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Chapter 19: Problem 54

In an electric motor, a coil with 100 turns of radius \(2.0 \mathrm{cm}\) can rotate between the poles of a magnet. The magnetic field strength is $0.20 \mathrm{T}\(. When the current through the coil is \)50.0 \mathrm{mA},$ what is the maximum torque that the motor can deliver?

Short Answer

Expert verified
Answer: The maximum torque that the motor can deliver is approximately 0.00126 Nm.

Step by step solution

01

Convert given values to SI units.

First, convert the given values to the SI units: the current I = 50.0 mA = 0.050 A, and the radius r = 2.0 cm = 0.02 m.
02

Calculate the area of the coil.

The area of the coil can be calculated using the formula for the area of a circle: A = πr², where r is the radius of the coil. With the radius given as 0.02 m, we can calculate the area as follows: A = π(0.02 m)² ≈ 0.00126 m².
03

Calculate the maximum torque.

Now we can calculate the maximum torque using the formula: Torque = N * I * A * B * sin(theta). With theta = 90 degrees, sin(theta) = 1. Using the values we have, we can calculate the torque: Torque = (100 turns)(0.050 A)(0.00126 m²)(0.20 T)(1) ≈ 0.00126 Nm. The maximum torque that the motor can deliver is approximately 0.00126 Nm.

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Most popular questions from this chapter

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