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\bullet An electron at point $A$ in Figure 20.61 has a speed $v_{0}$ of $1.41 \times 10^{6} \mathrm{m} / \mathrm{s}$ . Find (a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from $A$ to $B$ and (b) the time required for the electron to move from $A$ to $B$ . (c) What magnetic field would be needed if the particle were a proton instead of an electron?

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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

Rutgers, The State University of New Jersey

Simon Fraser University

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 electron at point A has…

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An electron moves with a s…

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An electron moves at 1.40 …

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An electron is moving at $…

so if you were to draw the trajectory of the electron, it's going to go through 1/2 circle. And so the velocity is always 10 tangent to the curve and perpendicular to the velocity is always the magnetic force. Ah, here it would be going town and the velocity would be going straight to obliterate. So here the magnetic field must be directed into the page. So ah, we can label this so again the magnetic fields always into the page. And we know that the forces equaling mass times acceleration and the only force ah, here would be the magnetic force. And in this case, the acceleration is centripetal. So we can say that the magnetic force equals the mass times the centripetal acceleration. Or we can say that the absolute value of the charge times the velocity times, the magnitude of the magnetic field times sign if I is going to be equal to m v squared, divided by our now here we know that fi is 90 degrees, so this term is going to be equal toe one. It could be eliminated, and we know that the magnitude of the magnetic field is going to be equal to the mass times the velocity divided by the not absolute value of the charge times the radius. So at this point for part A. When we're trying to find B, we can simply say that this is going to be equal to 9.11 times 10 to the negative 31st kilogram. And then the velocity of the electron is one point for one times 10 to the sixth meters per second. And this will be divided by ah, the charge of an elective electron 1.6 times 10 to the negative 19th cool ums and then times the radius of 0.5 meters. And we find that the magnitude of the magnetic field is equaling 1.61 times, 10 to the negative forthe Tesla's. So this will be your answer for party. Uh, that's draw that better? Yes, it will be an answer for a day. And then for part B. They're asking us for a time. We know that time is simply going to be equal to half a circumference length so pi r and then divided by v. The speed. So this would be pi Times are of 0.5 meters and then this will be divided by one point for one times 10 to the sixth meters per second. And the time is equaling 1.11 times 10 to the negative seventh seconds. So this would be your answer for part me and then for part, see ah, they're asking us to do the exact same calculations. Ah, two, rather, to find the magnetic field such that we have a proton instead of an election. So for a proton Ah, it's going to be opposite the magnetic fields, going to be opposite of the direction given that it was electron. So for a proton, we can say, um, the magnetic field must be directed out of the page, so that would be the direction of the magnetic field. And then to find the magnitude of the magnetic field, this would simply be equal to again envy over Q r. But now we are using the mass of a proton. So this would be 1.67 times 10 to the negative, 27th times the velocity of 1.41 times 10 to the sixth meters per second. And then this will be divided by 1.6 times 10 to the negative 19th cool arms and then times 100.5 meters and the magnitude of the magnetic field. That, given that we have a proton instead of an electron, would be 0.294 Tesla's. So this would be our answer for apart. See, that is the end of the solution. Thank you for watching.

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