Solve Current in Parallel Wires with Repulsion

In summary, the problem involves two parallel wires with a mass per unit length of 43 g/m, supported by 6.0 cm long strings and carrying the same current I. The wires repel each other at an angle of 16°. Using the equations sin(theta) = opposite/hyp and I^2 = [(2pi)(d)(mg/L)] / [(4pi)e-07], the magnitude of each current is found to be 132.64 A. However, after attempting to use a free body diagram and solving for I, the correct answer is determined to be 159.8 A.
  • #1
Touchme
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Homework Statement


Two long parallel wires, each with a mass per unit length of 43 g/m, are supported in a horizontal plane by 6.0 cm long strings, as shown in Figure P19.64. Each wire carries the same current I, causing the wires to repel each other so that the angle between the supporting strings is 16°.
Determine the magnitude of each current.

Homework Equations


sin (theta) = opposite/hyp

I^2 = [(2pi)(d)(mg/L)] / [(4pi)e-07]


The Attempt at a Solution



First I found d (the distance between the wires). Which is 2[sin(8)](0.06m) = 0.0167m.
Then I substitute the given into this equation: I^2 = [(2pi)(d)(mg/L)] / [(4pi)e-07]. I = sq.root of [(2pi)(0.0167)(0.043)(9.8)] / [(4pi)e-07]. the answer I got is 132.64 A. What did I do wrong?

I tried using free body diagram.
T = mg/sin8
Tcos8 = [(I^2)(L)((4pi)e-7)] / [(2pi)d]
(mgcos8)/(sin8) = [(I^2)(L)((4pi)e-7)] / [(2pi)d]
(mg/L)(7.115) = [(I^2)((4pi)e-7)] / [(2pi)0.0167]
I = 159.8 A

Still incorrect.
 

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  • #2
Ah shoot, I've made the most stupid mistake ever. I solved it.
 
  • #3


I would like to point out that the given information and equations may not be sufficient to accurately determine the magnitude of the current in this scenario. Additional information such as the tension in the strings and the distance between the wires at equilibrium would be needed to solve for the current using the equations provided. It is also important to consider the effects of gravity and the magnetic field on the wires in this setup. Further experimentation and calculations may be necessary to accurately determine the current in this parallel wire system.
 

1. What is the repulsion effect in parallel wires?

The repulsion effect in parallel wires refers to the phenomenon where two wires carrying electric current in the same direction will experience a repulsive force between them. This is due to the interaction between the magnetic fields generated by the currents in each wire.

2. How do you calculate the repulsive force between parallel wires?

The repulsive force between parallel wires can be calculated using the formula F = (μ0 * I1 * I2 * L) / (2π * d), where μ0 is the permeability of free space, I1 and I2 are the currents in the two wires, L is the length of the wires, and d is the distance between them.

3. What factors affect the strength of the repulsive force between parallel wires?

The strength of the repulsive force between parallel wires is affected by the distance between the wires, the magnitude of the currents in each wire, and the length of the wires. The permeability of free space also plays a role in determining the strength of the force.

4. How does the repulsion effect impact the current in parallel wires?

The repulsion effect can cause the currents in parallel wires to decrease, as the force between the wires pushes them away from each other. This can lead to a decrease in the overall current in the circuit. However, with proper design and placement of the wires, this effect can be minimized.

5. Are there any practical applications for the repulsion effect in parallel wires?

Yes, the repulsion effect in parallel wires has several practical applications. It can be used in circuit breakers to interrupt the flow of current, in high voltage power lines to prevent damage to the wires, and in maglev trains to create a levitation effect. It can also be used in various types of sensors and actuators.

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