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Two insulated wires, each \(2.40 \mathrm{~m}\) long, are taped together to form a two-wire unit that is \(2.40 \mathrm{~m}\) long. One wire carries a current of \(7.00 \mathrm{~A} ;\) the other carries a smaller current \(I\) in the opposite direction. The two-wire unit is placed at an angle of \(65.0^{\circ}\) relative to a magnetic field whose magnitude is \(0.360 \mathrm{~T}\). The magnitude of the net magnetic force experienced by the two-wire unit is \(3.13 \mathrm{~N}\). What is the current \(I ?\)

Short Answer

Expert verified
The current \( I \) is approximately \( 3.00 \ A \).

Step by step solution

01

Identify the Forces

The magnetic forces on the wires are given by the formula for the force on a current-carrying conductor in a magnetic field: \[ F = I L B \sin(\theta) \]. Here, \( L = 2.40 \ m \) is the length of the wire, \( B = 0.360 \ T \) is the magnetic field strength, and \( \theta = 65.0^\circ \) is the angle between the wire and the magnetic field. We need to calculate the forces on both wires and consider their directions.
02

Calculate Force for First Wire

The first wire carries a current of \(7.00 \ A\). Calculate the magnetic force using the formula: \[ F_1 = 7.00 \ A \times 2.40 \ m \times 0.360 \ T \times \sin(65.0^\circ) \]. First, calculate \( \sin(65.0^\circ) \), then multiply all values to get \( F_1 \).
03

Calculate Force for Second Wire

The second wire carries the unknown current \( I \) in the opposite direction. The force is: \[ F_2 = I \times 2.40 \ m \times 0.360 \ T \times \sin(65.0^\circ) \]. Calculate \( F_2 \) based on the unknown current \( I \).
04

Set up Net Force Equation

Since the wires are carrying current in opposite directions, the net force is the difference between the forces on the two wires: \[ |F_1 - F_2| = 3.13 \ N \]. Substitute the expressions for \( F_1 \) and \( F_2 \) to set up the equation to solve for \( I \).
05

Solve for Current I

Substitute the known values and solve the equation for \( I \). This involves isolating \( I \) on one side to find its value. Simplify to get \( I = |F_1 - 3.13 \ N| / (2.40 \ m \times 0.360 \ T \times \sin(65.0^\circ)) \). Calculate \( I \) with the solved numerical value of each component.
06

Calculate Numerical Answer

First, calculate the force \( F_1 = 7.00 \ A \times 2.40 \ m \times 0.360 \ T \times \sin(65.0^\circ) \). Use the known expression for \( \sin(65.0^\circ) \approx 0.9063 \). Solve: \( F_1 \approx 5.47 \ N \). Finally, solve for \( I \) using \( I = (5.47 \ N - 3.13 \ N)/(2.40 \ m \times 0.360 \ T \times 0.9063) \).
07

Derive Final Value of Current I

Simplify the expression to obtain the numerical value of \( I \). Calculate the difference \( 5.47 \ N - 3.13 \ N = 2.34 \ N \), and then solve for \( I \): \[ I = \frac{2.34 \ N}{2.40 \ m \times 0.360 \ T \times 0.9063} \]. After computation, the result is \( I \approx 3.00 \ A \).

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Key Concepts

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

Current-Carrying Conductor
When we talk about a current-carrying conductor, we are referring to a material, often a wire, through which electric current flows. In this context, the wire is part of an electrical circuit where electrons move in a specified direction, causing an electric current. The magnitude of this current is measured in amperes (A). Electric currents can produce a magnetic field around the wire, and this interaction is the basis of many electromagnetic devices.
In our problem, one wire carries a current of 7.00 A, while the other carries an unknown current, denoted as \(I\). The direction of the currents in these wires is opposite, meaning the forces that the magnetic field exerts on them will also be opposed. The total or net magnetic force acting on the overall system of wires must be considered to find the unknown \(I\). This scenario is a practical example of the fundamental relationships in electromagnetism.
Magnetic Field
A magnetic field is a region around a magnet, electric current, or changing electric field, where magnetic forces act on any other magnet or current present. The presence of a magnetic field can be represented with magnetic field lines, and the strength of this field is quantified in teslas (T).
In this exercise, the magnetic field has an intensity of 0.360 T. Its interaction with the wires carrying currents results in a magnetic force due to the magnetic field's influence on charges moving within the conductor. The direction and the magnitude of this force can be determined using the formula given by the equation \(F = I L B \sin(\theta)\), emphasizing the role of each factor in computing the exact force.
The angle \(\theta\) between the magnetic field and the direction of the current affects how strongly the field interacts with the current, which will be illustrated by the sine component in the force equation.
Sinusoidal Function
The sinusoidal function, often appearing in physics problems, describes the wave-like patterns associated with alternating currents and other oscillatory phenomena. In the formula for magnetic force, \(\sin(\theta)\) indicates how the angle \(\theta\) between the current's path and the magnetic field affects the resultant force.
Calculating \(\sin(65.0^\circ)\) gives a value of approximately 0.9063. This value is crucial because it scales the influence of the magnetic field on the wire, depending on their relative orientation. When the angle is such that the sine value is 1, the magnetic force is maximized since it means the field and current are perpendicular.
This relationship is significant because it dictates how effectively the magnetic field interacts with the wire, and it is central to determining the force exerted, hence crucial in solving problems like the one we've discussed.
Physics Problem Solving
Physics problem solving often involves identifying known and unknown quantities, applying relevant laws or formulas, and executing calculations in a logical order. This structured approach helps break down complex problems into manageable parts.
In this example, the net magnetic force provides a starting point, leading us to the discovery of the unknown current \(I\). By systematically applying the formula for the force on a current-carrying conductor and considering the superposition of forces—given that the currents are opposite—we derive the necessary equation to solve for \(I\).
  • Identify the forces acting on each wire.
  • Calculate known values like \(\sin(65.0^\circ)\).
  • Set up an equation to represent the net force condition.
  • Solve this equation, step by step, to find the unknown variable.
This disciplined methodology not only solves this specific problem but also illustrates the process of insight, planning, and calculation crucial in studying physics. Now with the answers in place, we observe a classic example of how physics principles can decode real-world phenomena.

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Most popular questions from this chapter

The electrons in the beam of a television tube have a kinetic energy of \(2.40 \times 10^{-15} \mathrm{~J}\). Initially, the electrons move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of \(2.00 \times 10^{-5} \mathrm{~T}\). (a) In what direction are the electrons deflected by this field component? (b) What is the acceleration of an electron in part (a)?

At for help with problems like this one. A very long, hollow cylinder is formed by rolling up a thin sheet of copper. Electric charges flow along the copper sheet parallel to the axis of the cylinder. The arrangement is, in effect, a hollow tube of current \(I\). Use Ampère's law to show that the magnetic field (a) is \(\mu_{0} I /(2 \pi r)\) outside the cylinder at a distance \(r\) from the axis and \((\mathrm{b})\) is zero at any point within the hollow interior of the cylinder. (Hint: For closed paths, use circles perpendicular to and centered on the axis of the cylinder.)

A wire of length \(0.655 \mathrm{~m}\) carries a current of \(21.0 \mathrm{~A}\). In the presence of a \(0.470-\mathrm{T}\) magnetic field, the wire experiences a force of \(5.46 \mathrm{~N}\). What is the angle (less than \(90^{\circ}\) ) between the wire and the magnetic field?

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 -T uniform magnetic field. The particle moves through onequarter 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 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?

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