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An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of $6.64 \times 10^{-27}$ kg traveling horizontally at 35.6 $\mathrm{km} / \mathrm{s}$ enters a uniform, vertical, 1.10 $\mathrm{T}$ magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.

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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

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

Lectures

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An alpha particle (a He nu…

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An alpha particle has a ma…

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An alpha particle consists…

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A positively charged parti…

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Predict/Calculate An alpha…

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$$\text { An alpha-particl…

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An alpha particle (the nuc…

02:54

A particle with mass $1.81…

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An alnha particle travels …

04:02

A particle with mass 1.81 …

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A proton (charge $+e,$ mas…

04:47

09:20

A cyclotron designed to ac…

02:10

A deuteron particle (the n…

05:40

A particle with charge $-5…

06:34

A cyclotron is to accelera…

So we want to find her party the radius of the path. So we can say that the radius of the path will simply be equal to the mass times the velocity divided by he absolutely of the charge times the magnitude of the magnetic field. This is going to be equal 26.64 times, 10 to the negative, 27th kilograms, times the velocity of 35,600 meters per second. This will be divided by ah two times 1.6 times 10 to the negative 19th cool arms and then times the magnitude of the magnetic field 1.10 Tesla's. And so we find that the radius is going to be 6.73 times 10 to the negative fourth and this will be meters. This will be the radius of the path. So this will be your answer for a party and then for part B. They're asking us to find Thie. They're asking us to find the direction of the magnetic force. So for a very short time interval, the displacement of the particle is in the direction of the of the velocity. So we can say that for a short delta T we can say that the displacement of the particle is in the direction of the velocity. So we can say that the magnetic force is going to be equal is rather we consider the magnetic force is going to be always perpendicular to the velocity, doing no work, so V is constant. So essentially, what it's saying is that the magnetic force, because the magnetic force is always perpendicular to the velocity, it has no bearing on the velocity. There's ah, it's not know component of the magnetic force is changing the velocity, and thus there is no acceleration the the magnetic forces doing no work on the particle. Therefore, again, there's no acceleration and velocity is constant, even though there's a magnetic force acting on the particle. That magnetic force, however, is acting in a is always perpendicular to velocity, so it's acting in a completely different direction. Eso again velocity is constant for part B. That is the end of the solution. Thank you for watching

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