Question

A pair of long, rigid metal rods, each of length 0.50 m, lie parallel to each other on a frictionless table. Their ends are connected by identical, very lightweight conducting springs with unstretched length l0 and force constant k(Fig.). When a current Iruns through the circuit consisting of the rods and springs, the springs stretch. You measure the distance xeach spring stretches for certain values of I. When I= 8.05 A, you measure that x= 0.40 cm. When I= 13.1 A, you find x= 0.80 cm. In both cases the rods are much longer than the stretched springs, so it is accurate to use Eq. (28.11) for two infinitely long, parallel conductors.

(a) From these two measurements, calculate l0 and k.

(b) If I= 12.0 A, what distance xwill each spring stretch?

(c) What current is required for each spring to stretch 1.00 cm?

Short answer

1. $l_0=0.833m$and$k=0.0656N/m$
2. x = 0.0071m
3. I=15.5 A

Step-by-step solution

Solving part (a) of the problem.

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,

$F=\frac{{\mu }_0{I}^{2}L}{2\pi r}$ (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,

$F=\frac{{\mu }_o{I}^{2}L}{2\pi \left(l_o+x\right)}$

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,

$2kx=\frac{{\mu }_o{I}^{2}L}{2\pi r}$

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,

$2kx=\frac{{\mu }_o{I}^{2}L}{2\pi \left(l_0+x\right)}$ (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,

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

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,

$\begin{array}{c}\frac{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(13.1\mathrm{A}{)}^2\right(0.50\mathrm{m})}{2\pi (0.0080\mathrm{m}+0.00833\mathrm{m})}=0.00105\mathrm{N}=2k\left(0.0080\mathrm{m}\right)\\ k=\frac{0.00105\mathrm{N}}{2\left(0.0080\mathrm{m}\right)}=0.0656\mathrm{N}/\mathrm{m}\\ k=0.0656\mathrm{N}/\mathrm{m}\end{array}$

Step 2: Solving part (b) of the problem.

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,

$\begin{array}{c}\frac{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(12.0\mathrm{A}{)}^2\right(0.50\mathrm{m})}{2\pi (x+0.00833\mathrm{m})}=2(0.0656\mathrm{N}/\mathrm{m})x\\ \frac{1.44\times {10}^{-5}\mathrm{N}\cdot \mathrm{m}}{x+0.00833\mathrm{m}}=(0.1312\mathrm{N}/\mathrm{m})x\\ x^2+\left(0.00833\mathrm{m}\right)x-1.097\times {10}^{-4}{\mathrm{m}}^2=0\end{array}$

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

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

we take the positive solution, which is,

x = 0.0071 m

Step 3: Solving part (c) of the problem.

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

$\begin{array}{c}\frac{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)I^2\left(0.50\mathrm{m}\right)}{2\pi (0.0100+0.00833\mathrm{m})}=2(0.0656\mathrm{N}/\mathrm{m})\left(0.0100\mathrm{m}\right)\\ I=\sqrt{\frac{4\pi (0.0100+0.00833\mathrm{m})(0.0656\mathrm{N}/\mathrm{m})\left(0.0100\mathrm{m}\right)}{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(0.50\mathrm{m}\right)}}\\ I=15.5\mathrm{A}\end{array}$