- 天同Lv 73 年前最愛解答
(a) Given PQ = L, hence PR = L.cos(40), and RQ = L.sin(40)
Area of loop = (1/2).(RQ).(PR) = (1/2).(L.sin(40)).(L.cos(40))
Magnetic moment of current loop = current x area = [I.(L^2)sin(40)cos(40)]/2
The direction of magnetic moment is given by the Right-Hand-Grip-Rule, and is pointing into the paper, i.e. in the -ve z-direction.
(b) Torque = [B.I.(L^2)sin(40)cos(40)]/2
The direction of torque is in the -ve y-direction.