Question

points C/10 21 y Notes Ask Your Consult Interactive Solution 21.51 to see how this problem can be solved s an area per turn of 3.40 x103 m2. The magnetic field is 0.24 T, and the current in the coll is 0.24 A. A brake shoe is pressed perpendicularly against the shaft normal force that the brake shoe exerts on the shaft? to keep the coil from turning. The coefficient of static friction between the shaft and the brake shoe is 0.74. The radius of the shaft is 0.010 m. What is the is ? 010 m. What is the magnitude of the minimum Section 21 6 3 5 6 8 9

Answer 1

Torque on the coil in the magnetic field is $\tau=NIAB\sin\theta$

Number of turns $N=401$ , Current through the coil $I=0.24\,A$ , Area of each turn is $A=3.40*10^{-3}\,m^2$ , magnetic field in the region is $B=0.24\,T$

Minimum frictional force that a brake shoe exerts on shaft is $\mu F$

Coefficient of static friction between shaft and brake shoe is $\mu=0.74$

Torque due to friction is $\tau=\mu F r$ ( is radius of shaft , $r=0.010\,m$ )

$\mu F r=NIAB\sin\theta$

$F=\frac{NIAB\sin\theta}{\mu r}$

When the plane of coil is lying along the field lines, $F=\frac{NIAB}{\mu r}$ ($\sin\theta=1$ )

$F=\frac{401*0.24*3.40*10^{-3}*0.24}{0.74*0.010}=10.6\,N$

Magnitude of minimum normal force that the brake shoe exerts on shaft is