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?

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

The torque exerted by the magnetic field on the coil is equal to the torque exerted by the static friction force.

Torque due to the magnetic field:
τ_magnetic = N * B * A * r

where N is the number of turns, B is the magnetic field, A is the area per turn, and r is the radius of the coil.

Torque due to static friction:
τ_friction = μ * F_normal * r

where μ is the coefficient of static friction, F_normal is the normal force, and r is the radius of the shaft.

Since the coil is not turning, the torques must be equal:

τ_magnetic = τ_friction

Therefore,
N * B * A * r = μ * F_normal * r

Simplifying the equation:
F_normal = (N * B * A) / μ

Now, we can plug in the given values:
N = 425 turns
B = 0.21 T
A = 3.00 * 10^-3 m^2
μ = 0.81

Calculating the magnitude of the minimum normal force:
F_normal = (425 * 0.21 * 3.00 * 10^-3) / 0.81
F_normal ≈ 2.623 N

Therefore, the magnitude of the minimum normal force that the brake shoe exerts on the shaft is approximately 2.623 N.