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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Figure 29-89 is an idealized schematic drawing of a rail gun. Projectile P sits between two wide rails of circular cross section; a source of current sends current through the rails and through the(conducting) projectile (a fuse is not used). (a) Let w be the distance between the rails, R the radius of each rail, and i the current. Show that the force on the projectile is directed to the right along the rails and is given approximately by $\stackrel{\mathbf{\to }}{\mathbf{F}}{\mathbf{=}}\frac{{\mathbf{i}}^{\mathbf{2}}{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{2}\mathbf{\pi }}{\mathbf{·}}{\mathbit{l}}{\mathbit{n}}\left(\frac{w+R}{R}\right)$(b) If the projectile starts from the left end of the rails at rest, find the speed v at which it is expelled at the right. Assume that I = 450 kA, w = 12 mm, R = 6.7 cm, L = 4.0 m, and the projectile mass is 10 g.

1. Force on the projectile is, $\stackrel{\to }{F}=\frac{{i}^{2}{\mu }_{0}}{2\pi }·\mathrm{ln}\left(\frac{w+R}{R}\right)\stackrel{^}{i}$
2. Speed of the projectile is, ${v}_{final}=2.3×{10}^{3}\text{m/s}$
See the step by step solution

## Step 1: Identification of the given data

1. Radius of circular cross-section of the rail is, $R=6.7 \text{cm}=6.7×{10}^{-2} \text{m}$.
2. Distance between the rails is, $w=12.0 \text{mm}=12×{10}^{-3} \text{m}$.
3. Length of the rail is, $L=4.0 \text{m}$.
4. Current through the rail is, $i=450×{10}^{3} \text{A}$.
5. Mass of the projectile is, $m=10 \text{g}=10×{10}^{-3} \text{kg}$.

## Step 2: Understanding the concept

As the projectile is moving in the presence of magnetic field due to the two rails (1 and 2), it is also carrying current. So small current element of projectile will experience force due to both rails.Net force on current element (${\mathbit{i}}{\mathbit{d}}\stackrel{\mathbf{\to }}{\mathbf{l}}$) will be the vector sum of individual forces due to rail 1 and 2.

Formula:

1. Force acing on current element kept in magnetic field, ${\mathbit{d}}\stackrel{\mathbf{\to }}{\mathbf{F}}{\mathbf{=}}{\mathbit{i}}\left(d\stackrel{\to }{l}×\stackrel{\to }{B}\right)$
2. Magnetic field due to long wire carrying current i , at distance r is

## Step 3: (a) Calculate force on the projectile

Consider a portion of projectile between $y&y+dy$ .We can find magnetic force on this segment due to both rails (1 and 2)

The net force on the segment is $\stackrel{\to }{F}=d\stackrel{\to }{{F}_{1}}+d\stackrel{\to }{{F}_{2}}$ , where $d\stackrel{\to }{{F}_{1}}$ and $d\stackrel{\to }{{F}_{2}}$ are the forces on portion of projectile due to rail 1 and 2 respectively.

$d\stackrel{\to }{F}=d\stackrel{\to }{{F}_{1}}+d\stackrel{\to }{{F}_{2}}=i\left(d\stackrel{\to }{l}×\stackrel{\to }{{B}_{1}}\right)+i\left(d\stackrel{\to }{l}×\stackrel{\to }{{B}_{2}}\right)$

We can find $B$ field due to rail 1 at distance $\left(2R+w-y\right)$ from the center of rail 1.We can use Ampere law for this, then we get

${B}_{1}=\frac{{\mu }_{0}i}{2\pi \left(2R+w-y\right)}\left(-\stackrel{^}{k}\right)$

Similarly, B field due to rail 2 at distance $\left(y\right)$ from the center of rail 2, is

${B}_{2}=\frac{{\mu }_{0}i}{2\pi y}\left(-\stackrel{^}{k}\right)$

And $d\stackrel{\to }{l}=-dy\cdot \stackrel{^}{j}$ using this we get

$\begin{array}{c}\mathrm{d}\stackrel{\to }{\mathrm{F}}=\mathrm{d}\stackrel{\to }{{\mathrm{F}}_{1}}+\mathrm{d}\stackrel{\to }{{\mathrm{F}}_{2}}\\ \mathrm{d}\stackrel{\to }{\mathrm{F}}=\mathrm{i}\left(\mathrm{dy}\cdot \left(-\stackrel{^}{\mathrm{j}}\right)×\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{\pi }\left(2\mathrm{R}+\mathrm{w}-\mathrm{y}\right)}\left(-\stackrel{^}{\mathrm{k}}\right)\right)+\mathrm{i}\left(\mathrm{dy}\cdot \left(-\stackrel{^}{\mathrm{j}}\right)×\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{\pi y}}\left(-\stackrel{^}{\mathrm{k}}\right)\right)\text{}\end{array}$

By using cross product rule $\stackrel{^}{j}×\stackrel{^}{k}=\stackrel{^}{i}$

$\begin{array}{c}\mathrm{d}\stackrel{\to }{\mathrm{F}}=\mathrm{i}\cdot \stackrel{^}{\mathrm{i}}\left(\frac{{\mathrm{\mu }}_{0}\mathrm{idy}}{2\mathrm{\pi }\left(2\mathrm{R}+\mathrm{w}-\mathrm{y}\right)}\right)\right)+\stackrel{^}{\mathrm{i}}\left(\frac{{\mathrm{\mu }}_{0}\mathrm{idy}}{2\mathrm{\pi y}}\right)\text{}\\ \mathrm{d}\stackrel{\to }{\mathrm{F}}=\mathrm{idy}\cdot \left(\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{\pi }\left(2\mathrm{R}+\mathrm{w}-\mathrm{y}\right)}+\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{\pi y}}\right)\stackrel{^}{\mathrm{i}}\text{}\end{array}$

Total force is obtained by integrating $d\stackrel{\to }{F}$ over $dy$

role="math" localid="1664355361241" $\begin{array}{c}\stackrel{\to }{\mathrm{F}}=\underset{\mathrm{R}}{\overset{\mathrm{R}+\mathrm{w}}{}}\mathrm{d}\stackrel{\to }{\mathrm{F}}=\underset{\mathrm{R}}{\overset{\mathrm{R}+\mathrm{w}}{}}\mathrm{idy}\cdot \left(\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{\pi }\left(2\mathrm{R}+\mathrm{w}-\mathrm{y}\right)}+\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{\pi y}}\right)\stackrel{^}{\mathrm{i}}\\ \stackrel{\to }{\mathrm{F}}=\underset{\mathrm{R}}{\overset{\mathrm{R}+\mathrm{w}}{}}\frac{{\mathrm{\mu }}_{0}{\mathrm{i}}^{2}}{2\mathrm{\pi }}\cdot \left(\frac{\mathrm{dy}}{\left(2\mathrm{R}+\mathrm{w}-\mathrm{y}\right)}+\frac{\mathrm{dy}}{\mathrm{y}}\right)\stackrel{^}{\mathrm{i}}\\ =\frac{{\mathrm{\mu }}_{0}{\mathrm{i}}^{2}}{2\mathrm{\pi }}\stackrel{^}{\mathrm{i}}\cdot \underset{\mathrm{R}}{\overset{\mathrm{R}+\mathrm{w}}{}}\left(\frac{\mathrm{dy}}{\left(2\mathrm{R}+\mathrm{w}-\mathrm{y}\right)}+\frac{\mathrm{dy}}{\mathrm{y}}\right)\end{array}$

" width="9" height="19" role="math" style="max-width: none; vertical-align: -4px;" localid="1664355443385">

Hence the force on the projectile is, role="math" localid="1664355437114" $\stackrel{\to }{\mathrm{F}}=\frac{{\mathrm{i}}^{2}{\mathrm{\mu }}_{0}}{2\mathrm{\pi }}\cdot \mathrm{ln}\left(\frac{\mathrm{w}+\mathrm{R}}{\mathrm{R}}\right)\stackrel{^}{\mathrm{i}}$

## Step 4: (b) Calculate speed of the projectile

To find the speed of projectile we can use work energy theorem

$\mathrm{d}\stackrel{\to }{\mathrm{F}}\cdot \mathrm{d}\stackrel{\to }{\mathrm{x}}=\mathrm{dk}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{final}}^{2}$

$\begin{array}{c}\mathrm{w}=\int \mathrm{dw}\\ =\int \stackrel{\to }{\mathrm{F}}\cdot \mathrm{d}\stackrel{\to }{\mathrm{x}}\\ =\underset{0}{\overset{\mathrm{L}}{}}\mathrm{dx}\frac{{\mathrm{i}}^{2}{\mathrm{\mu }}_{0}}{2\mathrm{\pi }}\cdot \mathrm{ln}\left(\frac{\mathrm{w}+\mathrm{R}}{\mathrm{R}}\right)\stackrel{^}{\mathrm{i}}\cdot \stackrel{^}{\mathrm{i}}\\ \mathrm{w}=\frac{{\mathrm{i}}^{2}{\mathrm{\mu }}_{0}}{2\mathrm{\pi }}\cdot \mathrm{ln}\left(\frac{\mathrm{w}+\mathrm{R}}{\mathrm{R}}\right)\mathrm{L}\end{array}$

$\frac{1}{2}m{v}_{final}^{2}=w$.

Which gives,

$\begin{array}{c}{v}_{final}=\sqrt{\left(\frac{2w}{m}\right)}={\left(\frac{2}{m}\cdot \frac{{i}^{2}{\mu }_{0}}{2\pi }\cdot \mathrm{ln}\left(\frac{w+R}{R}\right)L\right)}^{1/2}\\ {v}_{final}={\left(\frac{4}{m}\cdot \frac{{i}^{2}{\mu }_{0}}{4\pi }\cdot \mathrm{ln}\left(\frac{w+R}{R}\right)L\right)}^{1/2}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{v}_{final}={\left[\frac{4×\left(4\pi ×{10}^{-7}\text{\hspace{0.17em}Tm}/\text{A}\right)×{\left(450×{10}^{3}\text{\hspace{0.17em}A}\right)}^{2}}{4\pi ×10×{10}^{-3}\text{\hspace{0.17em}kg}}×4\text{\hspace{0.17em}m}×\mathrm{ln}\left(\frac{6.7\text{\hspace{0.17em}cm}+1.2\text{\hspace{0.17em}cm}}{6.7\text{\hspace{0.17em}cm}}\right)\right]}^{\frac{1}{2}}\\ =2.3×{10}^{3}\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence the speed of the projectile is, ${v}_{final}=2.3×{10}^{3}\text{\hspace{0.17em}m}/\text{s}$