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Chapter 27: Problem 37

A straight, 2.00 -m, \(150-\mathrm{g}\) wire carries a current in a region where the earth's magnetic field is horizontal with a magnitude of 0.55 gauss. (a) What is the minimum value of the current in this wire so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it? Does it seem likely that such a wire could support this size of current? (b) Show how the wire would have to be oriented relative to the earth's magnetic field to be supported in this way.

Short Answer

Expert verified
Minimum current: 13363.64 A; orientation must be perpendicular. Such high current is unlikely safe for a normal wire.

Step by step solution

01

Understand the Problem

We have a 2.00-meter long wire of 150 grams in a magnetic field of 0.55 gauss. We need to find the current required to make the magnetic force equal to the weight of the wire, considering only gravitational force acts apart from the magnetic force.
02

Convert Units

Convert the mass of the wire from grams to kilograms and the magnetic field from gauss to tesla. 1. Mass: \(150\,\text{g} = 0.150\,\text{kg}\)2. Magnetic field: \(0.55\,\text{gauss} = 0.55 \times 10^{-4}\,\text{T}\)
03

Calculate the Weight of the Wire

The weight of the wire is calculated using the formula for gravitational force: \(F_g = mg\). Here, \(m = 0.150\,\text{kg}\) and \(g = 9.81\,\text{m/s}^2\).\[F_g = 0.150 \times 9.81 = 1.4715\,\text{N}\]
04

Magnetic Force Formula

The magnetic force experienced by a wire is given by \(F_m = BIL \sin \theta\). To find the minimum current, we need the force to equal the weight: \(BIL \sin \theta = mg\). Assume \(\sin \theta = 1\) for maximum force.
05

Solve for Current (I)

Rearrange the formula to solve for current: \[I = \frac{mg}{BL}\]Substitute \(m = 0.150\,\text{kg}\), \(g = 9.81\,\text{m/s}^2\), \(B = 0.55 \times 10^{-4}\,\text{T}\), and \(L = 2.00\,\text{m}\):\[I = \frac{0.150 \times 9.81}{0.55 \times 10^{-4} \times 2.00}\approx 13363.64\,\text{A}\]
06

Orientation and Likelihood

The wire should be oriented perpendicular to the magnetic field to ensure \(\sin \theta = 1\) for maximum force. It is unlikely such a high current (approximately 13,364 Amperes) can be safely supported by a typical wire without damage due to excessive heating.

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Key Concepts

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

Current in a Wire
Current in a wire is a fascinating concept at the heart of electricity and magnetism. In a simple sense, it refers to the flow of electric charges through a conductor. The amount of current is measured in amperes (A), and it tells us how many charges are flowing past a point in the wire per second.

Here's how current typically works in a wire:
  • Electrons move through the material due to the applied voltage, creating a steady flow.
  • The size of the current depends on both the voltage applied and the resistance of the wire. A higher voltage or lower resistance means more current flows.
  • Ohm's Law, which is given by the formula \( I = V/R \), relates these quantities: \( I \) is current, \( V \) is voltage, and \( R \) is resistance.
Understanding the behavior of current in a wire, especially in interactions with magnetic fields, forms a critical part of electromagnetism. It helps in designing safe and efficient electrical systems.
Earth's Magnetic Field
Earth's magnetic field is an invisible force that extends from the Earth's interior out into space where it meets the solar wind. It acts much like a giant magnet with north and south poles.
  • The field is horizontal at the Earth's surface and affects compasses, which we often use to find our way.
  • Its strength is measured in units called gauss or tesla. For example, the typical strength of the Earth's magnetic field is about 0.55 gauss, which can also be expressed as \(0.55 \times 10^{-4}\,\text{T}\) in tesla.
This field arises from the motion of molten iron within the Earth's outer core, generating electric currents. These, in turn, produce magnetic fields. Understanding the Earth's magnetic field is crucial for navigation, as well as for understanding the protection it provides against charged particles from the sun.
Magnetic Field Orientation
Magnetic field orientation plays a critical role when we consider forces acting on a current-carrying wire in a magnetic field. The direction and orientation of this field determine how the magnetic force behaves.
  • The force on the wire is strongest when it's perpendicular to the magnetic field. This is due to the relationship in the formula \( F_m = BIL \sin \theta \), where \( \theta \) is the angle between the wire and the magnetic field. The sine function (\( \sin \theta \)) reaches its maximum value of 1 when \( \theta = 90^\circ \).
  • When the wire is aligned parallel or anti-parallel to the magnetic field, \( \theta = 0^\circ \) or \( 180^\circ \), the sine function returns 0, meaning no magnetic force acts in these orientations.
This concept is essential in applications ranging from electric motors to sensors, where precise control of the force exerted by magnetic fields is required. By manipulating the orientation of a wire or a current, we can change the magnitude and direction of the force exerted on it.

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Most popular questions from this chapter

Quark Model of the Neutron. The neutron is a particle with zero charge. Nonetheless, it has a nonzero magnetic moment with \(z\) -component \(9.66 \times\) \(10 ^ { - 27 } \mathrm { A } \cdot \mathrm { m } ^ { 2 } .\) This be explained by the internal structure of the neutron. A substantial body of evidence indicates that a neutron is composed of three fundamental particles called of three fundamental particles called quarks: an "up" (u) quark, of charge \(+ 2 e / 3 ,\) and two "down" \(( d )\) quarks, each of charge \(- e / 3 .\) The combination of the three quarks produces a net charge of \(2 e / 3 - e / 3 - e / 3 = 0\) . If the quarks are in motion, they can produce a nonzero magnetic moment. As a very simple model, suppose the \(u\) quark moves in a counterclockwise circular path and the \(d\) quarks move in a clock- wise circular path, all of radius \(r\) and all with the same speed \(v\) (Fig. P27.84). (a) Determine the current due to the circulation of the \(u\) quark. (b) Determine the magnitude of the magnetic moment due to the circulating \(u\) quark. (c) Determine the magnitude of the magnetic moment of the three-quark system. (Be careful to use the correct magnetic moment directions.) (d) With what speed \(v\) must the quarks move if this model is to reproduce the magnetic moment of the neutron? Use \(r = 1.20 \times 10 ^ { - 15 } \mathrm { m }\) (the radius of the neutron) for the radius of the orbits.

particle of mass 0.195 g carries a charge of \(-2.50 \times\) \(10^{-8} \mathrm{C} .\) The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

A straight, vertical wire carries a current of 1.20 A downward in a region between the poles of a large superconductingelectromagnet, where the magnetic field has magnitude \(B=\) 0.588 \(\mathrm{T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a \(1.00-\mathrm{cm}\) section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) \(30.0^{\circ}\) south of west?

An electromagnet produces a magnetic field of 0.550 \(\mathrm{T}\) in a cylindrical region of radius 2.50 \(\mathrm{cm}\) between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force is exerted on the wire?

A deuteron (the nucleus of an isotope of hydrogen) has a mass of \(3.34 \times 10^{-27} \mathrm{kg}\) and a charge of \(+e .\) The deuteron travels in a circular path with a radius of 6.96 \(\mathrm{mm}\) in a magnetic field with magnitude 2.50 \(\mathrm{T}\) (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?
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