A rod PQ of length 20.0 cm and mass 200 g is suspended horizontally from a fixed point O using two threads attached to its ends, each 20.0 cm long. A uniform magnetic field of strength 0.500 T exists around the wire PQ , pointing out of the page. The wire is connected to a battery such that when a switch is closed, a current of 2.0 A flows from P to Q . The wires linking PQ to the battery are loose and exert zero for

Step-by-step solution

Given mass of the rod m = 200 g = 0.2 kg and length L = 20.0 cm = 0.2 m . The weight of the rod acts downwards: W = mg = 0.2 9.8 = 1.96 N The current I = 2.0 A flows from P to Q (rightwards) and the magnetic field B = 0.500 T points out of the page. Using the right-hand rule, the magnetic force on the rod is directed downwards: F_m = ILB = 2.0 0.2 0.500 = 0.2 N Total downward force on the rod is: F_ net = W + F_m = 1.96 + 0.2 = 2.16 N Since OP = OQ = PQ = 20.0 cm , OPQ is an equilateral triangle. The angle each thr