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Chapter 5: Problem 32

(a) What is the direction of the force on a wire carrying a current due east in a location where the Earth's field is due north? Both are parallel to the ground. (b) Calculate the force per meter if the wire carries \(20.0 \mathrm{~A}\) and the field strength is \(3.00 \times 10^{-5} \mathrm{~T}\). (c) What diameter copper wire would have its weight supported by this force? (d) Calculate the resistance per meter and the voltage per meter needed.

Short Answer

Expert verified
(a) Upward; (b) \(6.00 \times 10^{-4} \text{ N/m}\); (c) \(0.15 \text{ mm}\); (d) \(5.1 \text{ m}\Omega\), \(0.102 \text{ V/m}\).

Step by step solution

01

Determine the Direction of Force (Part a)

The direction of the force on a wire carrying current in a magnetic field can be determined using the right-hand rule. Point your thumb in the direction of the current (due east), and your fingers in the direction of the magnetic field (due north). The palm of your hand will show the force's direction, which is upwards, perpendicular to both current and field.
02

Calculate Force Per Meter (Part b)

To calculate the force per meter on the wire in a magnetic field, use the formula \( F = I L B \sin(\theta) \), where \( I \) is the current, \( L \) is the length of the wire, \( B \) is the magnetic field strength, and \( \theta \) is the angle between current and magnetic field. Here, \( \theta = 90^\circ \) since they are perpendicular. Thus: \[ F = I B = 20.0 \times 3.00 \times 10^{-5} = 6.00 \times 10^{-4} \text{ N/m} \]
03

Determine Wire Diameter for Support (Part c)

The force supports the wire's weight, equal to \( \text{Weight per meter} = \rho \cdot (\pi (d/2)^2)g \), where \( \rho = 8.96 \times 10^3 \, \text{kg/m}^3 \) for copper and \( g = 9.81 \, \text{m/s}^2 \). Solving for diameter \( d \) using initial force per meter \( F = 6.00 \times 10^{-4} \, \text{N/m} \): \[ F = \rho \cdot \pi \left( \frac{d}{2} \right)^2 g \] \[ d = \sqrt{\frac{4F}{\pi \rho g}} \approx 0.15 \text{ mm} \]
04

Calculate Resistance Per Meter (Part d)

The resistance per meter is given by \( R = \frac{\rho_r}{A} \), where \( \rho_r = 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \) is the resistivity of copper, and \( A = \pi (d/2)^2 \) is the cross-sectional area. Given \( d = 0.15 \times 10^{-3} \text{ m} \), \[ A = \pi \left(\frac{0.075 \times 10^{-3}}{2}\right)^2 \] \[ R = \frac{1.68 \times 10^{-8}}{\pi (0.075 \times 10^{-3})^2} \approx 5.1 \text{ m}\Omega \]
05

Calculate Voltage Per Meter Needed (Part d)

Using Ohm's law \( V = IR \), where \( I = 20 \text{ A} \) and \( R = 5.1 \text{ m}\Omega \), find the voltage per meter: \( V = 20 \times 5.1 \times 10^{-3} = 0.102 \text{ V/m} \).

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Key Concepts

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

Right-Hand Rule
To understand the direction of the magnetic force on a current-carrying wire, we use the right-hand rule. If you imagine holding the wire in your right hand, align your thumb with the current's direction. In the given exercise, this is due east. Then orient your fingers to point towards the magnetic field, which is due north. Your palm, then, indicates the force's direction, which is upwards in this scenario.

The right-hand rule is a tool to visualize the interrelation between the current, magnetic field, and resulting force. Here, with the current going east and the magnetic field north, the resulting force is upwards due to the perpendicular interaction.
Magnetic Field
A magnetic field is an invisible area around a magnetic material or a current-carrying wire where magnetic forces can act. In our case, it extends from Earth and is directed to the north, parallel to the ground.
  • The magnitude of a magnetic field is measured in teslas (T).
  • In this exercise, the magnetic field strength given is \( 3.00 \times 10^{-5} \text{ T} \).

This field interacts with moving charges, such as those in a current, resulting in a magnetic force. The strength and direction of this field are essential for calculating the resulting force on the wire.
Current Direction
The flow of electric charge through a conductor, such as a copper wire, is known as current. The direction of current is conventionally taken to be the direction a positive charge would move. In this example, current flows due east in the wire.
  • Current is measured in amperes (A).
  • Here, the current flowing through the wire is \( 20.0 \text{ A} \).

The current direction is a crucial part of applying the right-hand rule and calculating the resultant force due to the magnetic field acting on the wire.
Force Calculation
To calculate the force exerted on a current-carrying wire within a magnetic field, we use the formula:\[ F = I L B \sin(\theta) \]where:
  • \( F \) is the force per meter on the wire.
  • \( I \) is the current through the wire (20.0 A).
  • \( L \) is the length of wire (considered as 1 meter for this per meter calculation).
  • \( B \) is the magnetic field strength (\( 3.00 \times 10^{-5} \text{ T} \)).
  • \( \theta \) is the angle between the current direction and magnetic field (\( 90^\circ \)), making \( \sin(\theta) = 1\).
Thus, the force per meter on the wire calculates to \( 6.00 \times 10^{-4} \text{ N/m} \), indicating how much force the wire experiences because of the interaction with the magnetic field.
Copper Wire Properties
Copper is a commonly used metal for wires due to its excellent conductive properties. It has specific characteristics relevant to electrical and magnetic calculations:
  • Density \( \rho = 8.96 \times 10^3 \text{ kg/m}^3 \)
  • Resistivity \( \rho_r = 1.68 \times 10^{-8} \Omega \cdot \text{m} \)

For the wire to support its weight through magnetic force, the diameter must be calculated using these properties. The calculated diameter is approximately \( 0.15 \text{ mm} \).

Additionally, with this diameter, the resistance per meter of the wire can be calculated to \( 5.1 \text{ m}\Omega \), and under the specified current, a voltage of \( 0.102 \text{ V/m} \) is needed per meter based on Ohm鈥檚 Law \( V = IR \). These properties ensure that copper wire affords optimal performance in electrical circuits.

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Most popular questions from this chapter

An electron moving at \(4.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) in a 1.25-T magnetic field experiences a magnetic force of \(1.40 \times 10^{-16} \mathrm{~N}\). What angle does the velocity of the electron make with the magnetic field? There are two answers. (a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than \(1.00 \times 10^{-12} \mathrm{~N}\). What is the greatest the charge can be if it moves at a maximum speed of \(30.0 \mathrm{~m} / \mathrm{s}\) in the Earth's field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.

A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the \(4.00 \mathrm{~cm}\) of wire in the field. What is the average field strength?

Is the Earth's magnetic field parallel to the ground at all locations? If not, where is it parallel to the surface? Is its strength the same at all locations? If not, where is it greatest?

(a) What is the force per meter on a lightning bolt at the equator that carries 20,000 A perpendicular to the Earth's \(3.00 \times 10^{-5}-\mathrm{T}\) field? (b) What is the direction of the force if the current is straight up and the Earth's field direction is due north, parallel to the ground?

(a) Find the charge on a baseball, thrown at \(35.0 \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's \(5.00 \times 10^{-5} \mathrm{~T}\) field, that experiences a 1.00-N magnetic force. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?
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