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

(i) 0
(ii) 6

Explanation

(i) 0 (ii) 6
(i)Let the direction of current in wire PQ is from P to Q and its magnitude be I.
The magnetic moment of the given loop is
`M=-I abhatk`
Torque on the loop due to magnetic force is
`tau_(1)=MxB=(-Iabhatk)xx(3hati+4hatk)B_(0)hati=-3IabB_(0)hatj`
Torque of weight of the loop about axis PQ is
`tau_(2)=rxxF=(a/2hati)xx(-mghatk)=(mga)/2hatj`
We see that when the current in the wire PQ is from P to Q `tau_(1)` and `tau_(2)` are in opposite directions, so they can cancel each other and the loop may remain in equilibrium. So, the direction of current I in wire PQ is from P to Q. Further for equilibrium of the loop
`|tau_(1)|=|tau_(2)|` or `3IabB_(0)=(mga)/2:I=(mg)/(6bB_(0))`

(ii) Magnetic force on wire RS is
`F=I(IxxB)=K[(-bhatj)xx{(3hati+4hatk)B_(0)}], F=IoB_(0)(3hatk-4hati)`