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Suppose a straight 1.00 -mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth's magnetic field \(\overrightarrow{\mathbf{B}},\) which is horizontal, perpendicular to the wire, and of magnitude \(5.0 \times 10^{-5} \mathrm{~T}\). What current would the wire carry? Does the answer seem feasible? Explain briefly.

Short Answer

Expert verified
The current is approximately 1.38 kA. This value is impractically high for a 1 mm copper wire.

Step by step solution

01

Understanding the Problem

We are to find the current that would make a copper wire float horizontally due to the Earth's magnetic field. The magnetic field is given as horizontal and perpendicular to the wire, with a magnitude of \( 5.0 \times 10^{-5} \text{ T} \). The wire's diameter is 1.00 mm.
02

Calculating the Magnetic Force

For the wire to float, the magnetic force \( F_m \) must equal the gravitational force \( F_g \) on the wire. The magnetic force is given by \( F_m = I \cdot L \cdot B \), where \( I \) is the current, \( L \) is the length of the wire, and \( B \) is the magnetic field.
03

Calculating the Gravitational Force

The gravitational force on the wire is \( F_g = m \cdot g \). The mass \( m \) of the wire can be found if we know the density \( \rho \) of copper and the volume \( V \) of the wire. The volume of the wire is given by \( V = \pi \cdot \left(\frac{d}{2}\right)^2 \cdot L \) where \( d = 1.00 \times 10^{-3} \text{ m} \). This gives \( m = \rho \cdot V \).
04

Applying Equilibrium of Forces

Set the gravitational force equal to the magnetic force: \( m \cdot g = I \cdot L \cdot B \). Substitute for \( m \): \( \rho \cdot V \cdot g = I \cdot L \cdot B \). Substitute for \( V \): \( \rho \cdot \pi \cdot \left(\frac{d}{2}\right)^2 \cdot L \cdot g = I \cdot L \cdot B \). Cancel \( L \) from both sides: \( I = \rho \cdot \pi \cdot \left(\frac{d}{2}\right)^2 \cdot g / B \).
05

Substitute Known Values

The density of copper \( \rho \) is approximately \( 8.96 \times 10^3 \text{ kg/m}^3 \), \( g = 9.81 \text{ m/s}^2 \), \( d = 1.00 \times 10^{-3} \text{ m} \), and \( B = 5.0 \times 10^{-5} \text{ T} \). Substitute these into the equation to find \( I \):\[I = \frac{8.96 \times 10^3 \cdot \pi \cdot \left( \frac{1.00 \times 10^{-3}}{2} \right)^2 \cdot 9.81}{5.0 \times 10^{-5}}\]Calculate \( I \) to find the current.
06

Calculate and Verify

Perform the calculation:\[I \approx \frac{8.96 \times 10^3 \cdot \pi \cdot 0.5 \times 10^{-6} \cdot 9.81}{5.0 \times 10^{-5}} = 1.38 \text{ kA}\]This current value, around 1.38 kA, seems very high. In practical terms, carrying such a high current in a 1 mm diameter wire is not feasible, as this would likely result in significant heating and damage to the wire.

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Key Concepts

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

Magnetic Force
When an electric current flows through a wire placed in a magnetic field, it experiences a force known as the magnetic force. This force is the reason a current-carrying conductor can move when in contact with a magnetic field. The strength of this force can be calculated using the formula \( F_m = I \cdot L \cdot B \), where:
  • \( F_m \) is the magnetic force.
  • \( I \) is the current flowing through the wire.
  • \( L \) is the length of the wire within the magnetic field.
  • \( B \) is the magnetic field strength.
If you have ever observed the movement of a compass needle in the presence of a nearby current-carrying wire, that is the magnetic force in action. The direction of this force is given by the "right-hand rule." If you point your thumb in the direction of current flow and your fingers in the direction of the magnetic field, your palm will face the direction of the force exerted on the wire.
Understanding the direction and magnitude of this force helps in designing electrical components that effectively interact with magnetic fields, such as electric motors.
Gravitational Force
The gravitational force is the attractive force that any two masses exert on each other. On Earth, this force is what pulls objects toward the ground. For a copper wire, or any object, this force can be calculated using the formula \( F_g = m \cdot g \), where:
  • \( F_g \) is the gravitational force.
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \) on Earth's surface.
To find the mass of the wire, it's necessary to determine its volume first and then use the material density. The volume of a cylinder (wire) is \( V = \pi \cdot \left(\frac{d}{2}\right)^2 \cdot L \), with \( d \) as the diameter and \( L \) as its length. Multiplying the volume by the density \( \rho \) of the wire material (copper in this case) gives the mass. This force must be overcome by another force, like the magnetic force, for the wire to "float" in equilibrium.
Equilibrium of Forces
Equilibrium in physics refers to the state where all the forces acting on an object are balanced, resulting in no net force and, hence, no acceleration. For the copper wire in this problem, equilibrium is achieved when the magnetic force perfectly balances the gravitational force, enabling the wire to "float."The condition for equilibrium can be expressed by setting the magnetic force equal to the gravitational force: \( F_m = F_g \). That leads to the equation:\[ m \cdot g = I \cdot L \cdot B \]By rearranging and simplifying, we find the current required for equilibrium:\[ I = \frac{\rho \cdot \pi \cdot \left(\frac{d}{2}\right)^2 \cdot g}{B} \]This equation reveals the current that must flow through the wire to keep it floating due to the Earth's magnetic field. Equilibrium of forces ensures that the wire does not fall under gravity or move away due to the magnetic force. Achieving such equilibrium in practice, especially with high currents and thin wires, may not be feasible, primarily due to the potential for overheating and material stress.

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Most popular questions from this chapter

A circular coil \(18.0 \mathrm{~cm}\) in diameter and containing twelve loops lies flat on the ground. The Earth's magnetic field at this location has magnitude \(5.50 \times 10^{-5} \mathrm{~T}\) and points into the Earth at an angle of \(66.0^{\circ}\) below a line pointing due north. If a 7.10-A clockwise current passes through the coil, determine \((a)\) the torque on the coil, and \((b)\) which edge of the coil rises up, north, east, south, or west.

(1I) What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic fields of magnitude \(8.8 \times 10^{3} \mathrm{V} / \mathrm{m}\) and \(7.5 \times 10^{-3} \mathrm{T}\) , respectively? What is the radius of the elec- tron orbit if the electric field is turned off?

(II) A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic field of magnitude 0.10 T. When the field is oriented normal to the slab's rectangular face, a Hall emf of 12 \(\mathrm{mV}\) is measured across the slab's width. The probe is then placed in a magnetic field of unknown magnitude \(B\) , and a Hall emf of 63 \(\mathrm{mV}\) is measured. Determine \(B\) assuming that the angle \(\theta\) between the unknown field and the plane of the slab's rectangular face is \((a) \theta=90^{\circ},\) and \((b) \theta=60^{\circ} .\)

(III) \(\mathrm{A} 3.40\) -g bullet moves with a speed of 155 \(\mathrm{m} / \mathrm{s}\) perpen- dicular to the Earth's magnetic field of \(5.00 \times 10^{-9} \mathrm{T}\) . If the bullet possesses a net charge of \(18.5 \times 10^{-9} \mathrm{C},\) by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.00 \(\mathrm{km} ?\)

A Hall probe used to measure magnetic field strengths consists of a rectangular slab of material (free-electron density \(n\) ) with width \(d\) and thickness \(t,\) carrying a current \(I\) along its length \(\ell\). The slab is immersed in a magnetic field of magnitude \(B\) oriented perpendicular to its rectangular face (of area \(\ell d\) ), so that a Hall emf \(\mathscr{E}_{\mathrm{H}}\) is produced across its width \(d\). The probe's magnetic sensitivity, defined as \(K_{\mathrm{H}}=\mathscr{E}_{\mathrm{H}} / I B,\) indicates the magnitude of the Hall emf achieved for a given applied magnetic field and current. A slab with a large \(K_{\mathrm{H}}\) is a good candidate for use as a Hall probe. (a) Show that \(K_{\mathrm{H}}=1\) /ent. Thus, a good Hall probe has small values for both \(n\) and \(t .(b)\) As possible candidates for the material used in a Hall probe, consider (i) a typical metal \(\left(n \approx 1 \times 10^{29} / \mathrm{m}^{3}\right)\) and (ii) a (doped) semiconductor \(\left(n \approx 3 \times 10^{22} / \mathrm{m}^{3}\right)\). Given that a semiconductor slab can be manufactured with a thickness of \(0.15 \mathrm{~mm}\), how thin (nm) should a metal slab be to yield a \(K_{H}\) value equal to that of the semiconductor slab? Compare this metal slab thickness with the 0.3 -nm size of a typical metal atom. (c) For the typical semiconductor slab described in part \((b),\) what is the expected value for \(\mathscr{E}_{\mathrm{H}}\) when \(I=100 \mathrm{~mA}\) and \(B=0.1 \mathrm{~T} ?\)

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