loop PQRS made from a uniform wire has length a, width b and mass m. It is free to rotate about the arm PQ, which remains hinged along a horizontal line taken as the y-axis (see figure). Take the vertically upward direction as the z-axis. A uniform magnetic field `vec(B) = (3 hat(i) + 4 hat(k)) B_(0)` exists in the region. The loop is held in the x-y plane and a current I is passed through it. The loop is now released and is found to stay in the horizontal position in equilibrium

(a) What is the direction of the current I in PQ?
(b) Find the magnetic force on the arm RS.
(c) Find the expression for I in terms of `B_(0)`, a, b and m

Answer

A, B, C, D

Explanation

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)`.