/*! 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 19 A wire with a mass of \(1.00 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A wire with a mass of \(1.00 \mathrm{~g} / \mathrm{cm}\) is placed on a horizontal surface with a coefficient of friction of \(0.200 .\) The wire carries a current of \(1.50 \mathrm{~A}\) eastward and moves horizontally to the north. What are the magnitude and the direction of the smalles vertical magnetic field that enables the wire to move in this fashion?

Short Answer

Expert verified
The magnitude of the smallest vertical magnetic field that would enable the wire to move horizontally North is \(1.31 \mathrm{~T}\), directed downwards (into the surface).

Step by step solution

01

Compute Frictional Force

First, calculate the frictional force acting underwater. The equation for this is \( F = \mu.m.g \) where \( \mu \) is the coefficient of friction, \( m \) represents mass, and \( g \) the acceleration due to gravity. On substitution we obtain that \( F = 0.200 \times 1.00 \times 9.8 = 1.96 \mathrm{~N} \)
02

Find Magnetic Force

The magnetic field must provide enough force to overcome friction. So, the magnetic force equals the frictional force. Therefore, it follows from the first step that the magnetic force \( F \) is equal to \( 1.96 \mathrm{~N} \).
03

Calculate Magnetic Field Strength

Now, use the formula \( F = B.I.L \) where \( F \) is the force due to the magnetic field, \( B \) is the magnetic field strength needed, \( I \) is the current passing through the wire, and \( L \) is the length of the wire. Here, we need to find \( B \) and the force \( F \), current \( I \) are known. Since the force is onto each unit length of the wire, therefore, the length \( L \) will be unity in this context and the formula becomes \( B = F/I \) Upon substitution, the value of \( B \) can be found as \( B = 1.96/1.5 = 1.31 \mathrm{~T} \)

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Key Concepts

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

Frictional Force
Frictional force is the resistance force that prevents two objects from sliding freely against each other. In the context of the wire on the horizontal surface, this force plays a crucial role in determining whether the wire remains stationary or begins to move. Frictional force depends on two main factors:
  • The coefficient of friction (\( \mu \)), which is a measure of how "sticky" or "slippery" the surfaces interacting are.
  • The normal force (often represented as the weight of the wire in the case of a horizontal surface), calculated by multiplying the mass of the object by the gravitational acceleration (\( g \approx 9.8 \text{ m/s}^2 \)).
To compute the frictional force, the formula \( F_f = \mu \cdot m \cdot g \) is used. Here, the frictional force acts as a barrier that the magnetic force must overcome for the wire to slide smoothly across the surface, meaning the magnetic force should be at least equal to the frictional force.
Magnetic Force
Magnetic force is essential in this exercise as the driving force that enables the wire to move despite the frictional resistance. When a current-carrying wire is placed in a magnetic field, it experiences a force called the magnetic force. This principle is described by the equation \( F = B \cdot I \cdot L \), where:
  • \( F \) is the magnetic force.
  • \( B \) is the magnetic field strength.
  • \( I \) is the current flowing through the wire.
  • \( L \) is the length of the wire.
In scenarios where the wire needs to move, such as this exercise, the magnetic force should at least equal the frictional force. This ensures that the magnetic field's influence is strong enough to counteract any resistive forces while allowing the wire to move freely and predictably in the desired direction.
Current and Wire
In physics, the relationship between current and a wire within a magnetic field illustrates fundamental electromagnetic principles. The current (\( I \)) flowing in the wire generates a magnetic field around it, and when subjected to an external magnetic field, a force is exerted on the wire. The direction of this force is determined using the right-hand rule.
The current in the wire, measured in amperes (A), influences the magnitude of the magnetic force exerted on the wire. If the current increases, the magnetic force also increases, assuming the external magnetic field remains constant. This direct relationship is captured in the formula \( F = B \cdot I \cdot L \).
Moreover, the physical orientation of the wire and the direction of the current play roles in determining the overall magnetic behavior of the system. In this exercise, the wire is arranged such that it moves due north while carrying a current eastward, indicating a perpendicular interaction with the magnetic field that generates enough magnetic force to overcome frictional resistance efficiently.

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Most popular questions from this chapter

A uniform horizontal wire with a linear mass density of \(0.50 \mathrm{~g} / \mathrm{m}\) carries a \(2.0-\mathrm{A}\) current. It is placed in a constant magnetic field with a strength of \(4.0 \times 10^{-3} \mathrm{~T}\). The field is horizontal and perpendicular to the wire. As the wire moves upward starting from rest, (a) what is its acceleration and (b) how long does it take to rise \(50 \mathrm{~cm}\) ? Neglect the magnetic field of Earth.

A \(0.200-\mathrm{kg}\) metal rod carrying a current of \(10.0 \mathrm{~A}\) glides on two horizontal rails \(0.500 \mathrm{~m}\) apart. What vertical mag netic field is required to keep the rod moving at a constant speed if the coefficient of kinetic friction between the rod and rails is \(0.100\) ?

A single-turn square loop of wire \(2.00 \mathrm{~cm}\) on a side carries a counterclockwise current of \(0.200 \mathrm{~A}\). The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30 turns per centimeter and carries a counterclockwise current of \(15.0 \mathrm{~A}\). Find the force on each side of the loop and the torque acting on the loop.

A straight wire carrying a \(3.0-\mathrm{A}\) current is placed in a uniform magnetic field of magnitude \(0.28 \mathrm{~T}\) directed perpendicular to the wire. (a) Find the magnitude of the magnetic force on a section of the wire having a length of \(14 \mathrm{~cm}\). (b) Explain why you can't determine the direction of the magnetic force from the information given in the problem.

A current \(I=15 \mathrm{~A}\) is directed along the positive \(x\) -axis and perpendicular to a magnetic field. A magnetic force per unit length of \(0.12 \mathrm{~N} / \mathrm{m}\) acts on the conductor in the negative \(y\) -direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes.

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