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 - current in loop `PQRS` is clockwise from `P` to
`QRS` (b) `oversetrarrF = BI_(0) b (3hatk -4hati) (C) I = (mg)/(6bB_(0))`
`oversetrarrB =(3hati +4hatk)B_(0)`
(a) `oversetrarr(tau)_(mg) +oversetrarrtau_(1) =0`
`oversetrarrtau_(1) = - oversetrarrtau_(mg) =0`
`I oversetrarrA xx oversetrarrB = - tau_(mg) hatj`
`iAB [hatn XX (3 hati + 4hatk) ] = tau_(mg) (-hatj)`
To satisfy equation `hatn = -hatk`
so current in loop must be in clockwise direction and current is from `P` to `Q`
`oversetrarrt_(1) = IAB [-hatk xx (3 hati + 4hatk)]`
`oversetrarrtau_(1) = 3I abB (-hatj)`
(b) `oversetrarrF_(RS) = I oversetrarrb xx oversetrarrB = Ib (-hatj) xx B_(0) (3hati + 4hatk)`
`= ib B_(0) (-4hati + 3hatk)`
(c) `oversetrarrtau_(mg) + oversetrarrttau_(1) =0 implies tau_(mg) = tau_(1) ..(1)`
`oversetrarrtau_(mg) = mg (a)/(2) (hatj)` and `oversetrarrtau_(1) = 3I abB (-hatj)`
From (1) and (2)
`(mga)/(2) =3I abB_(0) implies I =(mg)/(6bB_(0))`
.