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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?

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a) 80.8 $\mathrm{A}$b) $2.37 \times 10^{-3} \mathrm{N} / \mathrm{m}$

Physics 101 Mechanics

Physics 102 Electricity and Magnetism

Chapter 20

Magnetic Field and Magnetic Force

Motion Along a Straight Line

Motion in 2d or 3d

Electric Charge and Electric Field

Gauss's Law

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Electromagnetic Induction

Inductance

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

Lectures

18:38

In physics, electric flux is a measure of the quantity of electric charge passing through a surface. It is used in the study of electromagnetic radiation. The SI unit of electric flux is the weber (symbol: Wb). The electric flux through a surface is calculated by dividing the electric charge passing through the surface by the area of the surface, and multiplying by the permittivity of free space (the permittivity of vacuum is used in the case of a vacuum). The electric flux through a closed surface is zero, by Gauss's law.

04:28

A magnetic field is a mathematical description of the magnetic influence of electric currents and magnetic materials. The magnetic field at any given point is specified by both a direction and a magnitude (or strength); as such it is a vector field. The term is used for two distinct but closely related fields denoted by the symbols B and H. The term "magnetic field" is often used to refer to the B field. In a vacuum, B and H are the same, whereas in a material medium, B is a component of H. In the latter case, H is the "magnetic field strength", and B is the "magnetic flux".

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An electric bus operates b…

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(a) $A 0.750$ -m-long sect…

00:59

Two parallel wires are sep…

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(a) The hot and neutral wi…

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(a) A 0.750-m-long section…

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The wires which connect th…

08:35

A set of jumper cables use…

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The force per meter betwee…

so for party, they want to find the current. We know the power equals 65 horsepower and we know the electric potential equals 600 volts s o weaken. We know that the all We also know that the power is equaling the current times the electric potential so we can solve for the current, this is simply going to be equal to the power divided by the electric potential. Ah, this will be equal to 65 horsepower. There are 746 watts per horsepower and then divided by 600 volts. And we find that the current is going to be equal to 80.8 camp. So this would be your answer for party and then for part B. We're trying to find the attractive force between the wires per unit length. So we know that the force between the two wires is going to be equal tio the magnetic permeability in a vacuum or the primitive ity free space times l the induct in ce time's ice of one ice up to divided by two pi r And so we know that here are is going to be equal to 0.55 meters and again we're trying to find the ah, the attractive force between the wires per unit length. I said that this was thie inducting before because it was l. This is the length. My apologies. This is not the inductive. This's the length. So when we're trying to find the attractive force between the wires per unit length, we can actually divide by L. So we can say that eth over l again the attractive force between the wires per unit length would simply be mu not. I squared both the wires have the same, uh, current divided by two pi r. And so we can say that f divided by out is going to be equal to four times 10 to the negative. Seventh times, 80 point, eh? Oh, a squared will round to the very end and then divided by two pi times are of 20.55 meters, and so this is going to be equal to 2.4 times 10 to the negative third Newtons per meter. So this would be your answer for part B against the attractive force between the two wires per unit length. That is the end of the solution. Thank you for watching

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