Chapter 19, Problem 94P

(a)

To determine

The sign of the charge of the ions.

Answer to Problem 94P

The sign of the charge of the ions is $\text{positive}$ .

Explanation of Solution

The right hand rule states that in the case of a cross product, if the index finger of the right hand points in the direction of the first vector and the second vector points in the direction of the thumb, then the cross product will point in the direction of the thumb. Magnetic Lorentz force on a moving charge is the product of the charge and the cross product of the velocity of the charge and the magnetic field.

Right hand rule for the Lorentz force can be described as if the index finger of the right hand points in the direction of velocity of charge, the middle finger in the direction of the magnetic field, then the thumb will point in the direction of the magnetic force. In the given diagram, the velocity is directed to right, the magnetic field is directed into the page and the force is directed upward. This implies the ion is positive.

(b)

To determine

The accelerating plate which is positively charged.

Answer to Problem 94P

The positively charged accelerating plate is $\text{west plate}$ .

Explanation of Solution

In part (a), it is found that the ions are positively charged. Positive ions will be attracted by negatively charged plates and will be repelled by positively charged plates. In the given diagram, the ions are accelerated towards the right.

Since the ions are accelerated towards right, the east plate must be negatively charged. This implies the other accelerating plate must be positively charged. Thus the west plate is positively charged.

(c)

To determine

The deflection plate which is positively charged.

Answer to Problem 94P

The positively charged deflection plate is $\text{north plate}$ .

Explanation of Solution

Right hand rule for the Lorentz force can be described as if the index finger of the right hand points in the direction of velocity of charge, the middle finger in the direction of the magnetic field, then the thumb will point in the direction of the magnetic force. In a mass spectrometer, the correct velocity is selected by making the net force which is the sum of electric and magnetic force on the charge to be zero.

According to the right hand rule, the magnetic force on the positively charged ions is northward between the deflecting plates. To make the net force zero to select the correct velocity, the electric field must be southward. Since electric field points from positive plate to the negative plate, the north plate must be positively charged.

(d)

To determine

The correct values of $\Delta V_1$ and $\Delta V_2$ in order to count ¹²C⁺ ions.

Answer to Problem 94P

The value of $\Delta V_1$ to count ¹²C⁺ ions is $1.6\text{ kV}$ and that of $\Delta V_2$ is $320\text{ V}$ .

Explanation of Solution

Write the equation for the magnetic Lorentz force.

$\begin{array}{c}\overrightarrow{F}=q\left(\overrightarrow{v}\times B\right)\\ =qvB\sin \theta \end{array}$

Here, $\overrightarrow{F}$ is the magnetic Lorentz force, $q$ is the charge of the ion, $v$ is the velocity of the ion, $B$ is the magnetic field and $\theta$ is the angle between velocity and the magnetic field

The angle between the magnetic field and the velocity is $90°$ .

Substitute $90°$ for $\theta$ in the above equation.

$\begin{array}{c}\overrightarrow{F}=qvB\sin 90°\\ =qvB\end{array}$ (I)

The magnetic Lorentz force provides the centripetal acceleration required for the semi-circular motion of the ions.

Write the equation for the centripetal force.

${\overrightarrow{F}}_c=\frac{m{v}^2}{r}$ (II)

Here, ${\overrightarrow{F}}_c$ is the centripetal force, $m$ is the mass of the ion and $r$ is the radius of the path

Equate equations (I) and (II) and rewrite the equation for $v$ .

$\begin{array}{c}qvB=\frac{m{v}^2}{r}\\ v=\frac{qrB}{m}\end{array}$ (III)

Write the equation for a velocity selector.

$E=vB$

Here, $E$ is the magnitude of the electric field

Put equation (III) in the above equation.

$\begin{array}{c}E=\left(\frac{qrB}{m}\right)B\\ =\frac{qr{B}^2}{m}\end{array}$ (IV)

Write the equation for the potential difference across the deflecting plates.

$\Delta V_2=Ed$

Here, $\Delta V_2$ is the potential difference across the deflecting plates and $d$ is the separation between the plates

Put equation (IV) in the above equation.

$\begin{array}{c}\Delta V_2=\left(\frac{qr{B}^2}{m}\right)d\\ =\frac{qr{B}^{2}d}{m}\end{array}$ (V)

Apply the principle of conservation of energy for the given situation.

$\frac{1}{2}m{v}^2=q\Delta V_1$

Here, $\Delta V_1$ is the potential difference between the charged accelerating plates

Rewrite the above equation for $\Delta V_1$ .

$\Delta V_1=\frac{m{v}^2}{2q}$

Put equation (III) in the above equation.

$\begin{array}{c}\Delta V_1=\frac{m}{2q}{\left(\frac{qrB}{m}\right)}^2\\ =\frac{q{r}^2{B}^2}{2m}\end{array}$ (VI)

Conclusion:

Given that the radius of the path is $0.10\text{ m}$ , the value of the magnetic field is $0.20\text{ T}$ and the separation between the deflection plates is $1.00\text{ cm}$ . The charge of the ion is $1.60\times {10}^{-19}\text{ C}$ and the mass of the ¹²C⁺ ions is $1.993\times {10}^{-26}\text{ kg}$ .

Substitute $1.60\times {10}^{-19}\text{ C}$ for $q$ , $0.10\text{ m}$ for $r$ , $0.20\text{ T}$ for $B$ , $1.00\text{ cm}$ for $d$ and $1.993\times {10}^{-26}\text{ kg}$ for $m$ in equation (V) to find $\Delta V_2$ .

$\begin{array}{c}\Delta V_2=\frac{\left(1.60\times {10}^{-19}\text{ C}\right)\left(0.10\text{ m}\right){\left(0.20\text{ T}\right)}^2\left(1.00\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}{1.993\times {10}^{-26}\text{ kg}}\\ =\frac{\left(1.60\times {10}^{-19}\text{ C}\right)\left(0.10\text{ m}\right){\left(0.20\text{ T}\right)}^2\left(0.010\text{ m}\right)}{1.993\times {10}^{-26}\text{ kg}}\\ =320\text{ V}\end{array}$

Substitute $1.60\times {10}^{-19}\text{ C}$ for $q$ , $0.10\text{ m}$ for $r$ , $0.20\text{ T}$ for $B$ and $1.993\times {10}^{-26}\text{ kg}$ for $m$ in equation (VI) to find $\Delta V_1$ .

$\begin{array}{c}\Delta V_1=\frac{\left(1.60\times {10}^{-19}\text{ C}\right){\left(0.10\text{ m}\right)}^2{\left(0.20\text{ T}\right)}^2}{2\left(1.993\times {10}^{-26}\text{ kg}\right)}\\ =\frac{\left(1.60\times {10}^{-19}\text{ C}\right){\left(0.10\text{ m}\right)}^2{\left(0.20\text{ T}\right)}^2}{2\left(1.993\times {10}^{-26}\text{ kg}\right)}\\ =1.6\times {10}^3\text{ V}\\ =1.6\text{ kV}\end{array}$

Therefore, the value of $\Delta V_1$ to count ¹²C⁺ ions is $1.6\text{ kV}$ and that of $\Delta V_2$ is $320\text{ V}$ .

(e)

To determine

The correct values of $\Delta V_1$ and $\Delta V_2$ in order to count ¹⁴C⁺ ions.

Answer to Problem 94P

The value of $\Delta V_1$ to count ¹⁴C⁺ ions is $1.4\text{ kV}$ and that of $\Delta V_2$ is $280\text{ V}$ .

Explanation of Solution

Equation (VI) and (V) respectively can be used to determine $\Delta V_1$ and $\Delta V_2$ with the mass of ¹⁴C⁺ ions.

Conclusion:

Given that the mass of the ¹⁴C⁺ ions is $2.325\times {10}^{-26}\text{ kg}$ .

Substitute $1.60\times {10}^{-19}\text{ C}$ for $q$ , $0.10\text{ m}$ for $r$ , $0.20\text{ T}$ for $B$ and $2.325\times {10}^{-26}\text{ kg}$ for $m$ in equation (VI) to find $\Delta V_1$ .

$\begin{array}{c}\Delta V_1=\frac{\left(1.60\times {10}^{-19}\text{ C}\right){\left(0.10\text{ m}\right)}^2{\left(0.20\text{ T}\right)}^2}{2\left(2.325\times {10}^{-26}\text{ kg}\right)}\\ =\frac{\left(1.60\times {10}^{-19}\text{ C}\right){\left(0.10\text{ m}\right)}^2{\left(0.20\text{ T}\right)}^2}{2\left(2.325\times {10}^{-26}\text{ kg}\right)}\\ =1.4\times {10}^3\text{ V}\\ =1.4\text{ kV}\end{array}$

Substitute $1.60\times {10}^{-19}\text{ C}$ for $q$ , $0.10\text{ m}$ for $r$ , $0.20\text{ T}$ for $B$ , $1.00\text{ cm}$ for $d$ and $2.325\times {10}^{-26}\text{ kg}$ for $m$ in equation (V) to find $\Delta V_2$ .

$\begin{array}{c}\Delta V_2=\frac{\left(1.60\times {10}^{-19}\text{ C}\right)\left(0.10\text{ m}\right){\left(0.20\text{ T}\right)}^2\left(1.00\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}{2.325\times {10}^{-26}\text{ kg}}\\ =\frac{\left(1.60\times {10}^{-19}\text{ C}\right)\left(0.10\text{ m}\right){\left(0.20\text{ T}\right)}^2\left(0.010\text{ m}\right)}{2.325\times {10}^{-26}\text{ kg}}\\ =280\text{ V}\end{array}$

Therefore, the value of $\Delta V_1$ to count ¹⁴C⁺ ions is $1.4\text{ kV}$ and that of $\Delta V_2$ is $280\text{ V}$ .