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Chapter 28: Problem 31

The wires in a household lamp cord are typically 3.0 mm apart center to center and carry equal currents in opposite directions. If the cord carries direct current to a 100-W light bulb connected across a 120-V potential difference, what force per meter does each wire of the cord exert on the other? Is the force attractive or repulsive? Is this force large enough so it should be considered in the design of the lamp cord? (Model the lamp cord as a very long straight wire.)

Short Answer

Expert verified
The force is \(4.63 \times 10^{-5} \, \text{N/m}\) and attractive. It is too small to affect lamp cord design.

Step by step solution

01

Calculate the Current

First, we need to find the current flowing through the wires. Using the formula for power, \(P = IV\), where \(P = 100 \, \text{W}\) and \(V = 120 \, \text{V}\). Rearranging for current \(I\), we get \(I = \frac{P}{V} = \frac{100}{120} = 0.833 \, \text{A}\).
02

Calculate the Magnetic Force per Unit Length

The force per unit length \(F/L\) between two parallel wires carrying currents in opposite directions can be calculated using the formula: \[ F/L = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{d} \] where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\) is the permeability of free space, \(I_1\) and \(I_2\) are the currents (0.833 A) and \(d\) is the distance between wires (0.003 m).
03

Substitute and Calculate

Substituting the given values into the formula: \[ F/L = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{0.833 \times 0.833}{0.003} \] \[ F/L = 2 \times 10^{-7} \cdot \frac{0.693889}{0.003} = 4.63 \times 10^{-5} \, \text{N/m} \]
04

Determine the Nature of Force

Since the currents are in opposite directions, the force between the wires is attractive according to the Amp猫re's force law.
05

Consideration of Force in Design

The force calculated is \(4.63 \times 10^{-5} \, \text{N/m}\), which is relatively small. This force is not large enough to be a significant factor to consider in the design of a typical household lamp cord.

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Key Concepts

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

Current Calculation
In this exercise, we are dealing with the concept of electrical current, which is a flow of electric charge carried by moving electrons in a wire. To find the current, we use the relationship between power, voltage, and current given by the equation \(P = IV\), where:
  • \(P\) is the power in watts (W),
  • \(I\) is the current in amperes (A),
  • \(V\) is the voltage in volts (V).
For a 100-W light bulb connected to a 120-V source, the formula becomes \(I = \frac{P}{V}\). By substituting the given values, we calculate the current as \(I = \frac{100}{120} = 0.833 \text{ A}\).
This means each wire in the lamp cord carries an equal current of 0.833 A.
Amp猫re's Force Law
Amp猫re's force law is key to understanding how currents interact through magnetic forces. This law highlights that two parallel conductors carrying currents will exert forces on each other. Depending on the direction of the currents, the force can be attractive or repulsive.
- If the currents flow in the same direction, the force is attractive. - If they flow in opposite directions, the force is repulsive.
In our scenario, we have two currents of the same magnitude flowing in opposite directions, resulting in an attractive force. This means that the wires actually pull towards each other. Under practical conditions, Amp猫re鈥檚 force law explains many electromagnetic interactions we see in electrical circuits and devices.
Force per Unit Length
To calculate the force per unit length between two parallel wires, we use the specific formula for magnetic force per length \(\frac{F}{L}\):\[\frac{F}{L} = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{d}\]where:
  • \(\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}\) (T = tesla), is the permeability of free space,
  • \(I_1\) and \(I_2\) are the currents in amperes,
  • \(d\) is the distance between the wires in meters.
With \(I_1 = I_2 = 0.833 \text{ A}\) and \(d = 0.003 \text{ m}\), substituting into the formula gives:\[\frac{F}{L} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{0.833 \times 0.833}{0.003} = 4.63 \times 10^{-5} \text{ N/m}\]This force, although measurable, is quite small and does not significantly influence the design of everyday lamp cords, which are expected to handle small mechanical stresses.

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Most popular questions from this chapter

A long, straight wire lies along the \(z\)-axis and carries a 4.00-A current in the \(+z\)-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire centered at the origin: (a) \(x =\) 2.00 m, \(y = 0\), \(z =\) 0; (b) x = 0, \(y =\) 2.00 m, \(z =\) 0; (c) \(x =\) 2.00 m, \(y =\) 2.00 m, \(z =\) 0; (d) \(x\) = 0, \(y\) = 0, \(z\) = 2.00 m,

Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. (a) If you were unfortunate enough to be 5.0 m away from such a lightning bolt, how large a magnetic field would you experience? (b) How does this field compare to one you would experience by being 5.0 cm from a long, straight household current of 10 A?

Certain bacteria (such as \(Aquaspirillum\) \(magnetotacticum\)) tend to swim toward the earth's geographic north pole because they contain tiny particles, called magnetosomes, that are sensitive to a magnetic field. If a transmission line carrying 100 A is laid underwater, at what range of distances would the magnetic field from this line be great enough to interfere with the migration of these bacteria? (Assume that a field less than 5\(\%\) of the earth's field would have little effect on the bacteria. Take the earth's field to be 5.0 \(\times\) 10\(^{-5}\) T, and ignore the effects of the seawater.)

(a) How large a current would a very long, straight wire have to carry so that the magnetic field 2.00 cm from the wire is equal to 1.00 G (comparable to the earth's northward-pointing magnetic field)? (b) If the wire is horizontal with the current running from east to west, at what locations would the magnetic field of the wire point in the same direction as the horizontal component of the earth's magnetic field? (c) Repeat part (b) except the wire is vertical with the current going upward.

A wide, long, insulating belt has a uniform positive charge per unit area \(\sigma\) on its upper surface. Rollers at each end move the belt to the right at a constant speed \(v\). Calculate the magnitude and direction of the magnetic field produced by the moving belt at a point just above its surface. (\(Hint:\) At points near the surface and far from its edges or ends, the moving belt can be considered to be an infinite current sheet like that in Problem 28.73.)
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