Chapter 20, Problem 77GP

(a)

To determine

To Show: The frequency of the voltage is $\frac{Bq}{2\pi m}$ .

Answer to Problem 77GP

  $f=\frac{qB}{2\pi m}$

Explanation of Solution

Given info:

Magnetic field $=B$

Electric field $=E$

Voltage, $V=V_o\sin \left(2\pi ft\right)$

Formula used:

Lorentz force:

  $\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)+q\overrightarrow{E}$

Here, q is the charge.

Centripetal force:

  $F_c=\frac{m{v}^2}{r}$

Here, m is the mass, v is the velocity and r is the radius.

Calculation:

Lorentz force would balance the centripetal force in order to cause the particle to move in a circular orbit.

Therefore,

  $\frac{m{v}^2}{r}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)+q\overrightarrow{E}$

There is no electric field inside the Dees, therefore, $E=0$

  $\begin{array}{c}\frac{m{v}^2}{r}=qvB\sin {90}^o=qvB\\ \frac{mv}{r}=qB\end{array}$

Radius of the orbital:

  $r=\frac{mv}{qB}$

Orbital speed of the particle:

  $v=\frac{qBr}{m}$

Frequency of the voltage:

  $\begin{array}{c}f=\frac{1}{T}\\ =\frac{1}{\frac{2\pi r}{v}}\\ =\frac{v}{2\pi r}\end{array}$

Plugging in the value of $r$ :

  $\begin{array}{c}f=\frac{v}{2\pi \times \frac{mv}{qB}}\\ =\frac{qB}{2\pi m}\end{array}$

Conclusion:

Hence, frequency of the voltage, $f=\frac{qB}{2\pi m}$

(b)

To determine

To Show: Kinetics energy of the particle increases by $2q{V}_o$ .

Answer to Problem 77GP

  $\Delta K.E=2q{V}_o$

Explanation of Solution

Given info:

Magnetic field $=B$

Electric field $=E$

Voltage, $V=V_o\sin \left(2\pi ft\right)$

Initial speed of the particle, $v_o=0$

Formula used:

Formula of kinetics energy:

  $K.E=\frac{1}{2}m{v}^2$

Calculation:

Let, the speed of the particle be $v$ , when it passes through electric field and number of revolutions made by it be $n$ .

Kinetics energy build up from $2n$ passes through $V$ .

Increased in kinetics energy:

  $\begin{array}{c}\Delta K.E=\frac{1}{2}m{v}_o{}^2+2nqV\\ =0+2qV\left(n=1\text{ For each revolution}\text{.}\right)\\ =2q{V}_o\sin \left(2\pi ft\right)\\ \Delta K.E_{\max }=2q{V}_o\because \left(\sin \left(2\pi ft\right)=1\right)\end{array}$

Conclusion:

Hence, increase in kinetic energy: $2q{V}_o$

(c)

To determine

To Compute: Maximum kinetics energy of the proton.

Answer to Problem 77GP

  $K.E_{\max }=7.66\times {10}^{-12}\text{ J}$

Explanation of Solution

Given:

Radius of the cyclotron, $r=2.0\text{ m}$

Magnetic field strength, $B=0.50\text{ T}$

Charge on proton, $q=1.6\times {10}^{-19}\text{ C}$

Mass of the proton, $m=1.67\times {10}^{-27}\text{ kg}$

Formula used:

Velocity can be obtained using:

  $v=\frac{rqB}{m}$

Calculation:

Maximum kinetic energy:

  $\begin{array}{c}K.E_{\max }=\frac{1}{2}m{v}^2\\ =\frac{1}{2}\times m\times {\left(\frac{rqB}{m}\right)}^2\\ =\frac{1}{2}\frac{r^2{q}^2{B}^2}{m}\end{array}$

Plugging in the give values:

  $\begin{array}{c}K.E_{\max }=\frac{1}{2}\times \frac{{\left(2.0\times 1.6\times {10}^{-19}\times 0.5\right)}^2}{1.67\times {10}^{-27}}\\ =7.66\times {10}^{-12}\text{ J}\end{array}$

Conclusion:

Maximum kinetics energy: $7.66\times {10}^{-12}\text{ J}$