Chapter 22, Problem 76P

Review. Rail guns have been suggested for launching projectiles into space without chemical rockets. A tabletop model rail gun (Fig. P22.76) consists of two long, parallel, horizontal rails ℓ = 3.50 cm apart, bridged by a bar of mass m = 3.00 g that is free to slide without friction. The rails and bar have low electric resistance, and the current is limited to a constant I = 24.0 A by a power supply that is far to the left of the figure, so it has no magnetic effect on the bar. Figure P22.76 shows the bar at rest at the midpoint of the rails at the moment the current is established. We wish to find the speed with which the bar leaves the rails after being released from the midpoint of the rails. (a) Find the magnitude of the magnetic field at a distance of 1.75 cm from a single long wire carrying a current of 2.40 A. (b) For purposes of evaluating the magnetic field, model the rails as infinitely long. Using the result of part (a), find the magnitude and direction of the magnetic field at the midpoint of the bar. (c) Argue that this value of the field will be the same at all positions of the bar to the right of the midpoint of the rails. At other points along the bar, the field is in the same direction as at the midpoint, but is larger in magnitude. Assume the average effective magnetic field along the bar is five times larger than the field at the midpoint. With this assumption, find (d) the magnitude and (e) the direction of the force on the bar. (f) Is the bar properly modeled as a particle under constant acceleration? (g) Find the velocity of the bar after it has traveled a distance d = 130 cm to the end of the rails.

Figure P22.76

To determine

 The magnitude of the magnetic field.

Answer to Problem 76P

The magnitude of the magnetic field is $2.74\times {10}^{-4}\text{T}$

Explanation of Solution

Write the expression for the magnetic field for a conductor,

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

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space, $I$ is the current, and $r$ is the length of a conductor.

Conclusion:

Substitute $\text{4}\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$ , $24.0\text{A}$ for $I$ , and $\text{1}\text{.75 m}$ for $r$ in equation (I),

  $\begin{array}{c}B=\frac{\left(\text{4}\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(24.0\text{A}\right)}{2\pi \left(\text{1}\text{.75 m}\times \frac{{\text{10}}^{-2}\text{ m}}{\text{1cm}}\right)}\\ =2.74\times {10}^{-4}\text{T}\end{array}$

The magnitude of the magnetic field is $2.74\times {10}^{-4}\text{T}$

To determine

The magnitude and the direction of the magnetic field from the mid- point of the bar.

Answer to Problem 76P

The magnitude of the field at the mid-point of the bar is $2.74\times {10}^{-4}\text{T}$ and the direction is into the page.

Explanation of Solution

From the figure1 the current is diverted through the bar, here only half of each rails carriers currents, so the field produce by each rail are half of the infinitely long wire produces.

Conclusion:

Write the expression for the magnetic field produced by the conductor $AB$ at the point $C$ ,

    $B_{AB}=\frac{1}{2}B$        (I)

Here, $B_{AB}$ is the magnetic field, and $B$ is the total magnetic field.

Substitute $2.74\times {10}^{-4}\text{T}$ for $B$ in expression (I),

    $\begin{array}{c}{B}_{AB}=\frac{1}{2}\left(2.74\times {10}^{-4}\text{T}\right)\\ =1.37\times {10}^{-4}\text{T}\end{array}$

Write the expression for the magnetic field produced by the conductor $DE$ at the point $C$ ,

    $B_{DE}=\frac{1}{2}B$        (I)

Here, $B_{DE}$ is the magnetic field, and $B$ is the total magnetic field.

Substitute $2.74\times {10}^{-4}\text{T}$ for $B$ in expression (I),

    $\begin{array}{c}{B}_{DE}=\frac{1}{2}\left(2.74\times {10}^{-4}\text{T}\right)\\ =1.37\times {10}^{-4}\text{T}\end{array}$

 The total magnetic field at the point $C$ is,

    $\begin{array}{c}{B}_C=B_{AB}+B_{DE}\\ =1.37\times {10}^{-4}\text{T}+1.37\times {10}^{-4}\text{T}\\ =2.74\times {10}^{-4}\text{T}\end{array}$ 

Therefore, the magnitude of the field at the mid-point of the bar is $2.74\times {10}^{-4}\text{T}$ and the direction is into the page.

To determine

The reason for the value of the magnetic field will be same at all position of the bar to the right of the midpoint of the rails.

Answer to Problem 76P

The rail is long so the location of the bar does not depend upon the length of the rail to the right side.

Explanation of Solution

Here, it is assumed as the rail is infinitely long so, the length of the rail to the right of the bar does not depend upon the location of the bar.

Therefore the magnetic field will be same at all position of the bar to the right of the midpoint of the rails. 

Conclusion:

The rail is long so the location of the bar does not depend upon the length of the rail to the right side.

To determine

The magnitude of the force on the bar.

Answer to Problem 76P

 The magnitude of the force on the bar is $1.15\times {10}^{-3}\text{N}$ .

Explanation of Solution

Write the expression for the magnetic field in a wire,

  $\overrightarrow{F}=I\left(\overrightarrow{l}\times \overrightarrow{B}\right)$        (I)

Here, $\overrightarrow{F}$ is the centripetal force, $\overrightarrow{l}$ is the length of the particle, $I$ is the current of the particle, and  $\overrightarrow{B}$ is the magnetic field.

Conclusion:

Substitute $5\times 2.74\times {10}^{-4}\text{T}$ for $\overrightarrow{B}$ , $24.0\text{A}$ for $I$ , and $3.50\text{cm}\widehat{\text{k}}$ for $\overrightarrow{l}$ in the above equation (I) to find the value of $F$ .

    $\begin{array}{c}\overrightarrow{F}=\left(24.0\text{A}\right)\left(3.50\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\widehat{\text{k}}\right)\left(5\times 2.74\times {10}^{-4}\text{T}\right)\\ =1.15\times {10}^{-3}\text{N}\widehat{\text{i}}\end{array}$

The magnitude of the force on the bar is $1.15\times {10}^{-3}\text{N}\widehat{\text{i}}$ .

To determine

The direction of the force on the bar.

Answer to Problem 76P

The direction of the force on the bar is in positive $x$ direction.

Explanation of Solution

Write the expression for the magnetic field in a wire,

  $\overrightarrow{F}=I\left(\overrightarrow{l}\times \overrightarrow{B}\right)$        (I)

Here, $\overrightarrow{F}$ is the centripetal force, $\overrightarrow{l}$ is the length of the particle, $I$ is the current of the particle, and  $\overrightarrow{B}$ is the magnetic field.

Substitute $5\times 2.74\times {10}^{-4}\text{T}$ for $\overrightarrow{B}$ , $24.0\text{A}$ for $I$ , and $3.50\text{cm}\widehat{\text{k}}$ for $\overrightarrow{l}$ in the above equation (I) to find the value of $F$ .

    $\begin{array}{c}\overrightarrow{F}=\left(24.0\text{A}\right)\left(3.50\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\widehat{\text{k}}\right)\left(5\times 2.74\times {10}^{-4}\text{T}\right)\\ =1.15\times {10}^{-3}\text{N}\widehat{\text{i}}\end{array}$

Conclusion:

The force vector on the bar is $1.15\times {10}^{-3}\text{N}\widehat{\text{i}}$ , thus the direction of the force on the bar is in the positive $x$ direction.

To determine

Whether the bar is properly modeled as a particle under constant acceleration.

Answer to Problem 76P

Yes, the bar will move with constant acceleration of magnitude $0.384{\text{m/s}}^{\text{2}}$ .

Explanation of Solution

Write the expression to calculate the acceleration of bar,

    $a=\frac{F}{m}$        (I)

Here, $a$ is the acceleration, $F$ is the force , and $m$ is the mass of the bar.

Conclusion:

Substitute $1.15\times {10}^{-3}\text{N}$ for $F$ , and $3.00g$ for $m$ in the above equation (I) to find the value of $a$ .

    $\begin{array}{c}a=\frac{1.15\times {10}^{-3}\text{N}}{3.00g\times \left(\frac{{10}^{-3}\text{kg}}{1g}\right)}\\ =0.384{\text{m/s}}^{\text{2}}\end{array}$

Therefore, the bar will move with constant acceleration of magnitude $0.384{\text{m/s}}^{\text{2}}$

To determine

The velocity of the bar.

Answer to Problem 76P

The velocity of the bar is $0.998\text{m/s}$ .

Explanation of Solution

Write the equation for velocity of the bar,

  $v_f{}^2=v_i{}^2+2ad$        (I)

Here, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $d$  is the distance travelled by the bar.

Conclusion:

Substitute $0$ for $v_i$ , $130\text{cm}$ for $d$ , and $0.384{\text{m/s}}^{\text{2}}$ for $a$ in the above equation (I) to find the value of $a$ .

    $\begin{array}{c}{v}_f=0+\left(0.384{\text{m/s}}^{\text{2}}\right)\left(130\text{cm}\right)\\ =0.998\text{m/s}\end{array}$

The velocity of the bar is $0.998\text{m/s}$ .