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Chapter 21: Problem 38

A copper rod of length \(0.85 \mathrm{m}\) is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of \(k=75 \mathrm{N} / \mathrm{m} .\) A magnetic field with a strength of \(0.16 \mathrm{T}\) is oriented perpendicular to the surface of the table. (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is 12 A, by how much does each spring stretch?

Short Answer

Expert verified
(a) Left to right; (b) Each spring stretches by 10.88 mm.

Step by step solution

01

Analyze the forces involved

When a current flows through a conductor in a magnetic field, it experiences a force known as the Lorentz force. In this scenario, the copper rod sitting between the two springs will experience this force if a current is passed through it. The direction of the force is given by the right-hand rule, where the thumb points in the direction of current, fingers in the direction of the magnetic field, and the palm points in the direction of force.
02

Determine the direction of the current

According to the right-hand rule, the magnetic force due to the current should act outward, stretching the springs. To achieve this, the current should flow such that the force direction aligns with the spring stretching direction. If the magnetic field points downward (into the page), the current should flow from left to right along the rod to create an outward force on both springs.
03

Calculate the magnetic force on the rod

The magnetic force \( F \) on a current-carrying conductor of length \( L \) in a magnetic field \( B \) is given by the formula: \( F = BIL \), where \( I = 12 \text{ A} \) is the current, and \( L = 0.85 \text{ m} \) is the length of the rod. Substituting the values gives: \( F = 0.16 \times 12 \times 0.85 = 1.632 \text{ N}. \)
04

Relate magnetic force to spring stretch

The amount each spring stretches, \( x \), can be determined using Hooke's Law, \( F = kx \), where \( k = 75 \, \text{N/m} \) is the spring constant. From this, \( x = \frac{F}{2k} = \frac{1.632}{2 \times 75} = 0.01088 \text{ m} \). Therefore, each spring stretches by \( 0.01088 \text{ meters} \) or \( 10.88 \text{ mm}. \)

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Key Concepts

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

Magnetic Field
A magnetic field is an invisible force surrounding magnetic materials and moving electric charges. It is characterized by its strength and direction. In our exercise, the magnetic field is perpendicular to the table, with a strength of 0.16 T (teslas). This detail is crucial as it impacts the force experienced by the rod. The interaction of a magnetic field with a current in a conductor produces a force known as the Lorentz force. The magnetic field lines in this scenario extend downward through the table, affecting the copper rod's behavior. This perpendicular orientation is essential for understanding how the rod experiences force and how the direction of current flow influences it.
Hooke's Law
Hooke's Law is a principle of physics that describes the relationship between the force applied to a spring and the extension or compression of the spring. It is given by the equation: \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the extension or compression of the spring. In our exercise, this law helps determine how much the springs attached to the copper rod stretch when the current is applied. Each spring has a spring constant of 75 N/m. By finding the force exerted by the magnetic field on the rod, we use Hooke's Law to calculate how far the springs extend, revealing important details about the system's physical response.
Current in a Conductor
Current is the flow of electric charges, often electrons, through a conductor. In our context, the copper rod serves as this conductor, allowing a current of 12 A to flow through it. The direction of current is vital as it influences the direction of the Lorentz force exerted by the magnetic field. The movement of these charges in the rod generates an electromagnetic interaction with the magnetic field, leading to the application of force on the rod. Understanding the nature and direction of current flow is essential to predict how the rod will behave when the system is energized and how the springs attached to the rod will respond.
Right-Hand Rule
The right-hand rule is a handy mnemonic used to determine the direction of force exerted on a current-carrying conductor within a magnetic field. To apply the rule, you point your thumb in the direction of the current flow and your fingers in the direction of the magnetic field lines. Your palm will then naturally face the direction of the force experienced by the conductor. In the case of the copper rod, where the magnetic field points downwards and the desired outcome is for the force to stretch the springs outward, the current must flow from left to right. This understanding helps in visualizing and confirming the relationship between current direction and force orientation, an integral part of solving problems related to magnetic forces on conductors.

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Most popular questions from this chapter

A particle of mass \(6.0 \times 10^{-8} \mathrm{kg}\) and charge \(+7.2 \mu \mathrm{C}\) is traveling due east. It enters perpendicularly a magnetic field whose magnitude is \(3.0 \mathrm{T}\). After entering the field, the particle completes one-half of a circle and exits the field traveling due west. How much time does the particle spend traveling in the magnetic field?

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 \(8.2 \times 10^{-4} \mathrm{C}\).

Multiple-Concept Example 7 discusses how problems like this one can be solved. A \(+6.00 \mu \mathrm{C}\) charge is moving with a speed of \(7.50 \times 10^{4} \mathrm{m} / \mathrm{s}\) parallel to a very long, straight wire. The wire is \(5.00 \mathrm{cm}\) from the charge and carries a current of \(67.0 \mathrm{A}\) in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.

The drawing shows a parallel plate capacitor that is moving with a speed of \(32 \mathrm{m} / \mathrm{s}\) through a \(3.6-\mathrm{T}\) magnetic field. The velocity \(\overrightarrow{\mathbf{v}}\) is perpendicular to the magnetic field. The electric field within the capacitor has a value of \(170 \mathrm{N} / \mathrm{C},\) and each plate has an area of \(7.5 \times 10^{-4} \mathrm{m}^{2} .\) What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

A charge of \(-8.3 \mu \mathrm{C}\) is traveling at a speed of \(7.4 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is \(52^{\circ} .\) A force of magnitude \(5.4 \times 10^{-3} \mathrm{N}\) acts on the charge. What is the magnitude of the magnetic field?
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