Chapter 32, Problem 10P

To determine

(a) To Determine:

The magnetic field (B).

Answer to Problem 10P

Solution:

The magnetic field (B)is $\text{0}\text{.708}\text{T}$

Explanation of Solution

Given Info:

The radius of cyclotron is 1.0m.

The energy is 12 MeV or $12\times {10}^6\text{eV}$.

Formula Used:

Cyclotron is based on the principle that a charged particle can be accelerated to a very high energy with a small alternate electric field and with the help of a magnetic field.

Magnetic force= centripetal force

$\text{qvB=}\frac{{\text{mv}}^{\text{2}}}{\text{r}}$

$\text{v=}\frac{\text{qBr}}{\text{m}}$ (I)

• ‘q’ is the charge of deuteron

• ‘B’ is magnetic field

• ‘v’ is velocity of particle

• ’m’ is mass of the particle

• ‘r’ is the radius of particle

The kinetic energy of the charge particle is,

$\text{E=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

Substitute ‘v’ from equation (I)in the kinetic energy

$\begin{array}{l}\text{E=}\frac{\text{1}}{\text{2}}\text{m×}{\left(\frac{\text{qBr}}{\text{m}}\right)}^{\text{2}}\\ \text{E=}\frac{{\text{q}}^{\text{2}}{\text{B}}^{\text{2}}\text{r}}{\text{2m}}\end{array}$

The formula is used to find magnetic field,

$\text{B=}\sqrt{\frac{\text{2mE}}{{\text{q}}^{\text{2}}\text{r}}}$

Calculation:

Substitute for ‘E’ and 1.0m for ‘r’ to find ‘B’

$\text{B=}\sqrt{\frac{\text{2×(2×1}{\text{.67×10}}^{\text{-27}}\text{kg)×}\left({\text{12×10}}^{\text{6}}\text{eV}\right)\text{×}\left(\text{1}{\text{.6×10}}^{\text{-19}}\text{J/eV}\right)}{{\text{(1}{\text{.6×10}}^{\text{-19}}\text{C)}}^{\text{2}}\text{×1}\text{.0}}}$

$\text{B=0}\text{.708}\text{T}$ (II)

Thus, the magnetic field is $\text{0}\text{.708}\text{T}$

To determine

(b) To Determine:

The frequency (f).

Answer to Problem 10P

Solution:

The frequency (f)is $\text{5}{\text{.40×10}}^{\text{6}}\text{Hz}$

Explanation of Solution

Given Info:

The radius of cyclotron is 1.0m.

The energy is 12 MeV or.

Formula Used:

The period (T)is defined as,

$\text{T=}\frac{\text{2πr}}{\text{v}}$

$\text{T=}\frac{\text{2πr}}{\left(\frac{\text{qBr}}{\text{m}}\right)}$

$\text{T=}\frac{\text{2πm}}{\text{qB}}$

• ’T’ is period of revolution.

• ‘m’ is the mass of particle.

$\begin{array}{l}\text{f=}\frac{\text{1}}{\text{T}}\\ \text{f=}\frac{\text{qB}}{\text{2πm}}\end{array}$

Calculation:

Substitute 0.708 T for ‘B’ from equation (II)to find ‘f’

$\text{f=}\frac{\text{(1}{\text{.6×10}}^{\text{-19}}\text{C)×}\left(\text{0}\text{.708}\text{T}\right)}{\text{2×π×}\left(\text{2×1}{\text{.67×10}}^{\text{-27}}\text{kg}\right)}$

$\text{f=}\text{5}{\text{.40×10}}^{\text{6}}\text{Hz}$ (III)

Thus, the frequency is $\text{5}{\text{.40×10}}^{\text{6}}\text{Hz}$.

To determine

(c) To Determine:

The number of revolution (n)

Answer to Problem 10P

Solution:

The number of revolution is $\text{273}\text{rev}$.

Explanation of Solution

Given Info:

The radius of cyclotron is 1.0m.

The energy is 12 MeV or.

The potential difference is 22V or ${\text{22×10}}^{\text{3}}\text{V}$

Formula Used:

$\begin{array}{l}\text{E=2qVn}\\ \text{n=}\frac{\text{E}}{\text{2qV}}\end{array}$

• ’n’ is number of revolution.

• ‘V’ is the potential difference.

Calculation:

Substitute for ‘E’ and ${\text{22×10}}^{\text{3}}\text{V}$ for ‘V’ to find ‘n’,

$\text{n=}\frac{\left({\text{12×10}}^{\text{6}}\text{eV}\right)\text{×}\left(\text{1}{\text{.6×10}}^{\text{-19}}\text{J/eV}\right)}{\text{2×}\left(\text{1}{\text{.6×10}}^{\text{-19}}\text{C}\right)\text{×}\left({\text{22×10}}^{\text{3}}\text{V}\right)}$

$\text{n=273}\text{rev}$ (IV)

Thus, the number of revolution is $\text{273}\text{rev}$.

To determine

(d) To Determine:

The time taken to exit (t).

Answer to Problem 10P

Solution:

The time taken to exit (t)is $\text{5}{\text{.05×10}}^{\text{-5}}\text{sec}$.

Explanation of Solution

Given Info:

The radius of cyclotron is 1.0m.

The energy is 12 MeV or.

The potential difference is 22V or ${\text{22×10}}^{\text{3}}\text{V}$

Formula Used:

$\begin{array}{l}\text{f=}\frac{\text{n}}{\text{t}}\\ \text{t=}\frac{\text{n}}{\text{f}}\end{array}$

• ’t’ is time to exit.

• ‘n’ is number of revolution.

• ‘f’ s frequency of revolution

Calculation:

Substitute equation (III0 and (IV)to find ‘t’,

$\text{t =}\frac{\text{273}\text{rev}}{\left(\text{5}{\text{.40×10}}^{\text{6}}\text{Hz}\right)}$.

$\text{t =5}{\text{.05×10}}^{\text{-5}}\text{s}$ (V)

Thus, the time it takes to exit is $\text{5}{\text{.05×10}}^{\text{-5}}\text{s}$

To determine

(e) To Determine:

The distance (d)travel during time ’t’.

Answer to Problem 10P

Solution:

The distance travel (d)during time ‘t’ is $\text{857m}$.

Explanation of Solution

Given Info:

The radius of cyclotron is 1.0m.

The energy is 12 MeV or.

The potential difference is 22V or ${\text{22×10}}^{\text{3}}\text{V}$

Formula Used:

The formula used to find distance is,

$\begin{array}{l}\text{d=}\frac{\text{1}}{\text{2}}\text{×2πrn}\\ \text{d=πrn}\end{array}$

• ’d’ is distance.

• ‘n’ is number of revolution.

• ‘r’ is radius.

Calculation:

Substitute 273 rev for ‘n’ and 1.0 m for ‘’ to find ‘d’,

$\text{d=π×}\left(\text{1}\text{.0m}\right)\text{×}\left(\text{273rev}\right)$

$\text{d=857m}$ (VI):

Thus, distance travel during time ‘t’ is $\text{857m}$