What current will allow the wire to float ?

Click For Summary
To allow the upper wire to "float" and form an equilateral triangle with the lower wires, the magnetic force generated by the currents in the lower wires must equal the gravitational force acting on the upper wire. The gravitational force can be calculated using the linear mass density of 50 g/m, resulting in a force of 0.49 N/m. The magnetic force between the wires is given by the equation F/L = (μI)/(μr), where I is the current. After calculations, a current of 49 kA is suggested to balance the forces, indicating that a significant amount of current is required to counteract gravity. This highlights the challenge of achieving equilibrium in this setup.
Xaspire88
Messages
97
Reaction score
0
What current will allow the wire to "float"?

The linked image is a cross section through three long wires with linear mass density 50 g/m. They each carry equal currents in the directions shown. The lower two wires are 4.0 cm apart and are attached to a table. What current I will allow the upper wire to "float" so as to form an equilateral triangle with the lower wires?

The force that the upper wire feels is a combined force from the two other wires and the normal force of gravity. I am just not sure how to calculate the normal force of gravity with the linear mass density. Also the current that is needed will have to produce a magnetic field that combined equals the force downward on the wire from gravity, correct?

The equation for the force between two wires carrying identical currents with an unknown length is
F/L= (\muI)/(\mur)

For gravity

F= m*9.8m/s

i have a linear mass density of 50g/m.. I can take and divide both sides by L to get

F/L = (m*9.8m/s)/L but I'm afraid that doesn't help me.. Hmmm. Ill keep trying. Please any hints are helpful
 
Last edited:
Physics news on Phys.org
Please Help me out. I am missing something but i can't seem to connect the dots
 
If you have a linear density of M kg/m then you have a force of 9.8M N/m.
So F=9.8*0.05 N/m= 0.49N/m
 
so then i can set that force equal to the force between the wires and solve for I.
I get a value of 49kA when i do this.. this seems like a lot but it would also seem that an enormous amount of current would be needed to overcome the force of gravity. let me know what you think.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
4
Views
2K
Replies
5
Views
3K
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 8 ·
Replies
8
Views
4K