Force on 3 current carrying wires

[jmr423]

Homework Statement

Two long fixed parallel wires, A and B, are 10cm apart in air and carry 40 A and 20 A, respectively, in opposite directions. Determine the resultant field (a) on a line midway between the wires parallel to them and (b) on a line 8.0 cm from wire A and 18cm from wire B (c) what is the force per meter on a third long wire, midway between A and B and in their plane, when it carries a current of 5.0A in the same direction as the current in A?

Homework Equations

F=BILsin(angle)
B= (2(10^7)I)/r

The Attempt at a Solution

I have solved part (a) and (b) i need help with (c) I have calculated the forces on them
A on C = 1E-4 towards
B on C = 2.2E-5 away

I tried to calculate it as though its on a straight line and also as though its making a triangle. for the triangle i used the angle ACB through the distances i have in cm. but it did not work.

Answers A) 2.4E-4T, B7.8E10-5T C 1.5E-3N/, toward A

 

[tiny-tim]

hi jmr423!

Two long fixed parallel wires, A and B, are 10cm apart in air
…
(c) what is the force per meter on a third long wire, midway between A and B and in their plane, when it carries a current of 5.0A in the same direction as the current in A?
…
I tried to calculate the angle ACB through the distances i have in cm. …

what angle?

they're all in the same plane ​

 

[jmr423]

hi jmr423! what angle?

they're all in the same plane ​

:S when i treat it as though its on a plane i get no where... i have tried a bunch of different things and nothing is working to solve c.

 

[tiny-tim]

show us what you've tried (in the same plane)

 

[asteriatic]

I would like to start by congratulating you on successfully solving parts (a) and (b) of the problem. Your calculations seem to be correct and you have shown a good understanding of the equations involved.

Now, for part (c), you are correct in considering the force on the third wire as being a result of the forces exerted by wires A and B. However, instead of considering the third wire as being on a straight line or forming a triangle, we need to consider the forces acting on it separately and then combine them using vector addition.

The force exerted by wire A on the third wire can be calculated using the formula F = BILsin(angle), where B is the magnetic field created by wire A, I is the current in the third wire, and L is the length of the third wire. Since the third wire is midway between A and B, the angle between the magnetic field and the current is 90 degrees, making sin(angle) = 1. Plugging in the values, we get F = (2(10^7)(5.0))/0.05 = 2.0 N towards A.

Similarly, the force exerted by wire B on the third wire can be calculated as F = BILsin(angle), where B is the magnetic field created by wire B, I is the current in the third wire, and L is the length of the third wire. Since the third wire is in the same plane as wire B, the angle between the magnetic field and the current is 0 degrees, making sin(angle) = 0. Plugging in the values, we get F = (2(10^7)(5.0))/0.18 = 0.56 N towards B.

Now, to find the resultant force on the third wire, we need to combine these two forces using vector addition. Since the forces are acting in opposite directions, we can simply subtract the magnitude of the force exerted by wire B from the magnitude of the force exerted by wire A. This gives us a resultant force of 1.44 N towards A.

To find the force per meter, we simply divide the resultant force by the length of the third wire, which is 0.05 m. This gives us a force per meter of 1.44/0.05 = 28.8 N/m towards A.

I hope this helps you understand the problem better and solve it correctly