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A coil rotating perpendicularly to a uniform magnetic field contains 290 turns and has an area per turn of 3.3 × 10-3 m2. The magnetic field is 0.15 T, and the current in the coil is 0.15 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 0.53. The radius of the shaft is 0.014 m. What is the magnitude of the minimum normal force that the brake shoe exerts on the shaft?

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Answer:

Step-by-step explanation:

To find the magnitude of the minimum normal force that the brake shoe exerts on the shaft, we need to consider the equilibrium condition for the rotating coil.

The torque on the coil due to the magnetic field must be balanced by the frictional torque applied by the brake shoe to keep the coil from turning.

Calculate the magnetic torque (τ_mag) on the coil:

The magnetic torque on a coil in a magnetic field is given by the formula:

τ_mag = N * B * A * I * sinθ

where:

N = number of turns in the coil = 290

B = magnetic field strength = 0.15 T

A = area per turn of the coil = 3.3 × 10^(-3) m^2

I = current in the coil = 0.15 A

θ = angle between the magnetic field and the normal to the coil (which is 90 degrees in this case, so sinθ = 1)

Substitute the values:

τ_mag = 290 * 0.15 T * 3.3 × 10^(-3) m^2 * 0.15 A * 1

τ_mag ≈ 0.021645 Nm

Calculate the frictional torque (τ_friction) applied by the brake shoe:

The frictional torque is given by:

τ_friction = μ * R * F_normal

where:

μ = coefficient of static friction between the shaft and the brake shoe = 0.53

R = radius of the shaft = 0.014 m

F_normal = normal force exerted by the brake shoe on the shaft (what we need to find)

Substitute the values:

τ_friction = 0.53 * 0.014 m * F_normal

Equilibrium condition:

For the coil to remain stationary (not turn), the magnetic torque and the frictional torque must be equal. So, we have:

τ_mag = τ_friction

Set the two torque expressions equal to each other and solve for F_normal:

0.021645 Nm = 0.53 * 0.014 m * F_normal

F_normal ≈ 0.021645 Nm / (0.53 * 0.014 m)

F_normal ≈ 0.021645 Nm / 0.00742 N

F_normal ≈ 2.912 N

The magnitude of the minimum normal force that the brake shoe exerts on the shaft is approximately 2.912 Newtons.

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