Chapter 24, Problem 74P

(a)

To determine

The unit of the equation is watts.

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 $p=\frac{q^2{a}^2}{6\pi {\epsilon }_{\circ }{c}^3}$.

The formula to calculate the electromagnetic power is,

$p=\frac{q^2{a}^2}{6\pi {\epsilon }_{\circ }{c}^3}$ (1)

Here,

${\epsilon }_{\circ }$ is the emissivity of the free space.

$c$ is the speed of the light.

$q$ is the charge of the particle.

The dimension of ${\epsilon }_{\circ }$ is $\left[A^2{M}^{-1}{L}^{-3}{T}^4\right]$, the dimension of $q$ is $\left[AT\right]$, the dimension of $c$ is $\left[L{T}^{-1}\right]$ and the dimension of $a$ is $\left[L{T}^{-2}\right]$.

Substitute the dimensions $\left[A^2{M}^{-1}{L}^{-3}{T}^4\right]$ for ${\epsilon }_{\circ }$, $\left[AT\right]$ for $q$, $\left[L{T}^{-1}\right]$ for $c$ and $\left[L{T}^{-2}\right]$ for $a$ in the equation (1) to find the unit of $p$.

$\begin{array}{c}p=\frac{{\left[AT\right]}^2\times {\left[L{T}^{-2}\right]}^2}{\left[A^2{M}^{-1}{L}^{-3}{T}^4\right]\times {\left[L{T}^{-1}\right]}^3}\\ =\frac{A^2{L}^2{T}^{-2}}{A^2{M}^{-1}{T}^1}\\ =M{L}^2{T}^{-3}\end{array}$

The dimension $\left[M{L}^2{T}^{-3}\right]$ 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.

Answer to Problem 74P

The acceleration of the electron is $1.75\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The equation of the electromagnetic power is $p=\frac{q^2{a}^2}{6\pi {\epsilon }_{\circ }{c}^3}$.

The formula to calculate the acceleration is,

$a=\frac{qE}{m}$ (2)

Here,

$q$ is the charge of the electron.

$E$ is the magnitude of electric field.

$m$ is the mass of the electron.

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, $100\text{V}/\text{m}$ for $E$ and $9.1\times {10}^{-31}\text{kg}$ for $m$ in the above equation to find the value of $a$.

$\begin{array}{c}a=\frac{\left(1.6\times {10}^{-19}\text{C}\right)\left(100\text{V}/\text{m}\right)}{\left(9.1\times {10}^{-31}\text{kg}\right)}\\ =1.75\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Thus, the acceleration of the electron is $1.75\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}$.

Conclusion:

Therefore, the acceleration of the electron is $1.75\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}$.

(c)

To determine

The electromagnetic power radiated by the electron.

Answer to Problem 74P

The electromagnetic power radiated by the electron is $1.73\times {10}^{-24}\text{W}$.

Explanation of Solution

Given info: The equation of the electromagnetic power is $p=\frac{q^2{a}^2}{6\pi {\epsilon }_{\circ }{c}^3}$.

The expression for the electromagnetic power is,

$p=\frac{q^2{a}^2}{6\pi {\epsilon }_{\circ }{c}^3}$

Substitute $8.8\times {10}^{-12}$ for ${\epsilon }_{\circ }$, $1.6\times {10}^{-19}\text{C}$ for $q$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $1.75\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to find the value of $p$.

$\begin{array}{c}p=\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2{\left(1.75\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}\right)}^2}{6\pi \left(8.8\times {10}^{-12}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^3}\\ =1.73\times {10}^{-24}\text{W}\end{array}$

Thus, the electromagnetic power radiated by the electron is $1.73\times {10}^{-24}\text{W}$.

Conclusion:

Therefore, the electromagnetic power radiated by the electron is $1.73\times {10}^{-24}\text{W}$.

(d)

To determine

The electromagnetic power of the proton leaving a cyclotron.

Answer to Problem 74P

The electromagnetic power of the proton leaving a cyclotron is $2.08\times {10}^{-21}\text{W}$.

Explanation of Solution

Given info: The electric flux of the particle is $487{\text{Nm}}^{\text{2}}/\text{C}$. the power is radiated equally in all directions is $25.0\text{W}$.

Given info: The equation of the electromagnetic power is $p=\frac{q^2{a}^2}{6\pi {\epsilon }_{\circ }{c}^3}$.

The formula to calculate the acceleration is,

$a=\frac{q^2{B}^{2}r}{m^2}$ (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 $1.6\times {10}^{-19}\text{C}$ for $q$, $1.6\times {10}^{-27}\text{kg}$ for $m$, $0.500\text{m}$ for $r$ and $0.350\text{T}$ for $B$ in the equation (2) to find the value of $a$.

$\begin{array}{c}a=\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2{\left(0.350\text{T}\right)}^2\left(0.500\text{m}\right)}{{\left(1.6\times {10}^{-27}\text{kg}\right)}^2}\\ =6.1\times {10}^{14}\text{m}/{\text{s}}^{\text{2}}\end{array}$

The expression for the electromagnetic power is,

$p=\frac{q^2{a}^2}{6\pi {\epsilon }_{\circ }{c}^3}$

Substitute $8.8\times {10}^{-12}$ for ${\epsilon }_{\circ }$, $1.6\times {10}^{-19}\text{C}$ for $q$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $6.1\times {10}^{14}\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to find the value of $p$.

$\begin{array}{c}p=\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2{\left(6.1\times {10}^{14}\text{m}/{\text{s}}^{\text{2}}\right)}^2}{6\pi \left(8.8\times {10}^{-12}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^3}\\ =2.08\times {10}^{-21}\text{W}\end{array}$

Conclusion:

Therefore, the electromagnetic power of the proton leaving a cyclotron is $2.08\times {10}^{-21}\text{W}$.