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Chapter 24, Problem 74P

(a)

To determine

The unit of the equation is watts.

(a)

Expert Solution
Check Mark

Answer to Problem 74P

The unit of the equation on the right hand side is watts.

Explanation of Solution

Given info: The equation of the electromagnetic power is p=q2a26πεc3 .

The formula to calculate the electromagnetic power is,

p=q2a26πεc3 (1)

Here,

ε is the emissivity of the free space.

c is the speed of the light.

q is the charge of the particle.

The dimension of ε is [A2M1L3T4] , the dimension of q is [AT] , the dimension of c is [LT1] and the dimension of a is [LT2] .

Substitute the dimensions [A2M1L3T4] for ε , [AT] for q , [LT1] for c and [LT2] for a in the equation (1) to find the unit of p .

p=[AT]2×[LT2]2[A2M1L3T4]×[LT1]3=A2L2T2A2M1T1=ML2T3

The dimension [ML2T3] is of energy and the unit of energy is watts.

Conclusion:

Therefore, the unit of the equation on the right hand side is watts.

(b)

To determine

The acceleration of the electron.

(b)

Expert Solution
Check Mark

Answer to Problem 74P

The acceleration of the electron is 1.75×1013m/s2 .

Explanation of Solution

Given info: The equation of the electromagnetic power is p=q2a26πεc3 .

The formula to calculate the acceleration is,

a=qEm (2)

Here,

q is the charge of the electron.

E is the magnitude of electric field.

m is the mass of the electron.

Substitute 1.6×1019C for q , 100V/m for E and 9.1×1031kg for m in the above equation to find the value of a .

a=(1.6×1019C)(100V/m)(9.1×1031kg)=1.75×1013m/s2

Thus, the acceleration of the electron is 1.75×1013m/s2 .

Conclusion:

Therefore, the acceleration of the electron is 1.75×1013m/s2 .

(c)

To determine

The electromagnetic power radiated by the electron.

(c)

Expert Solution
Check Mark

Answer to Problem 74P

The electromagnetic power radiated by the electron is 1.73×1024W .

Explanation of Solution

Given info: The equation of the electromagnetic power is p=q2a26πεc3 .

The expression for the electromagnetic power is,

p=q2a26πεc3

Substitute 8.8×1012 for ε , 1.6×1019C for q , 3×108m/s for c and 1.75×1013m/s2 for a in the above equation to find the value of p .

p=(1.6×1019C)2(1.75×1013m/s2)26π(8.8×1012)(3×108m/s)3=1.73×1024W

Thus, the electromagnetic power radiated by the electron is 1.73×1024W .

Conclusion:

Therefore, the electromagnetic power radiated by the electron is 1.73×1024W .

(d)

To determine

The electromagnetic power of the proton leaving a cyclotron.

(d)

Expert Solution
Check Mark

Answer to Problem 74P

The electromagnetic power of the proton leaving a cyclotron is 2.08×1021W .

Explanation of Solution

Given info: The electric flux of the particle is 487Nm2/C . the power is radiated equally in all directions is 25.0W .

Given info: The equation of the electromagnetic power is p=q2a26πεc3 .

The formula to calculate the acceleration is,

a=q2B2rm2 (2)

Here,

q is the charge of the proton.

B is the magnetic field.

r is the radius leaving the cyclotron.

m is the mass of the proton.

Substitute 1.6×1019C for q , 1.6×1027kg for m , 0.500m for r and 0.350T for B in the equation (2) to find the value of a .

a=(1.6×1019C)2(0.350T)2(0.500m)(1.6×1027kg)2=6.1×1014m/s2

The expression for the electromagnetic power is,

p=q2a26πεc3

Substitute 8.8×1012 for ε , 1.6×1019C for q , 3×108m/s for c and 6.1×1014m/s2 for a in the above equation to find the value of p .

p=(1.6×1019C)2(6.1×1014m/s2)26π(8.8×1012)(3×108m/s)3=2.08×1021W

Conclusion:

Therefore, the electromagnetic power of the proton leaving a cyclotron is 2.08×1021W .

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Chapter 24 Solutions

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