91Ó°ÊÓ

Suppose we have pair of long, rigid metal rods, each of length 0.50 m, lying parallel to each other on a frictionless table as shown in the following figure, the ends of the two rods are connected by conducting springs with unstretched length $l_o$and force constant k. When a current of I runs through the rods, they will repel each other with a force of,

$F=\frac{{\mathrm{μ}}_0{I}^{2}L}{2\mathrm{π}r}$ (1)

where r is the distance between the wires, which is Simply the sum of the unstretched and stretched lengths of the springs, that is $r=l_o+x$, so,

$F=\frac{{\mathrm{μ}}_o{I}^{2}L}{2\mathrm{π}\left(l_o+x\right)}$

spring exerts with a Hook's force, that is kx. Where on each wire, this force must equal the force in (), note that we have two springs, so the net force is 2kx, from which we can write,

$2kx=\frac{{\mathrm{μ}}_o{I}^{2}L}{2\mathrm{π}r}$

each spring exerts with a Hooks force, that is kx. Where on each wire, this force must equal the force in (1), note that we have two springs, so the net force is 2kx, from which we can write,

$2kx=\frac{{\mathrm{μ}}_o{I}^{2}L}{2\mathrm{π}\left(l_0+x\right)}$ (2)

we have two cases with values for I and x, when I = 8.05 A, we measure x= 0.40 cm, and when I = 13.1 A, we measure x = 0.80 cm, substitute with these values into (2) we get,

$2k\left(0.40\mathrm{cm}\right)=\frac{{\mathrm{μ}}_o(8.05A{)}^{2}L}{2\mathrm{π}\left(l_o+0.40\right)}$ and $2k\left(0.08\mathrm{cm}\right)=\frac{{\mathrm{μ}}_o(13.1A{)}^{2}L}{2\mathrm{π}\left(l_o+0.80\right)}$

divide these two equations by each other we get,

$\frac{(13.1A{)}^2\left(l_o+0.80\right)}{(8.05A{)}^2\left(l_o+0.40\right)}=\frac{0.80}{0.40}=2.0$

now we can solve this equation for $l_0$as,

$\begin{array}{c}\frac{\left(l_0+0.40\mathrm{m}\right)}{\left(l_0+0.80\mathrm{m}\right)}=2{\left(\frac{8.05\mathrm{A}}{13.1\mathrm{A}}\right)}^2=0.755\\ l_0+0.40\mathrm{m}=0.755l_0+0.604\mathrm{m}\\ l_0=\frac{0.604\mathrm{m}−0.40\mathrm{m}}{0.245}=0.833\mathrm{m}\\ l_0=0.833\mathrm{m}\end{array}$

to find k we substitute into one of the equations in (3), where the length of the rods is =0.50 m, substitute into the second one in (3) to get,

$\begin{array}{c}\frac{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(13.1\mathrm{A}{)}^2\right(0.50\mathrm{m})}{2\mathrm{Ï€}(0.0080\mathrm{m}+0.00833\mathrm{m})}=0.00105\mathrm{N}=2k\left(0.0080\mathrm{m}\right)\\ k=\frac{0.00105\mathrm{N}}{2\left(0.0080\mathrm{m}\right)}=0.0656\mathrm{N}/\mathrm{m}\\ k=0.0656\mathrm{N}/\mathrm{m}\end{array}$

For I =12.0 A, we need to find x, substitute with this value and the calculated values in step 2 into (2), so we get,

$\begin{array}{c}\frac{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(12.0\mathrm{A}{)}^2\right(0.50\mathrm{m})}{2\mathrm{Ï€}(x+0.00833\mathrm{m})}=2(0.0656\mathrm{N}/\mathrm{m})x\\ \frac{1.44×{10}^{−5}\mathrm{N}â‹\dots \mathrm{m}}{x+0.00833\mathrm{m}}=(0.1312\mathrm{N}/\mathrm{m})x\\ x^2+\left(0.00833\mathrm{m}\right)x−1.097×{10}^{−4}{\mathrm{m}}^2=0\end{array}$

this is a quadratic equation that can be solved using the general law of the quadratic equation, as,

$\begin{array}{c}x=\frac{−0.00833\mathrm{m}Â\pm \sqrt{(0.00833\mathrm{m}{)}^2+4\left(1.097×{10}^{−4}{\mathrm{m}}^2\right)}}{2}\\ =−0.004165\mathrm{m}Â\pm 0.0112715\mathrm{m}\end{array}$

we take the positive solution, which is,

x = 0.0071 m

Finally, we need to find the current, that makes the springs stretch by = 1.00 cm, so we substitute into (2) to get,

$\begin{array}{c}\frac{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)I^2\left(0.50\mathrm{m}\right)}{2\mathrm{Ï€}(0.0100+0.00833\mathrm{m})}=2(0.0656\mathrm{N}/\mathrm{m})\left(0.0100\mathrm{m}\right)\\ I=\sqrt{\frac{4\mathrm{Ï€}(0.0100+0.00833\mathrm{m})(0.0656\mathrm{N}/\mathrm{m})\left(0.0100\mathrm{m}\right)}{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(0.50\mathrm{m}\right)}}\\ I=15.5\mathrm{A}\end{array}$