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$\bullet$ A 150 $\mathrm{V}$ battery is connected across two parallel metal plates of area 28.5 $\mathrm{cm}^{2}$ and separation 8.20 $\mathrm{mm} .$ A beam of alpha particles (charge $+2 e,$ mass $6.64 \times 10^{-27} \mathrm{kg} )$ is accelerated from rest through a potential difference of 1.75 $\mathrm{kV}$ and enters the region between the plates perpendicular to the electric field. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

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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

University of Michigan - Ann Arbor

University of Sheffield

McMaster 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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So we want the magnitude and the direction of the magnetic field such that the Alfa particle does not remain undefended through between the two plates. So we should first try to find the velocity of the Alfa particles through the conservation of energy So we can a so by the law of conservation of energy. If he's accelerated through, put through a electric potential, we can say that Q times we equals half M. V squared. So charge times the electric potential equals 1/2 times the mass times the velocity squared. We can solve for the velocity so this will be equals the square root of two times that charge times the electric potential divided by m the mass for in our foreign Alfa particle alpha particle has a charge of plus two e and then its mass is equaling 6.64 times, 10 to the negative 27th kilograms and then we have an electric potential want 1,750 volts. So to solve for velocity, we can say the velocity of the Alfa particle is equaling the square root of two times two times 1.6 times 10 to the negative 19th cool ums times 1,750 volts divided by mass. So divided by 6.64 times 10 to the negative 27th kilograms. And we're getting that The velocity of the Alfa particle is going to be equal to four 0.11 times 10 to the fifth power meters per second. So ah, let's This is not the answer, but just take note of this velocity. We know that the electric field between the plates is going to be equal to E, so he's going to be equal to the potential at B, divided by the distance between the two plates. So this will simply be equal to 150 volts. The potential at B divided by 0.0 820 meters, we find that the nine it'd of the electric field is 1,000 rather 18,300 of volts per meter. We need you know that the magnetic forced must cancel out the electric force because we want the alpha particle to not have any deflection. So this means that the magnitude of the magnetic field must equal the magnitude of the electric field divided by ah this feed TV, so this was simply be equal to 18,300 volts per meter and then divided by 4.11 times 10 to the fifth meters per second. And this is equaling 0.445 Tesla's. So this is the magnitude of thie magnetic field such that it counter acts, the electric force, and we know that magnetic field is going to be perpendicular to the electric field. So we can say that if we were to draw a diagram in the X direction, we would have the velocity and in the UAE direction, we would have the electric field. This would be the easy access, and then going into the page would be the magnetic field. So this would be the direction of the magnetic field would be into the page again. The magnitude being 0.445 Tesla's. In order to counsel out the electric force, we know that the magnetic field has to be perpendicular to the electric field. That is the end of the solution. Thank you for watching

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