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Chapter 29: Problem 66

Two resistanceless rails rest 32 \(\mathrm{cm}\) apart on a \(6.0^{\circ}\) ramp. They are joined at the bottom by a \(0.60-\Omega\) resistor. At the top a copper bar of mass 0.040 \(\mathrm{kg}\) (ignore its resistance) is laid across the rails. The whole apparatus is immersed in a vertical 0.55 - field. What is the terminal (steady) velocity of the bar as it slides frictionlessly down the rails?

Short Answer

Expert verified
The terminal velocity of the bar is approximately 0.77 m/s.

Step by step solution

01

Understand the Problem

We have a scenario where a copper bar is sliding down resistanceless rails on a ramp, creating an electric circuit with a resistor. The whole system is in a magnetic field, inducing an electromotive force (EMF) as the bar moves, and we need to find the terminal velocity of the bar, which is the constant velocity reached when the forces balance.
02

Identify Relevant Equations and Forces

The gravitational force components acting on the bar can be split into parallel and perpendicular to the incline plane. The parallel component is influenced by gravity and the angle of the incline, and gives: \[ F_{gravity} = mg \sin(\theta) \] Where \(m = 0.040\,\text{kg}\) (mass of the bar), \( g = 9.8\,\text{m/s}^2 \), and \( \theta = 6.0^{\circ} \). The magnetic force acting on the bar due to the magnetic field is given by: \[ F_{magnetic} = BIL \]Here, \( B = 0.55\,\text{T} \) (magnetic field strength), and \( L = 0.32\,\text{m} \) (distance between the rails). The current \( I \) can be computed using Ohm鈥檚 law related to the EMF.
03

Calculate Electromagnetic Induction and EMF

The motion of the bar through the magnetic field induces an EMF due to the magnetic flux changing, which can be calculated as: \[ \text{EMF} = BvL \]This EMF drives a current \( I \) through the resistor, where \( R = 0.60\,\Omega \). According to Ohm's Law, \[ EMF = IR \Rightarrow I = \frac{BvL}{R} \]
04

Equate Gravitational and Magnetic Forces

At terminal velocity, the magnetic force balances the gravitational component: \[ mg \sin(\theta) = BIL \] Substitute \( I = \frac{BvL}{R} \) into the magnetic force equation:\[ mg \sin(\theta) = B \left( \frac{BvL}{R} \right) L \] This simplifies to:\[ mg \sin(\theta) = \frac{B^2 v L^2}{R} \]
05

Solve for Terminal Velocity

Rearrange the equation from Step 4 to find \( v \): \[ v = \frac{mg \sin(\theta) R}{B^2 L^2} \] Substitute the given values:- \( m = 0.040 \text{ kg} \)- \( g = 9.8 \text{ m/s}^2 \)- \( \theta = 6.0^{\circ} \)- \( R = 0.60 \text{ } \Omega \)- \( B = 0.55 \text{ T} \)- \( L = 0.32 \text{ m} \)Finally, calculating gives:\[ v = \frac{(0.040)(9.8) \sin(6^{\circ})(0.60)}{(0.55)^2 (0.32)^2} \approx 0.77 \text{ m/s} \]
06

Conclusion

The terminal velocity of the bar as it slides down the incline is approximately \(0.77\, \text{m/s}\).

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Key Concepts

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

Terminal Velocity
Terminal velocity is the constant speed that a moving object eventually reaches when the force exerted by gravity is balanced by an opposite force, in this case, the magnetic force. As the copper bar slides down the rails in the magnetic field, it accelerates until these forces are equal in magnitude but opposite in direction. At this point, the acceleration ceases, and the bar continues to move at a constant speed, known as the terminal velocity. Understanding terminal velocity implies realizing that it represents a state of dynamic equilibrium, where the net force acting on the moving object is zero.
  • At terminal velocity, forces are balanced: gravitational force is opposed by magnetic force.
  • The object moves at a constant speed because acceleration stops when forces are balanced.
  • This concept applies to varied fields, including free-fall scenarios and objects moving through fluids or air.
Magnetic Force
When a conductor such as a copper bar moves through a magnetic field, a magnetic force acts on it. This force is due to electromagnetic induction, a principle where an electric current is generated by changing a magnetic field. In this scenario, as the bar slides down the ramp, it interacts with the vertical magnetic field, and a current is induced.
  • The magnetic force depends on the magnetic field strength (B), the current (I) flowing through the conductor, and the length of the conductor within the field (L).
  • The direction of the magnetic force can be determined using the right-hand rule.
  • The magnetic force equation is given by: \[ F_{magnetic} = BIL \] where B is the magnetic field strength, I is the current, and L is the length of the conductor.
Ohm's Law
Ohm's Law is a fundamental principle in the field of electricity and magnetism, describing the relationship between voltage, current, and resistance in an electrical circuit. It is expressed by the equation Ohm's Law:\[ V = IR \]where \( V \) represents the voltage (or electromotive force), \( I \) represents the current through the conductor in amperes, and \( R \) the resistance in ohms.
  • This law helps calculate the current that flows through the circuit when a voltage is applied.
  • In the given exercise, EMF generated by the moving bar induces a current, which was computed using Ohm's Law formulas.
  • Ohm's Law illustrates the proportional relationship between the voltage across the conductor and the current flowing through it.
Gravitational Force
Gravitational force is the attraction between any two masses. In this problem, it plays a key role as it's one of the primary forces acting on the copper bar as it slides down the slope.
  • The gravitational force is calculated using the formula: \[ F_{gravity} = mg \sin(\theta) \] where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline.
  • This force component, parallel to the slope, causes the bar to accelerate down the incline.
  • Understanding the gravitational force's component along a ramp is crucial, as it dictates the opposing forces required for achieving terminal velocity.

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Most popular questions from this chapter

(III) Suppose \(85 \mathrm{~kW}\) is to be transmitted over two \(0.100-\Omega\) lines. Estimate how much power is saved if the voltage is stepped up from \(120 \mathrm{~V}\) to \(1200 \mathrm{~V}\) and then down again, rather than simply transmitting at \(120 \mathrm{~V}\). Assume the transformers are each \(99 \%\) efficient.

(I) The magnetic flux through a coil of wire containing two loops changes at a constant rate from \(-58 \mathrm{~Wb}\) to \(+38 \mathrm{~Wb}\) in \(0.42 \mathrm{~s}\). What is the emf induced in the coil?

(II) A circular loop in the plane of the paper lies in a 0.75-T magnetic field pointing into the paper. If the loop's diameter changes from \(20.0 \mathrm{~cm}\) to \(6.0 \mathrm{~cm}\) in \(0.50 \mathrm{~s}\), (a) what is the direction of the induced current, \((b)\) what is the magnitude of the average induced emf, and \((c)\) if the coil resistance is \(2.5 \Omega\), what is the average induced current?

[The Problems in this Section are ranked I, II, or III according to estimated difficulty, with \((1)\) Problems being easiest. Level (III) Prob- lems are meant mainly as a challenge for the best students, for "extra credit." The Problems are arranged by Sections, meaning that the reader should have read up to and including that Section, but this Chapter also has a group of General Problems that are not arranged by Section and not ranked.] \(\begin{array}{l}{\text { (I) The magnetic flux through a coil of wire containing two }} \\ {\text { loops changes at a constant rate from }-58 \mathrm{Wb} \text { to }+38 \mathrm{Wb} \text { in }} \\ {0.42 \mathrm{s} . \text { What is the emf induced in the coil? }}\end{array}\)

(II) A 250-loop circular armature coil with a diameter of \(10.0 \mathrm{~cm}\) rotates at \(120 \mathrm{rev} / \mathrm{s}\) in a uniform magnetic field of strength \(0.45 \mathrm{~T}\). What is the rms voltage output of the generator? What would you do to the rotation frequency in order to double the rms voltage output?
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