/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Two pieces of the same wire have... [FREE SOLUTION] | 91Ó°ÊÓ

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Two pieces of the same wire have the same length. From one piece, a square coil containing a single loop is made. From the other, a circular coil containing a single loop is made. The coils carry different currents. When placed in the same magnetic field with the same orientation, they experience the same torque. What is the ratio \(I_{\text {square }} / I_{\text {circle }}\) of the current in the square coil to that in the circular coil?

Short Answer

Expert verified
The ratio \( \frac{I_{\text{square}}}{I_{\text{circle}}} = \frac{\pi}{4} \).

Step by step solution

01

Understanding Torque in a Magnetic Field

The torque \( \tau \) on a loop in a magnetic field is given by \( \tau = nIA \cdot B \cdot \sin\theta \), where \( n \) is the number of loops, \( I \) is the current, \( A \) is the area of the loop, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the field and the normal to the plane of the loop. Since \( n = 1 \) and \( \theta = 90^\circ \) (perpendicular), \( \sin\theta = 1 \). Thus, \( \tau = IAB \).
02

Expressing Lengths and Areas

Let \( L \) be the length of each piece of wire. For the square coil, one side is \( \frac{L}{4} \) and its area \( A_{\text{square}} = \left(\frac{L}{4}\right)^2 \). For the circular coil, the circumference is \( L = 2\pi r \), thus \( r = \frac{L}{2\pi} \). The area of the circular coil is \( A_{\text{circle}} = \pi r^2 = \pi \left(\frac{L}{2\pi}\right)^2 \).
03

Calculating Individual Areas

For the square: \[ A_{\text{square}} = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16} \].\For the circle: \[ A_{\text{circle}} = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{L^2}{4\pi} \].
04

Equating Torques

Set the torques equal: \( I_{\text{square}} A_{\text{square}} B = I_{\text{circle}} A_{\text{circle}} B \). Simplifying gives \( I_{\text{square}} \frac{L^2}{16} = I_{\text{circle}} \frac{L^2}{4\pi} \).
05

Solving for Current Ratio

Divide both sides by \( L^2B \) and solve for the current ratio: \( \frac{I_{\text{square}}}{I_{\text{circle}}} = \frac{\frac{1}{4}}{\frac{1}{\pi}} = \frac{\pi}{4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Current in Coils
When discussing current in coils, it is crucial to understand how it contributes to the torque a coil experiences in a magnetic field. The current (\( I \)) flowing through the coil interacts with the magnetic field to produce torque. This torque will tend to rotate the coil. The relationship is governed by the torque formula\( \tau = nIA \cdot B \cdot \sin\theta \), where:
  • \( n \) is the number of loops.
  • \( I \) is the current through the coil.
  • \( A \) is the area of the loop.
  • \( B \) is the magnetic field strength.
  • \( \theta \) is the angle between the magnetic field and the normal to the plane of the coil.
In this problem, only one loop is considered (\( n = 1 \)), and the field is perpendicular to the coil (\( \theta = 90^\circ \)), making\( \sin\theta = 1 \). As a result, the torque simplifies to\( \tau = IAB \). By equating the torques experienced by both the square and the circular coils, we gain insights into the relation of currents flowing in these two distinct coil shapes.
Square Coil
A square coil is made by shaping a wire into a square loop. It's important to think about its dimensions when determining its area.
The length of the wire, denoted as\( L \), is divided equally among the four sides of the square coil. Each side of the square, therefore, has a length of\( \frac{L}{4} \).

To calculate the area (\( A_{\text{square}} \)) of the square, we use the formula for the area of a square: \(A_{\text{square}} = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}\).This calculation is vital because the area impacts the magnetic torque generated in the coil.
The larger the area, the greater the torque for a given current and magnetic field strength.
Circular Coil
The circular coil is formed by bending the wire into a circular shape and is different from a square coil due to its distribution of length.
The total length of the wire remains\( L \), which serves as its circumference, expressed as\( L = 2\pi r \).Solving for the radius, we find\( r = \frac{L}{2\pi} \).

To find the area (\( A_{\text{circle}} \)) of the circular loop, we apply the formula for the area of a circle:\[A_{\text{circle}} = \pi r^2 = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{L^2}{4\pi}\].
Because of its shape, a circular coil can exhibit unique electromagnetic properties compared to a square coil. Its symmetrical shape often aids in certain applications, such as sensors or magnetic field generation.
Length and Area of Wire
In this context, the length and area of the wire play critical roles in determining the properties of both square and circular coils.
Though the total length (\( L \)) of the wire remains constant, it affects how the wire can be formed into different shapes and what makes up the total area of the coil loops.This length directly influences the sides of the square coil and the radius of the circular coil.

Understanding the area derived from the coil’s shape is crucial because it directly impacts the torque produced when an electric current flows through the coil within a magnetic field.

When comparing shapes like squares and circles, the area influences how much torque can be achieved for a given magnetic field.The analysis of this principle aids in applications where maximizing or controlling torque is essential, such as in electric motors and generators.

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Most popular questions from this chapter

A proton is projected perpendicularly into a magnetic field that has a magnitude of \(0.50 \mathrm{~T}\). The field is then adjusted so that an electron will follow the exact same circular path when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron? Verify that your answer is consistent with your answers to the Concept Questions.

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions A horizontal wire is hung from the \(W\) ceiling of a room by two massless strings. A uniform magnetic field is directed from the ceiling to the floor. When a current exists in the wire, the wire swings upward and makes an angle \(\phi\) with respect to the vertical, as the drawing shows. (a) How is the magnitude of the magnetic force related to the magnetic field, the length of the wire, and the current in the wire? (b) What is the direction of the magnetic force? (c) The wire is stationary, so it is in equilibrium. What are the forces that keep it in equilibrium? (d) What must be true about the sum of the forces in the horizontal direction and the sum of the forces in the vertical direction? Problem The wire has a length of \(0.20 \mathrm{~m}\) and a mass of \(0.080 \mathrm{~kg}\) and carries a current of 42 A. The magnitude of the magnetic field is \(0.070 \mathrm{~T}\). Find (a) the angle \(\phi\) and (b) the tension in each of the two strings.

A square coil of wire containing a single turn is placed in a uniform 0.25 -T magnetic field, as the drawing shows. Each side has a length of \(0.32 \mathrm{~m}\), and the current in the coil is 12 A. Determine the magnitude of the magnetic force on each of the four sides.

A charge is moving perpendicular to a magnetic field and experiences a force whose magnitude is \(2.7 \times 10^{-3} \mathrm{~N}\). If this same charge were to move at the same speed and the angle between its velocity and the same magnetic field were \(38^{\circ},\) what would be the magnitude of the magnetic force that the charge would experience?

Two isotopes of carbon, carbon- 12 and carbon- \(13,\) have masses of \(19.93 \times 10^{-27} \mathrm{~kg}\) and \(21.59 \times 10^{-27} \mathrm{~kg},\) respectively. These two isotopes are singly ionized \((+e)\) and each is given a speed of \(6.667 \times 10^{5} \mathrm{~m} / \mathrm{s}\). The ions then enter the bending region of a mass spectrometer where the magnetic field is \(0.8500 \mathrm{~T}\). Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

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