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

(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)`.