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 remasins hinged along a horizontal line taken as the y-axis (see figure). Take the vertically upward direction as the z-axis. A uniform magnetic fiedl `B=(3hati+3hatk)B_0` exists in the region. The lop is held in the xy-plane and a current `I` is passed through it. The loop in 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`.

Explanation

(a) Torque due to weight of coil,
`vectau=(a/2 hati)xx(-mghatk)=mga/2(hatj)`
For the equilibrium of loop, torque on it must be along negative y-axis. Let the magnetic moment of loop be `mu hat k`.As the loop lies in x-y plane, its magnetic moment vector (from right hand thumb rule) either points up or down.
Torque due to magnetic force, `vectau_B =vec mu xx vec B = mu hat k xx (3hat i+4hat k)B_0 = 3muB_0hat j`
If it is in negative direction, `vec mu` must point downward. So, the current in the coil must be from P to Q.

(b) Force acting on arm
`RS=I(vec I xx vec B)=I[(-bhatj)xx(3hati+4hatk)B_0]`
`=IB_0b(3hatk-4hatj)`
(c) In equilibrium `vectau_(gravity)+vec tau_B=0`
Hence, `3(abI)B_0=(mga)/2` or `I=(mg)/(6B_0b)`.