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An electron moves at $2.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ through a region in which there is a magnetic field of unspecified direction and magnitude $7.40 \times 10^{-2} \mathrm{T}$ (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

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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

University of Sheffield

McMaster University

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So here we know that the velocity of the particle is 2.5 times 10 to the six meters per second. We know that the magnetic field has a magnitude of 0.740 Tesla's and we know that the charge is going to be equal to negative e. So for party, when they're asking us for the minimum and maximum accelerations, we know that the magnetic force is the only force acting on the particle. This is going to be a so essentially the sum of forces. The net force will be equal to the mass times acceleration. However, more specifically, the magnetic forces equaling the absolute value of the charge times the velocity times, the magnitude of the magnetic field time sign. If I we know that when so automatically we can say that amen is going to be equal to zero meters per second squared and this is occurring when phi equals zero degrees. So when phi equals zero degrees, um, the entire equation becomes zero and we have a minimum acceleration of zero meters per second squared. A max occurs when five is equal to 90 degrees. So when this is equal in this term sign if eyes equal to one so a max, it's simply going to be equal to the theat salute. Value of the charge Times of a lost city times the magnitude of the magnetic field divided by the mass. So this will be 1.6 times 10 to the negative. 19th. Cool arms times. Two point five times 10 to the sixth meters per second and then times the magnitude of the magnetic field 0.740 Tesla's divided by the mass of a electron, so this would be a 9.11 times 10 to the negative 31st kilograms, and we find that the maximum acceleration is equaling 3.25 times 10 to the sixth 10th meters per second squared. So this would be your maximum. And again this occurs when Phi equals 90 degrees. And then this would be your minimum. When FIEs equaling zero degrees preferred be. They want us to find the angle at which the exploration is 1/4 of the maximum acceleration. So if we say that the acceleration is equaling Teo, the absolute value of the charge times the velocity times, the magnitude of the magnetic field. Time's sign five. Divided by the Mass. We can say that a should equal 1/4 of a max. Now a max simply equals. We know that a max simply equals the absolute value of the charge. Times of loss. Any time's the magnitude of the magnetic field divided by the mass. So essentially this becomes that sign of Fei simply equals one over four because that would be the factor that is being multiplied by the maximum acceleration in order for in order to get the acceleration that isthe 1/4 of the maximum. So we can say that Fi is simply equal to Arc sine of 0.25 and we find out that fives equaling 14.5 degrees such that again, the acceleration is 1/4 of the maximum acceleration. That is the end of the solution. Thank you for watching

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