The sign of the charge of the ions.

Question

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Answer 1

Question

Chapter 19, Problem 94P

(a)

To determine

The sign of the charge of the ions.

(a)

Expert Solution

Answer to Problem 94P

The sign of the charge of the ions is $positive$ .

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.

(b)

Expert Solution

Answer to Problem 94P

The positively charged accelerating plate is $west plate$ .

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.

(c)

Expert Solution

Answer to Problem 94P

The positively charged deflection plate is $north plate$ .

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 $ΔV1$ and $ΔV2$ in order to count ¹²C⁺ ions.

(d)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Explanation of Solution

Write the equation for the magnetic Lorentz force.

$F→=q(v→×B)=qvBsinθ$

Here, $F→$ 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 $θ$ is the angle between velocity and the magnetic field

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

Substitute $90°$ for $θ$ in the above equation.

$F→=qvBsin90°=qvB$ (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.

$F→c=mv2r$ (II)

Here, $F→c$ 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 $v$ .

$qvB=mv2rv=qrBm$ (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.

$E=(qrBm)B=qrB2m$ (IV)

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

$ΔV2=Ed$

Here, $ΔV2$ is the potential difference across the deflecting plates and $d$ is the separation between the plates

Put equation (IV) in the above equation.

$ΔV2=(qrB2m)d=qrB2dm$ (V)

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

$12mv2=qΔV1$

Here, $ΔV1$ is the potential difference between the charged accelerating plates

Rewrite the above equation for $ΔV1$ .

$ΔV1=mv22q$

Put equation (III) in the above equation.

$ΔV1=m2q(qrBm)2=qr2B22m$ (VI)

Conclusion:

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

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $1.993×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))1.993×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)1.993×10−26 kg=320 V$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $1.993×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=1.6×103 V=1.6 kV$

Therefore, the value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

(e)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹⁴C⁺ ions.

(e)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Explanation of Solution

Equation (VI) and (V) respectively can be used to determine $ΔV1$ and $ΔV2$ with the mass of ¹⁴C⁺ ions.

Conclusion:

Given that the mass of the ¹⁴C⁺ ions is $2.325×10−26 kg$ .

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $2.325×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=1.4×103 V=1.4 kV$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $2.325×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))2.325×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)2.325×10−26 kg=280 V$

Therefore, the value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

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Answer 2

Question

Chapter 19, Problem 94P

(a)

To determine

The sign of the charge of the ions.

(a)

Expert Solution

Answer to Problem 94P

The sign of the charge of the ions is $positive$ .

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.

(b)

Expert Solution

Answer to Problem 94P

The positively charged accelerating plate is $west plate$ .

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.

(c)

Expert Solution

Answer to Problem 94P

The positively charged deflection plate is $north plate$ .

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 $ΔV1$ and $ΔV2$ in order to count ¹²C⁺ ions.

(d)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Explanation of Solution

Write the equation for the magnetic Lorentz force.

$F→=q(v→×B)=qvBsinθ$

Here, $F→$ 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 $θ$ is the angle between velocity and the magnetic field

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

Substitute $90°$ for $θ$ in the above equation.

$F→=qvBsin90°=qvB$ (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.

$F→c=mv2r$ (II)

Here, $F→c$ 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 $v$ .

$qvB=mv2rv=qrBm$ (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.

$E=(qrBm)B=qrB2m$ (IV)

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

$ΔV2=Ed$

Here, $ΔV2$ is the potential difference across the deflecting plates and $d$ is the separation between the plates

Put equation (IV) in the above equation.

$ΔV2=(qrB2m)d=qrB2dm$ (V)

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

$12mv2=qΔV1$

Here, $ΔV1$ is the potential difference between the charged accelerating plates

Rewrite the above equation for $ΔV1$ .

$ΔV1=mv22q$

Put equation (III) in the above equation.

$ΔV1=m2q(qrBm)2=qr2B22m$ (VI)

Conclusion:

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

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $1.993×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))1.993×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)1.993×10−26 kg=320 V$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $1.993×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=1.6×103 V=1.6 kV$

Therefore, the value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

(e)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹⁴C⁺ ions.

(e)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Explanation of Solution

Equation (VI) and (V) respectively can be used to determine $ΔV1$ and $ΔV2$ with the mass of ¹⁴C⁺ ions.

Conclusion:

Given that the mass of the ¹⁴C⁺ ions is $2.325×10−26 kg$ .

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $2.325×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=1.4×103 V=1.4 kV$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $2.325×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))2.325×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)2.325×10−26 kg=280 V$

Therefore, the value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

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Answer 3

Question

Chapter 19, Problem 94P

(a)

To determine

The sign of the charge of the ions.

(a)

Expert Solution

Answer to Problem 94P

The sign of the charge of the ions is $positive$ .

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.

(b)

Expert Solution

Answer to Problem 94P

The positively charged accelerating plate is $west plate$ .

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.

(c)

Expert Solution

Answer to Problem 94P

The positively charged deflection plate is $north plate$ .

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 $ΔV1$ and $ΔV2$ in order to count ¹²C⁺ ions.

(d)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Explanation of Solution

Write the equation for the magnetic Lorentz force.

$F→=q(v→×B)=qvBsinθ$

Here, $F→$ 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 $θ$ is the angle between velocity and the magnetic field

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

Substitute $90°$ for $θ$ in the above equation.

$F→=qvBsin90°=qvB$ (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.

$F→c=mv2r$ (II)

Here, $F→c$ 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 $v$ .

$qvB=mv2rv=qrBm$ (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.

$E=(qrBm)B=qrB2m$ (IV)

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

$ΔV2=Ed$

Here, $ΔV2$ is the potential difference across the deflecting plates and $d$ is the separation between the plates

Put equation (IV) in the above equation.

$ΔV2=(qrB2m)d=qrB2dm$ (V)

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

$12mv2=qΔV1$

Here, $ΔV1$ is the potential difference between the charged accelerating plates

Rewrite the above equation for $ΔV1$ .

$ΔV1=mv22q$

Put equation (III) in the above equation.

$ΔV1=m2q(qrBm)2=qr2B22m$ (VI)

Conclusion:

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

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $1.993×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))1.993×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)1.993×10−26 kg=320 V$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $1.993×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=1.6×103 V=1.6 kV$

Therefore, the value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

(e)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹⁴C⁺ ions.

(e)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Explanation of Solution

Equation (VI) and (V) respectively can be used to determine $ΔV1$ and $ΔV2$ with the mass of ¹⁴C⁺ ions.

Conclusion:

Given that the mass of the ¹⁴C⁺ ions is $2.325×10−26 kg$ .

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $2.325×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=1.4×103 V=1.4 kV$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $2.325×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))2.325×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)2.325×10−26 kg=280 V$

Therefore, the value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

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Answer 4

Question

Chapter 19, Problem 94P

(a)

To determine

The sign of the charge of the ions.

(a)

Expert Solution

Answer to Problem 94P

The sign of the charge of the ions is $positive$ .

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.

(b)

Expert Solution

Answer to Problem 94P

The positively charged accelerating plate is $west plate$ .

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.

(c)

Expert Solution

Answer to Problem 94P

The positively charged deflection plate is $north plate$ .

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 $ΔV1$ and $ΔV2$ in order to count ¹²C⁺ ions.

(d)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Explanation of Solution

Write the equation for the magnetic Lorentz force.

$F→=q(v→×B)=qvBsinθ$

Here, $F→$ 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 $θ$ is the angle between velocity and the magnetic field

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

Substitute $90°$ for $θ$ in the above equation.

$F→=qvBsin90°=qvB$ (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.

$F→c=mv2r$ (II)

Here, $F→c$ 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 $v$ .

$qvB=mv2rv=qrBm$ (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.

$E=(qrBm)B=qrB2m$ (IV)

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

$ΔV2=Ed$

Here, $ΔV2$ is the potential difference across the deflecting plates and $d$ is the separation between the plates

Put equation (IV) in the above equation.

$ΔV2=(qrB2m)d=qrB2dm$ (V)

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

$12mv2=qΔV1$

Here, $ΔV1$ is the potential difference between the charged accelerating plates

Rewrite the above equation for $ΔV1$ .

$ΔV1=mv22q$

Put equation (III) in the above equation.

$ΔV1=m2q(qrBm)2=qr2B22m$ (VI)

Conclusion:

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

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $1.993×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))1.993×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)1.993×10−26 kg=320 V$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $1.993×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=1.6×103 V=1.6 kV$

Therefore, the value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

(e)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹⁴C⁺ ions.

(e)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Explanation of Solution

Equation (VI) and (V) respectively can be used to determine $ΔV1$ and $ΔV2$ with the mass of ¹⁴C⁺ ions.

Conclusion:

Given that the mass of the ¹⁴C⁺ ions is $2.325×10−26 kg$ .

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $2.325×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=1.4×103 V=1.4 kV$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $2.325×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))2.325×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)2.325×10−26 kg=280 V$

Therefore, the value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

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Answer 5

Chapter 19, Problem 94P

(a)

To determine

The sign of the charge of the ions.

(a)

Expert Solution

Answer to Problem 94P

The sign of the charge of the ions is $positive$ .

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.

(b)

Expert Solution

Answer to Problem 94P

The positively charged accelerating plate is $west plate$ .

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.

(c)

Expert Solution

Answer to Problem 94P

The positively charged deflection plate is $north plate$ .

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 $ΔV1$ and $ΔV2$ in order to count ¹²C⁺ ions.

(d)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Explanation of Solution

Write the equation for the magnetic Lorentz force.

$F→=q(v→×B)=qvBsinθ$

Here, $F→$ 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 $θ$ is the angle between velocity and the magnetic field

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

Substitute $90°$ for $θ$ in the above equation.

$F→=qvBsin90°=qvB$ (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.

$F→c=mv2r$ (II)

Here, $F→c$ 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 $v$ .

$qvB=mv2rv=qrBm$ (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.

$E=(qrBm)B=qrB2m$ (IV)

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

$ΔV2=Ed$

Here, $ΔV2$ is the potential difference across the deflecting plates and $d$ is the separation between the plates

Put equation (IV) in the above equation.

$ΔV2=(qrB2m)d=qrB2dm$ (V)

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

$12mv2=qΔV1$

Here, $ΔV1$ is the potential difference between the charged accelerating plates

Rewrite the above equation for $ΔV1$ .

$ΔV1=mv22q$

Put equation (III) in the above equation.

$ΔV1=m2q(qrBm)2=qr2B22m$ (VI)

Conclusion:

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

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $1.993×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))1.993×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)1.993×10−26 kg=320 V$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $1.993×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=1.6×103 V=1.6 kV$

Therefore, the value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

(e)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹⁴C⁺ ions.

(e)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Explanation of Solution

Equation (VI) and (V) respectively can be used to determine $ΔV1$ and $ΔV2$ with the mass of ¹⁴C⁺ ions.

Conclusion:

Given that the mass of the ¹⁴C⁺ ions is $2.325×10−26 kg$ .

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $2.325×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=1.4×103 V=1.4 kV$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $2.325×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))2.325×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)2.325×10−26 kg=280 V$

Therefore, the value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Answer 6

Chapter 19, Problem 94P

(a)

To determine

The sign of the charge of the ions.

(a)

Expert Solution

Answer to Problem 94P

The sign of the charge of the ions is $positive$ .

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.

Answer 7

Chapter 19, Problem 94P

(a)

To determine

The sign of the charge of the ions.

Answer 8

(a)

Expert Solution

Answer to Problem 94P

The sign of the charge of the ions is $positive$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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.

Answer 13

(b)

To determine

The accelerating plate which is positively charged.

(b)

Expert Solution

Answer to Problem 94P

The positively charged accelerating plate is $west plate$ .

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.

Answer 14

(b)

To determine

The accelerating plate which is positively charged.

Answer 15

(b)

Expert Solution

Answer to Problem 94P

The positively charged accelerating plate is $west plate$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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.

Answer 20

(c)

To determine

The deflection plate which is positively charged.

(c)

Expert Solution

Answer to Problem 94P

The positively charged deflection plate is $north plate$ .

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.

Answer 21

(c)

To determine

The deflection plate which is positively charged.

Answer 22

(c)

Expert Solution

Answer to Problem 94P

The positively charged deflection plate is $north plate$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

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.

Answer 27

(d)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹²C⁺ ions.

(d)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Explanation of Solution

Write the equation for the magnetic Lorentz force.

$F→=q(v→×B)=qvBsinθ$

Here, $F→$ 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 $θ$ is the angle between velocity and the magnetic field

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

Substitute $90°$ for $θ$ in the above equation.

$F→=qvBsin90°=qvB$ (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.

$F→c=mv2r$ (II)

Here, $F→c$ 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 $v$ .

$qvB=mv2rv=qrBm$ (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.

$E=(qrBm)B=qrB2m$ (IV)

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

$ΔV2=Ed$

Here, $ΔV2$ is the potential difference across the deflecting plates and $d$ is the separation between the plates

Put equation (IV) in the above equation.

$ΔV2=(qrB2m)d=qrB2dm$ (V)

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

$12mv2=qΔV1$

Here, $ΔV1$ is the potential difference between the charged accelerating plates

Rewrite the above equation for $ΔV1$ .

$ΔV1=mv22q$

Put equation (III) in the above equation.

$ΔV1=m2q(qrBm)2=qr2B22m$ (VI)

Conclusion:

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

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $1.993×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))1.993×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)1.993×10−26 kg=320 V$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $1.993×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=1.6×103 V=1.6 kV$

Therefore, the value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Answer 28

(d)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹²C⁺ ions.

Answer 29

(d)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the equation for the magnetic Lorentz force.

$F→=q(v→×B)=qvBsinθ$

Here, $F→$ 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 $θ$ is the angle between velocity and the magnetic field

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

Substitute $90°$ for $θ$ in the above equation.

$F→=qvBsin90°=qvB$ (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.

$F→c=mv2r$ (II)

Here, $F→c$ 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 $v$ .

$qvB=mv2rv=qrBm$ (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.

$E=(qrBm)B=qrB2m$ (IV)

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

$ΔV2=Ed$

Here, $ΔV2$ is the potential difference across the deflecting plates and $d$ is the separation between the plates

Put equation (IV) in the above equation.

$ΔV2=(qrB2m)d=qrB2dm$ (V)

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

$12mv2=qΔV1$

Here, $ΔV1$ is the potential difference between the charged accelerating plates

Rewrite the above equation for $ΔV1$ .

$ΔV1=mv22q$

Put equation (III) in the above equation.

$ΔV1=m2q(qrBm)2=qr2B22m$ (VI)

Conclusion:

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

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $1.993×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))1.993×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)1.993×10−26 kg=320 V$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $1.993×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(1.993×10−26 kg)=1.6×103 V=1.6 kV$

Therefore, the value of $ΔV1$ to count ¹²C⁺ ions is $1.6 kV$ and that of $ΔV2$ is $320 V$ .

Answer 34

(e)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹⁴C⁺ ions.

(e)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Explanation of Solution

Equation (VI) and (V) respectively can be used to determine $ΔV1$ and $ΔV2$ with the mass of ¹⁴C⁺ ions.

Conclusion:

Given that the mass of the ¹⁴C⁺ ions is $2.325×10−26 kg$ .

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $2.325×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=1.4×103 V=1.4 kV$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $2.325×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .

$ΔV2=(1.60×10−19 C)(0.10 m)(0.20 T)2(1.00 cm(1 m100 cm))2.325×10−26 kg=(1.60×10−19 C)(0.10 m)(0.20 T)2(0.010 m)2.325×10−26 kg=280 V$

Therefore, the value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Answer 35

(e)

To determine

The correct values of $ΔV1$ and $ΔV2$ in order to count ¹⁴C⁺ ions.

Answer 36

(e)

Expert Solution

Answer to Problem 94P

The value of $ΔV1$ to count ¹⁴C⁺ ions is $1.4 kV$ and that of $ΔV2$ is $280 V$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Equation (VI) and (V) respectively can be used to determine $ΔV1$ and $ΔV2$ with the mass of ¹⁴C⁺ ions.

Conclusion:

Given that the mass of the ¹⁴C⁺ ions is $2.325×10−26 kg$ .

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ and $2.325×10−26 kg$ for $m$ in equation (VI) to find $ΔV1$ .

$ΔV1=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=(1.60×10−19 C)(0.10 m)2(0.20 T)22(2.325×10−26 kg)=1.4×103 V=1.4 kV$

Substitute $1.60×10−19 C$ for $q$ , $0.10 m$ for $r$ , $0.20 T$ for $B$ , $1.00 cm$ for $d$ and $2.325×10−26 kg$ for $m$ in equation (V) to find $ΔV2$ .