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Chapter 28: Problem 29

A long, horizontal wire \(A B\) rests on the surface of a table and carries a current \(I\) . Horizontal wire \(C D\) is vertically above wire \(A B\) and is free to slide up and down on the two vertical metal guides \(C\) and \(D\) (Fig. 28.45\()\) . Wire \(C D\) is connected through the sliding contacts to another wire that also carries a current \(I,\) opposite in direction to the current in wire \(A B .\) The mass per unit length of the wire \(C D\) is \(\lambda\) . To what equilibrium height \(h\) will the wire \(C D\) rise, assuming that the magnetic force on it is due entirely to the current in the wire \(A B ?\)

Short Answer

Expert verified
The equilibrium height \( h \) is \( \frac{\mu_0 I^2}{2 \pi \lambda g} \).

Step by step solution

01

Understanding the Interaction of Currents

The magnetic force between two parallel wires carrying current is due to the magnetic field created by each wire and the force that acts on the other. If the currents are in opposite directions, the wires will repel each other.
02

Applying the Force between Parallel Currents Formula

The magnetic force per unit length between two parallel wires carrying currents is given by the formula: \[ F_{mag} = \frac{\mu_0 I_1 I_2}{2 \pi d} \]where \( \mu_0 \) is the permeability of free space, \( I_1 \) and \( I_2 \) are the currents, and \( d \) is the distance between the wires.
03

Determine the Distance for Equilibrium

The wire \( CD \) will rise to a height \( h \) where the magnetic force balances the gravitational force. The gravitational force per unit length on the wire \( CD \) is \( F_{gravity} = \lambda g \). For equilibrium: \[ \frac{\mu_0 I^2}{2 \pi h} = \lambda g \]Here, the gravitational force equals the magnetic force per unit length between the wires.
04

Solving for Height h

Rearrange the balance equation to solve for \( h \):\[ h = \frac{\mu_0 I^2}{2 \pi \lambda g} \]This formula gives us the equilibrium height \( h \) to which wire \( CD \) will rise.

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Key Concepts

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

Equilibrium Height
When we talk about equilibrium height in this context, we're referring to the height at which wire \( CD \) comes to rest vertically above wire \( AB \). This position is achieved when the upward magnetic force balances the downward gravitational force. The concept of equilibrium height is crucial in understanding how forces interact in physics.

In our exercise, the wire \( CD \) will rise because of the magnetic force generated by the current running through wire \( AB \). The equilibrium height \( h \) can be calculated using the formula: \[ h = \frac{\mu_0 I^2}{2 \pi \lambda g} \]Here:
  • \( \mu_0 \) is the magnetic permeability of free space.
  • \( I \) is the current flowing through each wire.
  • \( \lambda \) is the mass per unit length of wire \( CD \).
  • \( g \) is the acceleration due to gravity.
At this height, the forces are balanced, meaning \( CD \) won't rise further or fall. Understanding this equilibrium helps explain phenomena like magnetic levitation and stability in magnetic systems.
Parallel Currents
Parallel currents influence each other due to the magnetic fields they create. Magnetism arises because moving electric charges produce magnetic fields that exert forces on other moving charges. When two wires carry currents in the same direction, they attract each other. In our scenario, the currents are in opposite directions, resulting in a repulsive force.

The force between two parallel currents is expressed by:\[ F_{mag} = \frac{\mu_0 I_1 I_2}{2 \pi d} \]
  • \( \mu_0 \) is the magnetic constant or permeability of free space.
  • \( I_1 \) and \( I_2 \) are the currents in the wires.
  • \( d \) is the distance between the wires.
This repulsive interaction is what drives the wire \( CD \) upwards, potentially away from wire \( AB \). Understanding this principle is key in electromagnetic applications, from electrical engineering to physics experiments involving magnetic repulsion and attraction.
Gravitational Force
Gravity is the force that attracts objects towards the center of the Earth. In our exercise, the gravitational force acts downward on wire \( CD \). The wire's mass contributes to this force, calculated as the weight per unit length: \( F_{gravity} = \lambda g \). Here,
  • \( \lambda \) is the lineal mass density of wire \( CD \).
  • \( g \) denotes the gravitational acceleration, approximately \( 9.81 \, \text{m/s}^2 \) on Earth's surface.
This downward pull is constant and always acts in opposition to any upward force, like the magnetic force in this setup. For the system to reach equilibrium, this gravitational force must equal the magnetic force per unit length acting on \( CD \). Balancing these two forces is critical; it allows us to predict how high the wire will rise when released, giving insights into real-world systems, from simple physics demonstrations to complex machinery in industries.

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Most popular questions from this chapter

An alpha particle (charge + 2e) and an electron move in opposite directions from the same point, each with the speed of \(2.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) (Fig. \(28.32 ) .\) Find the magnitude and direction of the total magnetic field these charges produce at point \(P,\) which is 1.75 \(\mathrm{nm}\) from each of them.

A long, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is \(J .\) The current density, although symmetrical about the cylinder axis, is not constant but varies according to the relationship $$ \begin{array}{rlrl}{\overrightarrow{\boldsymbol{J}}} & {=\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{\boldsymbol{k}}} & {} & {\text { for } \boldsymbol{r} \leq \boldsymbol{a}} \\ {} & {=\mathbf{0}} & {} & {\text { for } \boldsymbol{r} \geq a}\end{array} $$ where \(a\) is the radius of the cylinder, \(r\) is the radial distance from the cylinder axis, and \(I_{0}\) is a constant having units of amperes. (a) Show that \(I_{0}\) is the total current passing through the entire cross section of the wire. (b) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \geq a\) . (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. (d) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \leq a\) . How do your results in parts \((b)\) and \((d)\) compare for \(r=a ?\)

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 1.40 \(\mathrm{m}\) long and 20.0 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

Two identical circular, wire loops 40.0 \(\mathrm{cm}\) in diameter each carry a current of 1.50 \(\mathrm{A}\) in the same direction. These loops are parallel to each other and are 25.0 \(\mathrm{cm}\) apart. Line \(a b\) is normal to the plane of the loops and passes through their centers. A proton is fired at 2400 \(\mathrm{km} / \mathrm{sperpendicular}\) to line \(a b\) from a point midway between the centers of the loops. Find the magnitude and direction of the magnetic force these loops exert on the proton just after it is fired.

A long, straight wire lies along the \(y\) -axis and carries a current \(I=8.00\) A in the \(-y\) -direction (Fig. 28.39\()\) . In addition to the magnetic field due to the current in the wire, a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) with magnitude \(1.50 \times 10^{-6} \mathrm{T}\) is in the \(+x\) -direction \(\mathrm{What}\) is the total field (magnitude and direction) at the following points in the \(x z\) -plane: \((a) x=0, z=1.00 \mathrm{m}\) (b) \(x=1.00 \mathrm{m}, z=0 ;(\mathrm{c}) x=0\) \(z=-0.25 \mathrm{m} ?\)
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