Chapter 19, Problem 52P

(a)

To determine

The direction of the magnetic field of the wire at the position of the proton.

Answer to Problem 52P

The direction of the magnetic field of the wire at the position of the proton is along the negative y-direction.

Explanation of Solution

Given Info: The wire is lying on a horizontal table in the xy-plane. The wire carries a current of $1.20\text{μΑ}$ and the current is flowing in the positive x-direction. The proton is travelling in the negative x-direction. The speed of the proton is $2.30\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}$ and it travels a distance $d$ above the wire.

Explanation:

The direction of the magnetic field can be determined using the right hand rule 2.

Right hand rule 2: when the thumb of the right hand points in the direction of the current, the fingers of the right hand naturally curl in the direction of the magnetic field.

Since the current is flowing in the positive x-direction and the proton is travelling in the negative x-direction at a distance a distance $d$ above the wire, using the right hand rule 2, the direction of the magnetic field at the position of the proton will be along negative y-direction.

Conclusion: The direction of the magnetic field of the wire at the position of the proton is along the negative y-direction.

(b)

To determine

The direction of the magnetic force acting on the proton.

Answer to Problem 52P

The direction of the magnetic force on the proton is along the positive z-direction.

Explanation of Solution

Given Info: The wire is lying on a horizontal table in the xy-plane. The wire carries a current of $1.20\text{μΑ}$ and the current is flowing in the positive x-direction. The proton is travelling in the negative x-direction. The speed of the proton is $2.30\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}$ and it travels a distance $d$ above the wire.

Explanation:

The direction of the magnetic force can be determined using the right hand rule.

Right hand rule 1: When the fingers of the right hand points in the direction of the velocity vector and curls in the direction of the magnetic field moving through the smallest angle, the thumb points in the direction of the magnetic force.

The current is flowing in the positive x-direction and the proton is travelling in the negative x-direction at a distance a distance $d$ above the wire, use the right hand rule 1. Point the fingers of the right and in the negative x-direction and try to curl in the direction of negative y-direction which is the direction of the magnetic field from the result of section (a), the thumb will point in the positive z-direction. Hence the direction of the magnetic force on the proton is along the positive z-direction.

Conclusion: The direction of the magnetic force on the proton is along the positive z-direction.

(c)

To determine

The reason why the direction of the proton’s motion does not change.

Answer to Problem 52P

The direction of the motion of the proton cannot be changed as there is no net force on proton.

Explanation of Solution

Given Info: The wire is lying on a horizontal table in the xy-plane. The wire carries a current of $1.20\text{μΑ}$ and the current is flowing in the positive x-direction. The proton is travelling in the negative x-direction. The speed of the proton is $2.30\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}$ and it travels a distance $d$ above the wire.

Explanation:

For the change of magnitude of velocity or the direction change of any particle, there should be net force acting on the particle. Without the presence of any net force, the direction of any particle or the magnitude of the velocity cannot be changed

The proton travels with a constant velocity. Hence the net force on the proton is zero. So the direction of the proton cannot be changed.

Conclusion: The direction of the motion of the proton cannot be changed as there is no net force on proton.

(d)

To determine

A symbolic expression for $d$ in terms of acceleration due to gravity $g$, the proton mass $m$, the speed of the proton $v$, charge $q$, and the current $I$.

Answer to Problem 52P

The expression for $d$ is $\frac{qv{\mu }_{0}I}{2\pi mg}$.

Explanation of Solution

Given Info: The wire is lying on a horizontal table in the xy-plane. The wire carries a current of $1.20\text{μΑ}$ and the current is flowing in the positive x-direction. The proton is travelling in the negative x-direction. The speed of the proton is $2.30\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}$ and it travels a distance $d$ above the wire.

Explanation:

The z-component of the force on proton is zero. Hence considering upward as the positive z-direction,

$F_m-F_g=0$      (1)

• $F_m$ is the magnetic force on the proton which acts along the positive z-direction
• $F_g$ is the gravitational force which acts downward hence negative z-direction

Since the direction of the velocity of the proton and the direction of the magnetic field are perpendicular to each other the magnetic force on the proton will be,

$F_m=qvB$

• $q$ is the charge of the proton
• $v$ is the velocity of the proton
• $B$ is the magnetic field

Hence from (1),

$qvB-mg=0$

• $m$ is the mass of the proton
• $g$ is the acceleration due to gravity

The magnetic field will be,

$B=\frac{mg}{qv}$      (2)

The magnetic field in terms of the current and distance can be written as,

$B=\frac{{\mu }_{0}I}{2\pi d}$      (3)

• $I$ is the current
• $d$ is the distance

From (1) and (2),

$\frac{mg}{qv}=\frac{{\mu }_{0}I}{2\pi d}$

Hence the distance $d$ will be,

$d=\frac{qv{\mu }_{0}I}{2\pi mg}$

Conclusion: The expression for $d$ is $\frac{qv{\mu }_{0}I}{2\pi mg}$.

(e)

To determine

The numeric answer for the distance $d$.

Answer to Problem 52P

The numeric value for the distance $d$ is $5.40\text{cm}$.

Explanation of Solution

Given Info: The wire is lying on a horizontal table in the xy-plane. The wire carries a current of $1.20\text{μΑ}$ and the current is flowing in the positive x-direction. The proton is travelling in the negative x-direction. The speed of the proton is $2.30\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}$ and it travels a distance $d$ above the wire.

Explanation:

From the result of the section (d) the expression for the distance $d$ is,

$d=\frac{qv{\mu }_{0}I}{2\pi mg}$

Substitute $1.60\times {10}^{-19}\text{C}$ for $q$, $2.30\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}$ for $v$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}$ for ${\mu }_0$, $1.20\text{μΑ}$ for $I$, $1.67\times {10}^{-27}\text{kg}$ for $m$ and $9.80\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$ to determine the distance $d$,

$\begin{array}{c}d=\frac{\left(1.60\times {10}^{-19}\text{C}\right)\left(2.30\times {10}^4\text{m}\cdot {\text{s}}^{\text{-1}}\right)\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(1.20\times {10}^{-6}\text{Α}\right)}{2\pi \left(1.67\times {10}^{-27}\text{kg}\right)\left(9.80\text{m}\cdot {\text{s}}^{\text{-2}}\right)}\\ =5.40\text{cm}\end{array}$

Conclusion: The numeric value for the distance $d$ is $5.40\text{cm}$.