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Chapter 21: Problem 40

A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of \(0.20 \mathrm{m}\) and a mass of \(0.080 \mathrm{kg} .\) A uniform magnetic field of magnitude 0.070 T is directed from the ceiling to the floor. When a current of \(I=42\) A exists in the wire, the wire swings upward and, at equilibrium, makes an angle \(\phi\) with respect to the vertical, as the drawing shows. Find (a) the angle \(\phi\) and (b) the tension in each of the two strings.

Short Answer

Expert verified
(a) \( \phi \approx 36.87^\circ \), (b) Tension in each string is \(0.49 \ \text{N}\).

Step by step solution

01

Analyzing Forces on the Wire

The forces acting on the wire are gravity, the magnetic force, and the tension in the strings. The gravitational force is vertically downward and given by \( F_g = mg \), where \( m = 0.080 \ \text{kg} \) and \( g = 9.8 \ \text{m/s}^2 \). The magnetic force, due to the current in the magnetic field, is given by \( F_B = I \cdot L \cdot B \), where \( I = 42 \ \text{A} \), \( L = 0.20 \ \text{m} \), and \( B = 0.070 \ \text{T} \). These forces are balanced by the components of the tension in the strings.
02

Calculating Magnetic Force

Calculate the magnetic force using the formula \( F_B = I \cdot L \cdot B \). Substituting the given values: \( F_B = 42 \ \text{A} \cdot 0.20 \ \text{m} \cdot 0.070 \ \text{T} = 0.588 \ \text{N} \).
03

Calculating Gravitational Force

Find the gravitational force using \( F_g = mg \). Here, \( F_g = 0.080 \ \text{kg} \cdot 9.8 \ \text{m/s}^2 = 0.784 \ \text{N} \).
04

Establishing Equilibrium Conditions

At equilibrium, the wire makes an angle \( \phi \) with the vertical. The vertical component of tension balances the gravitational force, and the horizontal component balances the magnetic force. Therefore, \( T \cdot \cos(\phi) = F_g \) and \( T \cdot \sin(\phi) = F_B \). Where \( T \) is the tension in the wire.
05

Calculating the Angle \( \phi \)

Using the relationship \( \tan(\phi) = \frac{F_B}{F_g} \), substitute the calculated forces: \( \tan(\phi) = \frac{0.588}{0.784} = 0.75 \). Thus, \( \phi = \arctan(0.75) \approx 36.87^\circ \).
06

Finding the Tension in Each String

Since the tension is equally distributed between the two strings, first find the total tension \( T \) from \( \cos(\phi) \) as \( T = \frac{F_g}{\cos(\phi)} \). Therefore, \( T = \frac{0.784}{\cos(36.87^\circ)} = \frac{0.784}{0.8} = 0.98 \ \text{N} \). Each string thus supports half of this tension: \( \frac{0.98 \ \text{N}}{2} = 0.49 \ \text{N} \).

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Key Concepts

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

Equilibrium of Forces
In the scenario where a wire is hung using strings and subjected to multiple forces, it's important to understand the concept of equilibrium of forces. When an object is in equilibrium, it is in a state of balance. This means that the sum of all the forces acting on the object equals zero, preventing any acceleration. In this problem, the forces acting on the wire are:
  • The gravitational force, pulling the wire downward.
  • The magnetic force, exerted by the magnetic field.
  • The tension from the strings, which counteracts these forces.
The equilibrium is achieved when:
  • The sum of vertical forces equals zero. The vertical component of tension balances out the gravitational force.
  • The sum of horizontal forces also equals zero. The horizontal component of tension offsets the magnetic force.
Understanding this balance of forces is crucial. It allows us to calculate unknowns, such as the angle at which the wire stabilizes and the tension in the supporting strings.
Tension in Strings
Tension is a force that acts along the length of an object, like strings or ropes, when they are pulled at either end. In this exercise, the tension occurs in the strings supporting the wire.Strings experience tension that ensures the wire remains suspended, even when external forces like gravity and magnetic force act upon it. For the wire in our scenario, the tension is distributed equally across both strings. To solve for tension, understand that it can be separated into vertical and horizontal components:
  • The vertical component (\( T \cdot \cos(\phi) \)) balances the gravitational force.
  • The horizontal component (\( T \cdot \sin(\phi) \)) opposes the magnetic force.
By using trigonometry, we can determine these components and thus solve for the total tension. Calculating each component provides insight into how forces interact and maintain the wire in its equilibrium position.
Magnetic Field Influence
Magnetic forces arise from the interaction between magnetic fields and electric currents. In the context of this exercise, we have a wire with current in a magnetic field which results in a magnetic force acting on the wire.The direction and magnitude of the magnetic force are crucial. They depend on:
  • The amount of current (\( I \)) flowing through the wire.
  • The length of the wire within the magnetic field (\( L \)).
  • The strength of the magnetic field (\( B \)).
The force is given by the equation: \( F_B = I \, L \, B \). Here, the force direction follows the right-hand rule, which states that if you point your thumb in the direction of the current and your fingers in the direction of the magnetic field, your palm faces the direction of the force.In the exercise, this magnetic force acts horizontally, contributing to the wire's swing and the establishment of equilibrium. Understanding how magnetic fields influence forces helps solve various physics-related problems, especially those involving electric currents and magnetic interactions.

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Most popular questions from this chapter

A charged particle enters a uniform magnetic field and follows the circular path shown in the drawing. (a) Is the particle positively or negatively charged? Why? (b) The particle's speed is \(140 \mathrm{m} / \mathrm{s},\) the magnitude of the magnetic field is \(0.48 \mathrm{T}\), and the radius of the path is \(960 \mathrm{m}\). Determine the mass of the particle, given that its charge has a magnitude of \(8.2 \times 10^{-4} \mathrm{C}\).

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?

The two conducting rails in the drawing are tilted upward so they each make an angle of \(30.0^{\circ}\) with respect to the ground. The vertical magnetic field has a magnitude of \(0.050 \mathrm{T}\). The \(0.20-\mathrm{kg}\) aluminum rod (length \(=\) \(1.6 \mathrm{m}\) ) slides without friction down the rails at a constant velocity. How much current flows through the rod?

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{kg}\) is traveling due east with a speed of \(85 \mathrm{m} / \mathrm{s}\) and enters a \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{s},\) at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

A wire carries a current of 0.66 A. This wire makes an angle of \(58^{\circ}\) with respect to a magnetic field of magnitude \(4.7 \times 10^{-5}\) T. The wire experiences a magnetic force of magnitude \(7.1 \times 10^{-5} \mathrm{N} .\) What is the length of the wire?
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