Question

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 1

Correct Answer - B
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=-Iabhatk`
Torque on the loop due to magnetic force is
`tau_1=MxxB`
`=(-Iabhatk)xx(3hati+4hatk)B_0hati`
`=-3IabB_0hatj`
Torque on weight of the loop about axis `PQ` is
`tau_2=rxxF=(a/whati)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)`
Magnetic force on wire `RS` is
`F=I((IxxB)`
`=I[(_bhatj)xx{(3hati+4hatk)B_0}]`
`F=IbB_0(3hatk-4hati)`