91Ó°ÊÓ

Terminal Speed. A bar of length \(L=0.8 \mathrm{m}\) is free to shide without friction on horizontal rails, as shown in Fig. 29.48 . There is a uniform magnetic ficld \(B=1.5 \mathrm{T}\) directed into the plane of the figure. At one end of the rails there is a battery with emf \(\mathcal{E}=12 \mathrm{V}\) and a switch. The bar has mass 0.90 \(\mathrm{kg}\) and resistance \(5.0 \Omega,\) and all other resistance in the circuit can be ignored. The switch is closed at time \(t=0\) . (a) Sketch the speed of the bar as a function of time. (b) Just aner the switch is closed, what is the acceleration of the bar? (c) What is the acceleration of the bar when its speed is 2.0 \(\mathrm{m} / \mathrm{s} ?\) (d) What is the terminal speed of the bar?

Step by step solution

01

Understand the Problem

We are dealing with a bar moving on rails in a magnetic field with an induced electromotive force (emf). We need to calculate the acceleration under different conditions and find the terminal speed of the bar.

02

Calculate Initial Current

When the switch is closed, the current passing through the circuit can be calculated using Ohm’s Law: \[ I = \frac{\mathcal{E}}{R} = \frac{12}{5} = 2.4 \, \text{A} \]

03

Calculate Magnetic Force

The magnetic force on the bar due to the magnetic field is given by: \[ F_B = ILB = 2.4 \times 0.8 \times 1.5 = 2.88 \, \text{N} \]

04

Calculate Initial Acceleration

Using Newton's second law, the initial acceleration can be calculated as: \[ a = \frac{F_B}{m} = \frac{2.88}{0.9} = 3.2 \, \text{m/s}^2 \]

05

Calculate Induced emf due to Motion

When the bar moves, an emf is induced given by: \[ \mathcal{E}_{\text{induced}} = BLv \] where \(v\) is the velocity of the bar.

06

Expression for Terminal Speed

At terminal speed, the total emf in the circuit becomes zero because externally applied emf becomes equal to induced emf: \[ \mathcal{E} = \mathcal{E}_{\text{induced}} \, \Rightarrow \, 12 = 1.5 \times 0.8 \times v_t \] Solving for \(v_t\) gives:\[ v_t = 10 \, \text{m/s} \]

07

Calculate Acceleration at 2 m/s

Using the modified expression for emf: \[ 12 - BLv = IR \] Substituting values into the force: \[ F_B = \left( \frac{12 - 1.5 \times 0.8 \times 2}{5} \right) 1.5 \times 0.8 \] Calculate the current, then the force, then the acceleration: \[ a = \frac{1.92}{0.9} = 2.13 \, \text{m/s}^2 \]

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Key Concepts

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

Magnetic Force

Magnetic force acts on a current-carrying conductor when it is placed in a magnetic field. This force is essential in understanding the movement of the bar in this problem. The magnetic force can be calculated using the formula \[ F_B = ILB \]where:

• \(I\) is the current flowing through the conductor,
• \(L\) is the length of the conductor,
• \(B\) is the magnetic field strength.

In the exercise, the magnetic field and current interact, resulting in a force that propels the bar with a calculated magnitude of 2.88 N. This force is what causes the initially stationary bar to start moving.
The greater the current and the stronger the magnetic field, the larger the force that can be exerted on the bar.

Ohm's Law

Ohm's Law is a fundamental principle used to calculate the flow of electric current through a circuit. It states that the current \(I\) in a circuit is directly proportional to the electromotive force (emf) \(\mathcal{E}\) and inversely proportional to the resistance \(R\):\[ I = \frac{\mathcal{E}}{R} \]In this exercise, we used Ohm’s Law to find the initial current passing through the bar after the switch is closed. With \(\mathcal{E} = 12 \, \text{V}\) and \(R = 5 \, \text{Ω}\), the initial current is calculated as 2.4 A. This current, in the presence of a magnetic field, results in a magnetic force that moves the bar.

Terminal Velocity

Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. In this context, it's where the speeding bar on the rails reaches a state where net force and thus acceleration is zero. This occurs because the induced electromotive force created by the moving bar in the magnetic field eventually matches the applied emf from the battery.
At terminal speed, we have:\[ \mathcal{E} = B L v_t \]Solving this equation enables us to determine how fast the bar moves at terminal speed, resulting in \(v_t = 10 \, \text{m/s}\).
This means the net force acting on the bar becomes zero, ceasing further acceleration, indicating the maximum constant speed of the bar.

Electromotive Force

Electromotive force (emf) is the energy provided per charge by an energy source. It is what drives the current around a circuit. In this exercise, there are two sources of emf to consider: the battery emf and the induced emf from the bar's motion.The battery provides a voltage of \(\mathcal{E} = 12 \, \text{V}\), driving the initial current that produces movement. As the bar moves, it cuts through the magnetic lines, inducing an additional emf given by the equation:\[ \mathcal{E}_{\text{induced}} = B L v \]The induced emf opposes the battery emf as per Lenz's Law, reducing the current and thus magnetic force over time. At terminal velocity, these become equal, balancing forces in the circuit.