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)from `P` to `Q`
(b)`IbB_(c)(3hatk-4hati)`
( c) `(mg)/(6bB_(0))`

Explanation

(a) and (c )
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 `vecM=-lab hatk`
Torque on the loop due to magnetic force is:
`vectau_(1)=vecMxxvecB`
`=(-lab hatk)xx{(3 hati+4 hat k)B_(0)}`
`=-3labB_(0) hatj`
Torque of weight of the loop about axis `PQ` is
`vectau_(2)=vecrxxvecF`
`=(a/2 hati)xx(-mg hatk)`
`=(mga)/2 hatj`
We see that when the current in the wire `PQ` is from `P` to `Q` ,`tau_(1)` and `tau_(2)` are in opposite direction so they can cancel each other and the loop may remain in equilibrium.So the direction of current `l` in the wire `PQ` is from `P` to `Q`. Further for equilibrium of the loop.
`|vectau_(1)|=|vectau_(2)|`
or `3labB=(mga)/2`
`l=(mg)/(6bB_(0))`
(b)Magnetic force on wire `RS` is:
`F=I(vecexxB)`
`=l[(-b hatj)xx{(3 hatj)+4hatk)B_(0)}]`
or `vecF=IbB_(0)(3hatk-4hatj)`