Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 22, Problem 16P

(a)

To determine

The frequency of the proton.

(a)

Expert Solution
Check Mark

Answer to Problem 16P

The frequency of the proton is. 7.66×107s1 .

Explanation of Solution

Given Info: The accelerating voltage is 600V , the outer radius is 0.350m and the magnitude of magnetic field is 0.800T .

The formula for the frequency is,

f=qB2πm

Here,

B is the magnitude of magnetic field.

m is the mass of proton.

q is the charge of proton.

V is the accelerating voltage.

Substitute 0.800T for B , 1.6×1027kg for m , 1.6×1019C for q and 600V for V in above equation to find f .

f=(1.6×1019C)(0.800T)2π(1.6×1027kg)=7.66×107s1

Thus, the frequency of the proton is. 7.66×107s1 .

Conclusion:

Therefore, the frequency of the proton is. 7.66×107s1 .

(b)

To determine

The exit speed of the proton.

(b)

Expert Solution
Check Mark

Answer to Problem 16P

The exit speed of the proton is. 2.68×107m/s .

Explanation of Solution

Given Info: The accelerating voltage is 600V , the outer radius is 0.350m and the magnitude of magnetic field is 0.800T .

The formula for the speed is,

v=qBrm

Here,

r is the radius of the outermost orbit.

Substitute 0.800T for B , 1.6×1027kg for m , 1.6×1019C for q and 0.350m for r in above equation to find v .

v=(1.6×1019C)(0.800T)(0.350m)(1.6×1027kg)=2.68×107m/s

Thus, the exit speed of the proton is. 2.68×107m/s .

Conclusion:

Therefore, the exit speed of the proton is. 2.68×107m/s .

(c)

To determine

The maximum kinetic energy.

(c)

Expert Solution
Check Mark

Answer to Problem 16P

The maximum kinetic energy is 3.75MeV .

Explanation of Solution

Given Info: The accelerating voltage is 600V , the outer radius is 0.350m and the magnitude of magnetic field is 0.800T .

The formula for the kinetic energy is,

KEmax=12mv2

Substitute 1.67×1027kg for m and 2.68×107m/s for v in above equation to find KEmax .

KEmax=12(1.67×1027kg)(2.68×107m/s)2=5.99×1013J×(6.2×1012MeV)1J=3.75MeV

Thus, the maximum kinetic energy is 3.75MeV .

Conclusion:

Therefore, the maximum kinetic energy is 3.75MeV .

(d)

To determine

The number of revolutions.

(d)

Expert Solution
Check Mark

Answer to Problem 16P

The number of revolutions are 3.13×103revolutions .

Explanation of Solution

Given Info: The accelerating voltage is 600V , the outer radius is 0.350m and the magnitude of magnetic field is 0.800T .

The formula for the number of revolutions is,

N=Emax2qV

Substitute 600V for V , 1.6×1019C for q and 4.4×1013J for KEmax in above equation to find N .

N=(4.4×1013J)2(1.6×1019C)(600V)=3.13×103revolutions

Thus, the number of revolutions are 3.13×103revolutions .

Conclusion:

Therefore, the number of revolutions are 3.13×103revolutions .

(e)

To determine

The time of acceleration.

(e)

Expert Solution
Check Mark

Answer to Problem 16P

The time of acceleration is 2.57×104s .

Explanation of Solution

Given Info: The accelerating voltage is 600V , the outer radius is 0.350m and the magnitude of magnetic field is 0.800T .

The formula for the number of revolutions is,

T=2πmqB

Substitute 0.800T for B , 1.6×1027kg for m and , 1.6×1019C for q in above equation to find T .

T=2π(1.6×1027kg)(1.6×1027kg)(0.800T)=2.57×104s

Thus, the time of acceleration is 2.57×104s .

Conclusion:

Therefore, the time of acceleration is 2.57×104s .

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Chapter 22 Solutions

Principles of Physics: A Calculus-Based Text

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