Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 39, Problem 41PQ
To determine

The magnitude of velocity of an electron in the laboratory frame.

Expert Solution & Answer
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Answer to Problem 41PQ

The magnitude of velocity of the electron measured in the laboratory frame is 8.50×107m/s.

Explanation of Solution

Write the expression for the magnitude of velocity measured from the laboratory frame.

    v=vx2+vy2+vz2                                                                                                    (I)

Here, v is the resultant velocity, vx is the velocity transformation in the x direction, vy is the velocity transformation in the y direction, and vz the velocity transformation in the z direction.

Write the expression for velocity components in each direction,

    vx=vx+vrel[1+vrel×vxc2]                                                                                                    (II)

    vy=vyγ[1+vrel×vxc2]                                                                                                (III)

    vz=vzγ[1+vrel×vxc2]                                                                                                (IV)

Here, vrel is the relative velocity, c is the speed of light, γ is the Lorentz constant, vx is the inverse velocity transformation in x direction, vy is the inverse velocity transformation in y direction and vz is the inverse velocity transformation in z direction.

Write the expression for Lorentz constant,

    γ=11(vrelc)2                                                                                                       (V)

Conclusion:

Substitute 2.00×107m/s for vrel and 3.00×108m/s for c in equation (V) to find γ.

    γ=11(2.00×107m/s3.00×108m/s)2=11(0.06)2=10.998=1.002

Substitute 5.00×107m/s for vx, 2.00×107m/s for vrel and 3.00×108m/s for c in equation (III) to find vx.

    vx=5.00×107m/s+2.00×107m/s[1+5.00×107m/s×2.00×107m/s(3.00×108)2m2/s2]=7.00×107m/s[1+10159×1016]=7.00×107m/s1.011=6.92×107m/s

Substitute 4.00×107m/s for vy, 1.002 for γ, 5.00×107m/s for vx, 2.00×107m/s for vrel and 3.00×108m/s for c in equation (IV) to find vy.

    vy=4.00×107m/s1.002[1+2.00×107m/s×5.00×107m/s(3.00×108)2m2/s2]=4.00×107m/s1.002[1.011]=4.00×107m/s1.013=3.95×107m/s

Substitute 3.00×107m/s for vz, 1.002 for γ, 5.00×107m/s for vx, 2.00×107m/s for vrel and 3.00×108m/s for c in equation (V) to find vz.

    vz=3.00×107m/s1.002[1+2.00×107m/s×5.00×107m/s(3.00×108)2m2/s2]=3.00×107m/s1.002[1.011]=3.00×1071.013=2.96×107m/s

Substitute 6.92×107m/s for vx, 3.95×107m/s for vy and 2.96×107m/s for vz in the equation (I) to obtain the velocity of the electron.

    v=vx2+vy2+vz2=(6.92×107m/s)2+(3.95×107m/s)2+(2.96×107m/s)2=72.2505×1014m2/s2=8.50×107m/s

Therefore, the magnitude of velocity of electron is 8.50×107m/s.

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Chapter 39 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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