91Ó°ÊÓ

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, $m=10 \text{g}=10×{10}^{-3} \text{kg}$.

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 ($\mathit{i}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{l}}$) 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,$\mathit{d}\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathit{i}\left(d\stackrel{→}{l}×\stackrel{→}{B}\right)$
2. Magnetic field due to long wire carrying current i, at distance r is

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$\stackrel{→}{F}=d\stackrel{→}{F_1}+d\stackrel{→}{F_2}$, where$d\stackrel{→}{F_1}$ and$d\stackrel{→}{F_2}$ are the forces on portion of projectile due to rail 1 and 2 respectively.

$d\stackrel{→}{F}=d\stackrel{→}{F_1}+d\stackrel{→}{F_2}=i(d\stackrel{→}{l}×\stackrel{→}{B_1})+i(d\stackrel{→}{l}×\stackrel{→}{B_2})$

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

$B_1=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}(2R+w−y)}(−\widehat{k})$

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

$B_2=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}y}(−\widehat{k})$

And $d\stackrel{→}{l}=−dyâ‹\dots \widehat{j}$using this we get

$\begin{array}{c}\mathrm{d}\stackrel{→}{\mathrm{F}}=\mathrm{d}\stackrel{→}{{\mathrm{F}}_1}+\mathrm{d}\stackrel{→}{{\mathrm{F}}_2}\\ \mathrm{d}\stackrel{→}{\mathrm{F}}=\mathrm{i}(\mathrm{dy}â‹\dots (−\widehat{\mathrm{j}})×\frac{{\mathrm{μ}}_0\mathrm{i}}{2\mathrm{Ï€}(2\mathrm{R}+\mathrm{w}−\mathrm{y})}(−\widehat{\mathrm{k}}))+\mathrm{i}\left(\mathrm{dy}â‹\dots (−\widehat{\mathrm{j}})×\frac{{\mathrm{μ}}_0\mathrm{i}}{2\mathrm{Ï€²â}}(−\widehat{\mathrm{k}})\right)\end{array}$

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

$\begin{array}{c}\mathrm{d}\stackrel{→}{\mathrm{F}}=\mathrm{i}â‹\dots \widehat{\mathrm{i}}\left(\frac{{\mathrm{μ}}_0\mathrm{idy}}{2\mathrm{Ï€}(2\mathrm{R}+\mathrm{w}−\mathrm{y})}\right))+\widehat{\mathrm{i}}\left(\frac{{\mathrm{μ}}_0\mathrm{idy}}{2\mathrm{Ï€²â}}\right)\\ \mathrm{d}\stackrel{→}{\mathrm{F}}=\mathrm{idy}â‹\dots \left(\frac{{\mathrm{μ}}_0\mathrm{i}}{2\mathrm{Ï€}(2\mathrm{R}+\mathrm{w}−\mathrm{y})}+\frac{{\mathrm{μ}}_0\mathrm{i}}{2\mathrm{Ï€²â}}\right)\widehat{\mathrm{i}}\end{array}$

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

role="math" localid="1664355361241" $\begin{array}{c}\stackrel{→}{\mathrm{F}}=\underset{\mathrm{R}}{\overset{\mathrm{R}+\mathrm{w}}{}}\mathrm{d}\stackrel{→}{\mathrm{F}}=\underset{\mathrm{R}}{\overset{\mathrm{R}+\mathrm{w}}{}}\mathrm{idy}â‹\dots \left(\frac{{\mathrm{μ}}_0\mathrm{i}}{2\mathrm{Ï€}(2\mathrm{R}+\mathrm{w}−\mathrm{y})}+\frac{{\mathrm{μ}}_0\mathrm{i}}{2\mathrm{Ï€²â}}\right)\widehat{\mathrm{i}}\\ \stackrel{→}{\mathrm{F}}=\underset{\mathrm{R}}{\overset{\mathrm{R}+\mathrm{w}}{}}\frac{{\mathrm{μ}}_0{\mathrm{i}}^2}{2\mathrm{Ï€}}â‹\dots \left(\frac{\mathrm{dy}}{(2\mathrm{R}+\mathrm{w}−\mathrm{y})}+\frac{\mathrm{dy}}{\mathrm{y}}\right)\widehat{\mathrm{i}}\\ =\frac{{\mathrm{μ}}_0{\mathrm{i}}^2}{2\mathrm{Ï€}}\widehat{\mathrm{i}}â‹\dots \underset{\mathrm{R}}{\overset{\mathrm{R}+\mathrm{w}}{}}\left(\frac{\mathrm{dy}}{(2\mathrm{R}+\mathrm{w}−\mathrm{y})}+\frac{\mathrm{dy}}{\mathrm{y}}\right)\end{array}$

$$" width="9" height="19" role="math">F→=i2μ02π⋅lnw+RRi^

Hence the force on the projectile is,role="math" localid="1664355437114" $\stackrel{→}{\mathrm{F}}=\frac{{\mathrm{i}}^2{\mathrm{μ}}_0}{2\mathrm{Ï€}}â‹\dots \ln \left(\frac{\mathrm{w}+\mathrm{R}}{\mathrm{R}}\right)\widehat{\mathrm{i}}$

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

$\mathrm{d}\stackrel{→}{\mathrm{F}}â‹\dots \mathrm{d}\stackrel{→}{\mathrm{x}}=\mathrm{dk}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{final}}^2$

$\begin{array}{c}\mathrm{w}=∫\mathrm{dw}\\ =∫\stackrel{→}{\mathrm{F}}â‹\dots \mathrm{d}\stackrel{→}{\mathrm{x}}\\ =\underset{0}{\overset{\mathrm{L}}{}}\mathrm{dx}\frac{{\mathrm{i}}^2{\mathrm{μ}}_0}{2\mathrm{Ï€}}â‹\dots \ln \left(\frac{\mathrm{w}+\mathrm{R}}{\mathrm{R}}\right)\widehat{\mathrm{i}}â‹\dots \widehat{\mathrm{i}}\\ \mathrm{w}=\frac{{\mathrm{i}}^2{\mathrm{μ}}_0}{2\mathrm{Ï€}}â‹\dots \ln \left(\frac{\mathrm{w}+\mathrm{R}}{\mathrm{R}}\right)\mathrm{L}\end{array}$

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

Which gives,

$\begin{array}{c}{v}_{final}=\sqrt{\left(\frac{2w}{m}\right)}={\left(\frac{2}{m}â‹\dots \frac{i^2{\mathrm{μ}}_0}{2\mathrm{Ï€}}â‹\dots \ln \left(\frac{w+R}{R}\right)L\right)}^{1/2}\\ v_{final}={\left(\frac{4}{m}â‹\dots \frac{i^2{\mathrm{μ}}_0}{4\mathrm{Ï€}}â‹\dots \ln \left(\frac{w+R}{R}\right)L\right)}^{1/2}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{v}_{final}={\left[\frac{4×(4\mathrm{Ï€}×{10}^{−7}\text{â€Í¿³¾}/\text{A})×{(450×{10}^3\text{‼î})}^2}{4\mathrm{Ï€}×10×{10}^{−3}\text{ k²µ}}×4\text{″¾}×\ln \left(\frac{6.7\text{ c³¾}+1.2\text{ c³¾}}{6.7\text{ c³¾}}\right)\right]}^{\frac{1}{2}}\\ =2.3×{10}^3\text{″¾}/\text{s}\end{array}$

Hence the speed of the projectile is,$v_{final}=2.3×{10}^3\text{″¾}/\text{s}$