Question

1. A rod (length = 10 cm) moves on two horizontal frictionless conducting rails, as shown below. The magnetic field in the region is directed perpendicularly to the plane of the rails and is uniform and constant. If a constant force of 0.60 N moves the bar at a constant velocity of 2.0 m/s, show that the current through the 12-2 load resistor is 0.32 A. 1222

Answer 1

Given, Length of Rod l= 10 cm =o.1 m Size constant force, f = 0.6ON relocity, v. 2 = 2 m/s Resistance, R = 122 I a current Magnetic force, F = BIl F =) I = Be Induced emf, {=Blue from ohm's law, E = IR » La wled ( BIN R R I Blu R 2 from 2 f D Bev R Be ? v=FIR >> B2 e FOR 3) B » B FIR eru >> B = 0.6 x 12 0:12 * 2 > B = B = (360 Т. I a from, BIV R (360x 0,1x2 126 -0.32 I = 0.32 A

so current through the 122 load resistor 0.32 A (proved). is 2) for primary, و کم 200 V622 v For secondary N2=50 for ideal transformer, N2 1 ک؟ N 3 » ₂a v » a 20% X 622 Grinen, Load resistance V2 = 155.5 volt RL = 72 current in Secon 155.5 Secondary , I 2 2 2 2 2 2 7 RMS curr rrent, Irms v2 155.5 2x7 - 15.7A Į Irms - 15.7 so, The rms. current in secondary is (Proved) 15.7A