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\(\bullet\) An electric bus operates by drawing current from two parallel overhead cables, at a potential difference of \(600 \mathrm{V},\) and spaced 55 \(\mathrm{cm}\) apart. When the power input to the bus's motor is at its maximum power of \(65 \mathrm{hp},(\) a) what current does it draw and (b) what is the attractive force per unit length between the cables?

Short Answer

Expert verified
The bus draws a current of approximately 80.82 A and the force per unit length between the cables is about 7.44 mN/m.

Step by step solution

01

Convert Horsepower to Watts

The problem states that the maximum power of the bus's motor is \(65\, \mathrm{hp}\). First, let's convert horsepower to watts, since 1 horsepower is equivalent to approximately 746 watts. Thus, \(65\, \mathrm{hp} = 65 \times 746\, \mathrm{W}\).
02

Calculate the Power in Watts

Multiply the number of horsepower by 746 to convert to watts:\[65\, \mathrm{hp} = 65 \times 746 = 48490\, \mathrm{W}.\]
03

Use Power Formula to Find Current

The power \(P\) is related to voltage \(V\) and current \(I\) by the equation:\[P = V \times I.\]Rearrange for current:\[I = \frac{P}{V}\]Substitute the known values:\[I = \frac{48490}{600} = 80.82\, \mathrm{A}.\]
04

Calculate Magnetic Force per Unit Length

The attractive force per unit length between two parallel wires carrying currents \(I\) separated by distance \(d\) is given by:\[f/l = \frac{\mu_0 I^2}{2\pi d},\]where \(\mu_0 = 4\pi \times 10^{-7} \mathrm{N/A}^2\) is the permeability of free space. First, convert the separation distance to meters (55 cm = 0.55 m) and then substitute the values:\[f/l = \frac{(4\pi \times 10^{-7}) \times (80.82)^2}{2\pi \times 0.55}.\]
05

Simplify and Solve for Force per Unit Length

Simplify the equation from Step 4:\[f/l = \frac{4\pi \times 10^{-7} \times 6521.47}{1.1}\]\[f/l \approx \frac{8.18 \times 10^{-3}}{1.1}\]\[f/l \approx 7.44 \times 10^{-3} \mathrm{N/m}.\]

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Key Concepts

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

Electric Current
Electric current is the flow of electric charge through a conductor, such as a wire or a cable. This current is typically carried by moving electrons, and it is measured in amperes (A). In electrical circuits, the current is driven by a potential difference, or voltage, across the circuit.
The current in a circuit is calculated using the formula:
  • \(I = \frac{P}{V}\)
The bus in our problem is powered by converting 65 horsepower to watts, resulting in a power of 48490 watts. By dividing the power by the voltage of 600 volts, we find that the bus draws a current of approximately 80.82 amperes. Understanding this relationship helps us know how much electricity flows through and powers devices.
Power Conversion
Power conversion involves changing power from one form to another. In the context of electric buses, it's essential to convert mechanical power (expressed in horsepower) to electrical power (expressed in watts).
This conversion uses the fact that 1 horsepower is equivalent to 746 watts. So, when you have an electric motor rated in horsepower, you must convert it to watts for calculations related to electrical circuits.
  • The conversion formula is: \(1 ext{ hp} = 746 ext{ W}\).
In our example, we changed 65 horsepower to 48490 watts, allowing us to use voltage and other electrical formulas effectively. This step is vital for understanding how much energy an electric device uses and efficiently designing power systems.
Magnetic Force
Magnetic force between two currents plays a fundamental role in electric circuits, especially when wires run parallel to each other carrying currents. Two parallel wires with currents running through them exert an attractive or repulsive force on each other due to their magnetic fields. This force can be calculated using the formula:
  • \(\frac{f}{l} = \frac{\mu_0 I^2}{2\pi d}\)
where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( d \) is the distance between the wires.
This concept shows us how currents can create magnetic effects, influencing the design and operation of circuits. Using this understanding, we found in our exercise that the force per unit length between the cables is approximately 7.44 mN/m.
Parallel Wires
Parallel wires are a common arrangement in electrical circuits, especially when supplying power over long distances. When wires run parallel and carry currents, they influence each other through their magnetic fields.
  • The influence is directly related to the currents they carry and the distance between them.
  • The closer the wires, the stronger the interaction.
In our case, the wires are 0.55 meters apart. Knowing how to calculate and consider these interactions is important, especially in designing systems to prevent unwanted interference and ensure safety and efficiency.
Understanding the behavior of parallel wires helps in planning how electric power lines are set up to optimize space while minimizing magnetic force effects between conductive pathways.

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Most popular questions from this chapter

\(\bullet\) Between the poles of a powerful magnet is a cylindrical uniform magnetic field with a diameter of 3.50 \(\mathrm{cm}\) and a strength of 1.40 \(\mathrm{T}\) . A wire carries a current through the center of the field at an angle of \(65.0^{\circ}\) to the magnetic field lines. If the wire experiences a magnetic force of \(0.0514 \mathrm{N},\) what is the current flowing in it?

\(\bullet\) A beam of protons is accelerated through a potential dif- ference of 0.745 \(\mathrm{kV}\) and then enters a uniform magnetic field traveling perpendicular to the field. (a) What magnitude of field is needed to bend these protons in a circular arc of diameter 1.75 \(\mathrm{m} ?\) (b) What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

\(\bullet\) Singly ionized (onne electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 \(\mathrm{V} / \mathrm{m}\) and the magnetic field is 0.0315 \(\mathrm{T}\) . The ions next enter a uniform magnetic field of magnitude 0.0175 \(\mathrm{T}\) that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is \(17.5 \mathrm{cm},\) what is their mass?

\(\bullet\) A beam of protons traveling at 1.20 \(\mathrm{km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field in a direction perpendicular to its original direction (Fig. \(20.60 ) .\) The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

\(\cdot\) A solenoid contains 750 coils of very thin wire evenly wrapped over a length of 15.0 \(\mathrm{cm} .\) Each coil is 0.800 \(\mathrm{cm}\) in diameter. If this solenoid carries a current of \(7.00 \mathrm{A},\) what is the magnetic field at its center?

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