Skip to main content
Ch. 28 - Sources of Magnetic Field
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 27, Problem 61

A long horizontal wire carries a current of 42 A. A second wire, made of 1.00-mm-diameter copper wire and parallel to the first, is kept in suspension magnetically 5.0 cm below (Fig. 28–60). (a) Determine the magnitude and direction of the current in the lower wire. (b) Is the lower wire in stable equilibrium? (c) Repeat parts (a) and (b) if the second wire is suspended 5.0 cm above the first due to the first’s magnetic field.
Diagram showing two parallel wires: one carrying 42 A current and the other 5 cm below, with unknown current direction indicated.

Verified step by step guidance
1
Step 1: Understand the problem setup. The first wire carries a current of 42 A, and the second wire is suspended magnetically either below or above the first wire. The magnetic force between the wires must balance the gravitational force acting on the lower wire. The goal is to determine the current in the lower wire and analyze its equilibrium state.
Step 2: Calculate the gravitational force per unit length acting on the lower wire. Use the formula for the mass per unit length of the wire: \( \text{mass per unit length} = \rho \cdot \frac{\pi d^2}{4} \), where \( \rho \) is the density of copper (approximately \( 8.96 \times 10^3 \, \text{kg/m}^3 \)) and \( d \) is the diameter of the wire (1.00 mm). Then, multiply by \( g \) (gravitational acceleration, \( 9.8 \, \text{m/s}^2 \)) to find the gravitational force per unit length.
Step 3: Use Ampère's law and the formula for the magnetic force per unit length between two parallel wires: \( F/L = \frac{\mu_0 I_1 I_2}{2 \pi r} \), where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T·m/A} \)), \( I_1 \) and \( I_2 \) are the currents in the wires, and \( r \) is the distance between the wires (5.0 cm). Set the magnetic force equal to the gravitational force to solve for \( I_2 \), the current in the lower wire.
Step 4: Determine the direction of the current in the lower wire. Use the right-hand rule for magnetic fields and forces. For the wires to repel each other (if the lower wire is suspended below the first), the currents must flow in opposite directions. If the lower wire is suspended above the first, the currents must flow in the same direction.
Step 5: Analyze the stability of the equilibrium. Consider whether a small displacement of the lower wire would result in restoring forces or further displacement. If the wires are suspended below each other, the equilibrium is stable because the magnetic force increases with decreasing distance. If the wires are suspended above each other, the equilibrium is unstable because the magnetic force decreases with increasing distance.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
8m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Magnetic Field Due to a Current

A current-carrying wire generates a magnetic field around it, described by the right-hand rule. The direction of the magnetic field lines forms concentric circles around the wire, with the strength of the field decreasing with distance from the wire. For a long straight wire, the magnetic field (B) at a distance (r) from the wire can be calculated using the formula B = (μ₀I)/(2πr), where μ₀ is the permeability of free space and I is the current.
Recommended video:
Guided course
12:33
Magnetic Field Produced by Straight Currents

Force on a Current-Carrying Wire in a Magnetic Field

When a current-carrying wire is placed in a magnetic field, it experiences a force. This force can be calculated using the formula F = I(L × B), where F is the force, I is the current, L is the length of the wire in the magnetic field, and B is the magnetic field vector. The direction of the force is given by the right-hand rule, indicating that it is perpendicular to both the current direction and the magnetic field.
Recommended video:
Guided course
09:57
Magnetic Force on Current-Carrying Wire

Equilibrium of Forces

An object is in equilibrium when the net force acting on it is zero. In the context of the lower wire, this means that the magnetic force acting on it due to the upper wire must balance the gravitational force acting downward. If the magnetic force is greater than the gravitational force, the wire will rise; if less, it will fall. Stability of equilibrium can be assessed by considering how the system responds to small displacements from its equilibrium position.
Recommended video:
Related Practice
Textbook Question

Helmholtz coils are two identical circular coils having the same radius 𝑅 and the same number of turns N, separated by a distance equal to the radius 𝑅 and carrying the same dc current I in the same direction. (See Fig. 28–61.) They are used in scientific instruments to generate nearly uniform magnetic fields. (They can be seen in the photo, Fig. 27–19.) (a) Determine the magnetic field B at points 𝓍 along the line joining their centers. Let 𝓍 = 0 at the center of one coil, and 𝓍 = 𝑅 at the center of the other. (b) Show that the field midway between the coils is particularly uniform by showing that dB/d𝓍 = 0 and d²B/d𝓍² = 0 at the midpoint between the coils. (c) If 𝑅 = 10.0 cm, N = 85 turns and I = 3.0 A, what is the field at the midpoint between the coils, 𝓍 = 𝑅/2?

2
views
Textbook Question

You want to get an idea of the magnitude of magnetic fields produced by overhead power lines. You estimate that a transmission wire is about 12 m above the ground. The local power company tells you that the lines operate at 145 kV and provide a maximum of 45 MW to the local area. Estimate the maximum magnetic field you might experience walking under one such power line, and compare to the Earth’s field. [For an ac current, values are rms, and the magnetic field will be changing.]

1
views
Textbook Question

In Fig. 28–57 the top wire is 1.00-mm-diameter copper wire and is suspended in air due to the two magnetic forces from the bottom two wires. The current is 35.0 A in each of the two bottom wires. Calculate the required current in the suspended wire (M).

1
views
Textbook Question

Three long parallel wires are 3.5 cm from one another. (Looking along them, they are at three corners of an equilateral triangle.) The current in each wire is 9.50 A, but its direction in wire M is opposite to that in wires N and P (Fig. 28–57). Determine the magnetic force per unit length on each wire due to the other two.


1
views
Textbook Question

(III) A square loop of wire, of side d, carries a current I. (a) Determine the magnetic field B at points on a line (call it the 𝓍 axis) perpendicular to the plane of the square which passes through the center of the square (Fig. 28–56). Express B as a function of 𝓍, the distance from the center of the square. (b) For 𝓍 ≫ d, does the square appear to be a magnetic dipole? If so, what is its dipole moment?


1
views
Textbook Question

A set of Helmholtz coils (see Problem 62, Fig. 28–61) have a radius 𝑅 = 10.0 cm and are separated by a distance 𝑅 = 10.0 cm . Each coil has 85 loops carrying a current I = 2.0 A. Graph B as a function of 𝓍.

1
views