A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end. A magnetic field of magnitude B = 0.265 T points out of the page. (a) If the rails are separated by 23.5 cm and the speed of the rod is 50.6 cm/s, what is the magnitude of the emf generated in volts? (b) If the rod has a resistance of 25.1 O and the rails and connector have negligible resistance, what is the current in amperes in the rod? (c) At what rate is energy being transferred to thermal energy?
Explanation / Answer
Given: (a) Applied magnetic field = B = 0.265 T Speed of the Metal rod = V = 50.6 cm /s = 0.506 m/s Distance between rails = L = 23.5 cm = 0.235m It is known by the formula , we have induced emf in the rails is : e = - BL V = (0.265)(0.506)(0.235 ) = 0.0315 volt (Magnitude ) (b) Resistence of the rod = R = 25.1 thus, current induced in the rails is : I = e / R = (0.0315)/ (25.1) = 0.001255 A (c) Rate of energy transforred to thermal energy : P = I2 R = (0.001255)2 (25.1) = 3.956 x 10-5 watt (c) Rate of energy transforred to thermal energy : P = I2 R = (0.001255)2 (25.1) = 3.956 x 10-5 watt