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Magnetism (CurrentCarrying Wires) 
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#1
Feb2210, 09:52 PM

P: 49

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, nonconducting, 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)=(u_{0}I_{1}I_{2})/2Pi r 3. The attempt at a solution I tried to solve using the equations above for K, and does not work. 


#2
Feb2310, 05:19 AM

P: 624

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



#3
Feb2410, 01:27 AM

P: 49

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=1.26e6 (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? 


#4
Feb2410, 02:07 AM

P: 7

Magnetism (CurrentCarrying Wires)
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



#5
Feb2410, 11:18 AM

P: 49

So what I have is correct? Can you check?



#7
Feb2410, 10:04 PM

P: 49

Thank you. :)



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