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Chapter 19: Problem 53

In an electric motor, a circular coil with 100 turns of radius $2.0 \mathrm{~cm}$ can rotate between the poles of a magnet. When the current through the coil is \(75 \mathrm{~mA},\) the maximum torque that the motor can deliver is \(0.0020 \mathrm{~N} \cdot \mathrm{m} .\) (a) What is the strength of the magnetic field? (b) Is the torque on the coil clockwise or counterclockwise as viewed from the front at the instant shown in the figure?

Short Answer

Expert verified
a) The strength of the magnetic field is approximately 0.212 T. b) The torque on the coil is counterclockwise as viewed from the front at the instant shown in the figure.

Step by step solution

01

List the given values

We have the following information: - Number of turns in the coil (n) = 100 turns - Radius of the coil (r) = 2.0 cm = 0.02 m (converted to meters) - Current through the coil (I) = 75 mA = 0.075 A (converted to amperes) - Maximum torque that the motor can deliver (τ_max) = 0.0020 N·m
02

Find the magnetic field strength

We'll use the formula for the torque on a current loop in a magnetic field: τ = n * I * A * B * sinθ Where: τ is the torque, n is the number of turns in the coil, I is the current through the coil, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the magnetic field and the plane of the coil. At the maximum torque (θ = 90°), sinθ = 1, and the formula becomes: τ_max = n * I * A * B We need to find the area (A) of the coil. The coil has a circular shape, so: A = π * r^2 = π * (0.02)^2 = 0.001256 m^2 Now we can solve for B: B = τ_max / (n * I * A) B = 0.0020 N·m / (100 * 0.075 A * 0.001256 m^2) B ≈ 0.212 T
03

Determine the direction of the torque

To find the direction of the torque, we'll use the right-hand rule. This rule states that if you grab the coil with your right hand, with your fingers pointing in the direction of the current, your thumb will point in the direction of the torque. Assembling the loop with the fingers around the loop in the direction of the current, the thumb points outward from the front of the coil. Therefore, the torque on the coil is counterclockwise, as viewed from the front at the instant shown in the figure. #Answer# (a) The strength of the magnetic field is approximately 0.212 T. (b) The torque on the coil is counterclockwise as viewed from the front at the instant shown in the figure.

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