Chapter 30, Problem 30.66AP

Review. Rail guns have been suggested for launching projectiles into space without chemical rockets. A tabletop model rail gun (Fig. P29.42) 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 P29.42 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 P29.42

To determine

The magnitude of the magnetic field at a distance of $1.75\text{cm}$ from a single long wire.

Answer to Problem 30.66AP

The magnitude of the magnetic field at a distance of $1.75\text{cm}$ from a single long wire is $2.74\times {10}^{-4}\text{T}$.

Explanation of Solution

Given info: The length of the rails is $3.50\text{cm}$, the mass of bar is $3.00\text{g}$, value of current is $24.0\text{A}$.

Formula to calculate the magnetic field is

$B=\frac{{\mu }_{0}I}{2\pi r}$ (1)

Here,

$B$ is the magnetic field.

${\mu }_0$ is the absolute permeability.

$I$ is the current flow in the wire.

$r$ is the distance of the point.

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $24.0\text{A}$ for $I$, and $1.75\text{cm}$ for $r$ to find $B$.

$\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\times 24.0\text{A}}{2\pi \times 1.75\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}\\ =2.74\times {10}^{-4}\text{T}\end{array}$

Conclusion:

Therefore, the magnitude of the magnetic field at a distance of $1.75\text{cm}$ from a single long wire is $2.74\times {10}^{-4}\text{T}$.

To determine

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

Answer to Problem 30.66AP

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

Explanation of Solution

Given info: The length of the rails is $3.50\text{cm}$, the mass of bar is $3.00\text{g}$, value of current is $24.0\text{A}$.

The diagram that represents the given situation is shown below.

Figure (I)

Since the current is diverted through the bar, only half of each rails carries currents, so the field produce by each rail is half of the infinitely long wire produces.

Write the expression for the magnetic field produce by conductor AB at point C is,

$B_{\text{AB}}=\frac{1}{2}B$

Substitute $2.74\times {10}^{-4}\text{T}$ for $B$ in above equation.

$\begin{array}{c}{B}_{\text{AB}}=\frac{1}{2}\times 2.74\times {10}^{-4}\text{T}\\ \text{=1}\text{.37}{10}^{-4}\text{T}\end{array}$

Write the expression for the magnetic field produce by conductor DE at point C is,

$B_{\text{DE}}=\frac{1}{2}B$

Substitute $2.74\times {10}^{-4}\text{T}$ for $B$ in above equation.

$\begin{array}{c}{B}_{\text{DE}}=\frac{1}{2}\times 2.74\times {10}^{-4}\text{T}\\ \text{=1}\text{.37}{10}^{-4}\text{T}\end{array}$

Formula to calculate the total magnetic field at point C is,

$B_{\text{C}}=B_{\text{AB}}+B_{\text{DE}}$

Substitute $\text{1}\text{.37}{10}^{-4}\text{T}$ for $B_{\text{AB}}$ and $B_{\text{DE}}$ in above equation.

$\begin{array}{c}{B}_{\text{C}}=\text{1}\text{.37}{10}^{-4}\text{T+1}\text{.37}{10}^{-4}\text{T}\\ =2.74\times {10}^{-4}\text{T}\end{array}$

Conclusion:

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

To determine

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

Answer to Problem 30.66AP

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

Explanation of Solution

Given info: The length of the rails is $3.50\text{cm}$, the mass of bar is $3.00\text{g}$, value of current is $24.0\text{A}$.

The assumption makes that the rails are infinitely long so, the length of the rail to the right of the bar does not depend upon the location of the bar. Hence the magnetic field will be same at all position of the bar to the right of the midpoint of the rails.

Conclusion:

Therefore, due to infinite length of the rails makes the field same at all position of the midpoint of the bar.

To determine

The magnitude of the force on the bar.

Answer to Problem 30.66AP

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

Explanation of Solution

Given info: The length of the rails is $3.50\text{cm}$, the mass of bar is $3.00\text{g}$, value of current is $24.0\text{A}$.

Formula to calculate the force on the bar is,

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

Here,

$I$ is the value of current.

$\overrightarrow{\text{l}}$ is the length vector.

$\overrightarrow{\text{B}}$ is the field vector.

Substitute $24.0\text{A}$ for $I$, $3.50\text{cm}\widehat{\text{k}}$ for $\overrightarrow{\text{l}}$ and $\left(5\times 2.74\times {10}^{-4}\text{T}\right)$ for $\overrightarrow{\text{B}}$ to find $\overrightarrow{\text{F}}$

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

Conclusion:

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

To determine

The direction of the force on the bar.

Answer to Problem 30.66AP

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

Explanation of Solution

Given info: The length of the rails is $3.50\text{cm}$, the mass of bar is $3.00\text{g}$, value of current is $24.0\text{A}$.

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

Conclusion:

Therefore, the direction of the force on the bar is in positive x-direction.

To determine

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

Answer to Problem 30.66AP

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

Explanation of Solution

Given info: The length of the rails is $3.50\text{cm}$, the mass of bar is $3.00\text{g}$, value of current is $24.0\text{A}$.

The length of the bar, value of current and field producer is constant so, the force exerted on the bar is constant that gives uniform acceleration of the bar.

Formula to calculate the acceleration of the bar is,

$a=\frac{F}{m}$

Substitute $1.15\times {10}^{-3}\text{N}$ for $F$ and $3.00\text{g}$ for $m$ to find $a$.

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

Conclusion:

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

To determine

The velocity of the bar.

Answer to Problem 30.66AP

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

Explanation of Solution

Given info: The length of the rails is $3.50\text{cm}$, the mass of bar is $3.00\text{g}$, value of current is $24.0\text{A}$.

Formula to calculate the velocity of the bar is,

$v_{\text{f}}^2=v_{\text{i}}^2+2ad$

Here,

$v_{\text{f}}$ is the final velocity of the bar.

$v_{\text{i}}$ is the initial velocity of the bar.

$d$ is the distance travelled by the bar.

Substitute $0$ for $v_{\text{i}}$, $0.384\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $130\text{cm}$ for $d$ to find $v_{\text{f}}$.

$\begin{array}{c}{v}_{\text{f}}=0+2\left(0.384\text{m}/{\text{s}}^{\text{2}}\right)\left(130\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =0.998\text{m}/\text{s}\end{array}$

Conclusion:

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