Question

The coil in the figure below contains 433 turns and has an area per turn of 3.30 10-3 m2. The magnetic field is 0.24 T, and the current in the coil is 0.24 A. A brake shoe is pressed perpendicularly against the shaft to keep the coil from turning. The coefficient of static friction between the shaft and the brake shoe is 0.72. The radius of the shaft is 0.010 m. What is the magnitude of the minimum normal force that the brake shoe exerts on the shaft? N

Answer 1

Answer: I given data Number of turns N = 433; Area perturn A=3.30x103 m Magnetic field B=0.245, Current I= 0.24 A, Radius of shaft R=0.am 1 . Coefficent of static friction les=0.72 Torque due to friction T=fs R. T = lg Fuk : THE NIA B Sino 6:0=98] l's FNR = NIA SINO B FN = NIAB USR ..433X 0.24x(3.30x10}) (0.24) 0.72x0.01 .: Fu = 11:43N