The coil in the figure below contains 425 turns and has an area per turn of 3.00 10-3 m2. The magnetic field is 0.21 T, and the current in the coil is 0.27 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 0.81. The radius of the shaft is 0.011 m. What is the magnitude of the minimum normal force that the brake shoe exerts on the shaft?

M=M(fr)

M=p⒨Bsinα
sinα=sin90=1,

M=NIAB
M(fr)=F(fr)R=μF⒩R
NIAB =μF⒩R
F⒩= NIAB/μR=
=425•0.27•3•10⁻³•0.21/0.81•0.011=
=8.11 N

To find the magnitude of the minimum normal force that the brake shoe exerts on the shaft, we need to consider the torque exerted by the coil and balance it with the static friction force on the shaft.

The torque exerted by the coil can be calculated using the formula:

τ = NIAB sinθ

Where:
N = number of turns in the coil (425)
I = current in the coil (0.27 A)
A = area per turn of the coil (3.00 * 10^-3 m^2)
B = magnetic field (0.21 T)
θ = angle between the magnetic field and the normal to the coil (90°)

Plugging in the values:

τ = (425)(0.27)(3.00 * 10^-3)(0.21)(sin90°)

Since sin90° = 1, the torque simplifies to:

τ = (425)(0.27)(3.00 * 10^-3)(0.21)

Now we can use the torque equation to calculate the static friction force:

Frictional torque = Static friction force x radius of the shaft

τ = Ff * r

Where:
τ = torque (calculated above)
r = radius of the shaft (0.011 m)

Plugging in the values:

(425)(0.27)(3.00 * 10^-3)(0.21) = Ff * 0.011

Solving for Ff (static friction force):

Ff = (425)(0.27)(3.00 * 10^-3)(0.21) / 0.011

Ff ≈ 7.19 N

Therefore, the magnitude of the minimum normal force that the brake shoe exerts on the shaft is approximately 7.19 N.

To find the magnitude of the minimum normal force that the brake shoe exerts on the shaft, we can use the concept of static friction.

First, let's determine the magnetic moment (µ) of the coil. The magnetic moment is given by the formula:

µ = NIA

Where:
N = number of turns in the coil = 425
I = current in the coil = 0.27 A
A = area per turn = 3.00e-3 m^2

Substituting the values into the formula, we get:

µ = 425 * 0.27 * 3.00e-3 = 0.34425 A*m^2

The torque exerted on the coil due to the magnetic field is given by the formula:

τ = µ * B

Where:
B = magnetic field = 0.21 T

Substituting the values into the formula, we get:

τ = 0.34425 * 0.21 = 0.0723 N*m

Since the coil is prevented from turning by the brake shoe, the static friction between the shaft and the brake shoe provides an equal and opposite torque to balance the torque due to the magnetic field. The torque due to static friction is given by the formula:

τ_friction = r * F_friction

Where:
r = radius of the shaft = 0.011 m
F_friction = static frictional force

Substituting the values into the formula, we get:

0.0723 N*m = 0.011 m * F_friction

Solving for F_friction, we find:

F_friction = 0.0723 N*m / 0.011 m

F_friction = 6.57 N

Therefore, the magnitude of the minimum normal force that the brake shoe exerts on the shaft is 6.57 N.