Magnetism (Current-Carrying Wires)

Charanjit

1. The problem statement, all variables and given/known data
Two straight wires, each with a resistance of 0.170 ohm and a length of 3.90m, are lying parallel to each other on a smooth horizontal table. Their ends are connected by identical, non-conducting, light springs, each spring having an unstretched length of 1.08 cm. A wire of negligible resistance connects the wires at one end. When a switch is closed to connect a battery with a voltage of 49.0 V between the other ends of the wires, the wires move apart and come to rest with a separation of 1.57 cm

Question: Find the force constants of the springs.



2. Relevant equations

F=-kx
(F/L)=(u₀I₁I₂)/2Pi r



3. The attempt at a solution

I tried to solve using the equations above for K, and does not work.

Fightfish

Did you modify the k in the spring equation to take into account the presence of both springs in a parallel arrangement?

Charanjit

Yes I did. This is what I got:
F=-2kx
(F/L)=(uI^2)/(2pi a)

k=(uI^2L)/(4piax)

I tried solving this, but got wrong answers. What I did:
u=1.26e-6 (Constant given to use)
I=144.1176 (Combining resistors in series and finding current)
L=3.90 (Length of the wire)
a=0.0157 (Seperation of the wires)
x=0.0049 (stretch of the spring)

Thats what I use, and I got 106 N/m. Can you verify that please?

dineshnaveen

due to passing of current the magnetic field is created in the rods hence we know the formula F=vlb from that u can find the force and from F=kx u can find the spring constant

Charanjit

So what I have is correct? Can you check?

Redbelly98

I agree, good job.

Charanjit already knew that!

Charanjit

Thank you. :)