Why Does the Force Calculation on Parallel Wires Yield Zero?

In summary, the conversation discusses a problem involving two parallel wires carrying currents of 40.0 A and being 20.0 cm apart. The speaker attempts to calculate the force per unit length exerted on one wire by the other using the formula Bwire=4*10^-7(constant)*I/2piR, but realizes that something is missing. They also mention using the formula F/L=IBsin(theta) and discuss their attempts to find the correct solution.
  • #1
myersb05
14
0
I felt like this problem was simple. We submit our answers in WebAssign however, and I can not find the correct solution.

Two parallel wires are 20.0 cm apart, and each carries a current of 40.0 A.

(a) If the currents are in the same direction, find the force per unit length exerted on one of the wires by the other.

(b) Repeat the problem with the currents in opposite directions.


I know that Bwire=4*10^-7(constant)*I/2piR
So I calculated Bwire by using 4*10^-7*40/2pi*.2 and got a value of .000013 for each of the wires because their distances apart and currents are equal.

I also know that F/L=IBsin(theta). I plugged my current of 40 amps and my newly calculated B wire in. I thought that there is no angle between two parallel wires so using 0, my values for force came out as 0. After finding that was not correct, I disregarded the sin part of the equation thinking that it would probably be 1 or -1. I came up with a F/L value of .00052 N/m which is not correct. Where am I going wrong?
 
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  • #2
Hi myersb05,

myersb05 said:
I felt like this problem was simple. We submit our answers in WebAssign however, and I can not find the correct solution.

Two parallel wires are 20.0 cm apart, and each carries a current of 40.0 A.

(a) If the currents are in the same direction, find the force per unit length exerted on one of the wires by the other.

(b) Repeat the problem with the currents in opposite directions.


I know that Bwire=4*10^-7(constant)*I/2piR
So I calculated Bwire by using 4*10^-7*40/2pi*.2 and got a value of .000013 for each of the wires because their distances apart and currents are equal.

I don't believe this formula is quite right. Do you see what's missing?
 
  • #3



Dear student,

Thank you for sharing your thought process and calculations. It seems like you have a good understanding of the equations involved in solving this problem. However, there are a few things to consider that may help you find the correct solution.

Firstly, when calculating the magnetic field (B) due to a current-carrying wire, it is important to consider the distance (R) between the two wires. In this case, the distance between the wires is 20 cm, or 0.2 m. However, in your calculation, you used 0.2 m as the radius (R) instead of the distance between the wires. This may have led to an incorrect value for B.

Secondly, when calculating the force per unit length, the angle between the current and the magnetic field should be taken into account. In this case, since the wires are parallel, the angle between the current and the magnetic field is 90 degrees. Therefore, the correct equation to use would be F/L = I*B*sin(90). As you mentioned, the sine of 90 degrees is either 1 or -1, depending on the direction of the current. This leads to a force per unit length of 0.00052 N/m for both wires, regardless of the direction of the current.

In summary, to find the correct solution, make sure to use the correct distance between the wires when calculating the magnetic field, and take into account the angle between the current and the magnetic field when calculating the force per unit length. I hope this helps. Keep up the good work in your studies.

Best,
 

Related to Why Does the Force Calculation on Parallel Wires Yield Zero?

What is the force on parallel wires?

The force on parallel wires is the attractive or repulsive force between two parallel conductors carrying electric currents. The force is caused by the magnetic fields produced by the flowing currents interacting with each other.

How is the force on parallel wires calculated?

The force on parallel wires can be calculated using the following equation: F = (μ0 * I1 * I2 * L) / (2πd), where μ0 is the permeability constant, I1 and I2 are the currents in the wires, L is the length of the wires, and d is the distance between them.

What factors affect the force on parallel wires?

The force on parallel wires is affected by the currents in the wires, the distance between them, and the permeability of the medium surrounding the wires. It is also affected by the direction of the currents and the orientation of the wires.

Is the force on parallel wires always attractive?

No, the force on parallel wires can be either attractive or repulsive depending on the direction of the currents in the wires. If the currents are flowing in the same direction, the force is attractive. If the currents are flowing in opposite directions, the force is repulsive.

What are some real-life applications of the force on parallel wires?

The force on parallel wires has several practical applications, including electromagnets, electric motors, and generators. It is also used in transmission lines to transfer electrical energy over long distances and in electronic devices such as speakers and headphones.

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