How Can a Magnetic Field Make a Copper Wire Levitate?

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SUMMARY

The discussion focuses on calculating the magnetic field required to levitate a 1.0 cm copper wire carrying a current of 50.0 A. The correct approach involves determining the wire's mass using its volume and density, with the formula for the volume of a cylinder, v = πr²h. The final calculations yield a magnetic field strength of 0.14 T directed out of the page when the force is upward. The participants emphasize the importance of using the right-hand rule to determine the magnetic field direction.

PREREQUISITES
  • Understanding of magnetic fields and forces on current-carrying wires
  • Familiarity with the right-hand rule for determining magnetic field direction
  • Knowledge of volume calculation for cylindrical objects
  • Basic physics concepts including mass, weight, and force equations
NEXT STEPS
  • Study the right-hand rule in detail for magnetic field direction determination
  • Learn about the Lorentz force law and its applications
  • Explore the properties of magnetic fields generated by current-carrying conductors
  • Investigate the principles of electromagnetic levitation and its practical applications
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in the practical applications of magnetic fields in technology and engineering.

Mitchtwitchita
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Homework Statement



A 1.0 cm copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic field that produces on it an upward force exactly equal in magnitude to the wire's weight, causing the wire to "levitate." What are the field's direction and magnitude.

Homework Equations



B = [(Mu)I]/[2(pi)r]

The Attempt at a Solution



I don't really know how to get this one started. Am I supposed to determine the weight of the copper wire by the area and the atomic weight? Can somebody please help me out with this one?
 
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Mitchtwitchita said:
I don't really know how to get this one started. Am I supposed to determine the weight of the copper wire by the area and the atomic weight? Can somebody please help me out with this one?
You are supposed to find the mass from the mass density and the volume. Also you need to use the equation for the force on a current-carrying wire in a magnetic field. The equation that you posted gives the magnetic field due to a very long wire. Not the same thing.
 
I don't know how to find the mass through the mass density and the volume. I assume the volume would be that of a cylinder?
 
Mitchtwitchita said:
I don't know how to find the mass through the mass density and the volume. I assume the volume would be that of a cylinder?
Yes, what expression gives the volume of a cylinder?
 
Last edited:
m = dv
d = 8.94 g/cm^3
v = (pi)r^2h
=(3.14)(0.50 cm)^2(100cm)
=78.5 cm^3

Therefore, m = dv
=(8.94 g/cm^3)(78.5 cm^3)
=7.01 x 10^2 g
=0.701 kg?
 
F = ma
=(0.701 kg)(9.81 m/s^2)
=6.88 N

B = F/IL
=(6.88 N)/(50 A)(1.0 m)
=0.14 T

Does this answer look correct to you?
 
Your mass calculation is incorrect. You have the wrong radius and the wrong length for the wire. Your calculation for the B field is the correct approach. However, you still need to specify the direction of the field. Hint: Use the right hand rule.
 
Oops, I wrote the wrong units for the length of the wire. I apologize, it should read 1.0 m. Let me make another attempt at it.
 
v = (pi)r^2h
=(3.14)(0.050 cm)^2(100 cm)
=0.785 cm^3

m = dv
=(8.94 g/cm^3)(0.785)
=7.02 g
=7.02 x 10^-3 kg

F = ma
=(7.02 x 10^-3)(9.81 m/s^2)
=6.88 x 10^-2 N

Therefore, B would equal 1.4 x 10^-3 T, and the direction of the field would be coming out of the page if it means the force is up in the field of the page.
 
  • #10
Mitchtwitchita said:
... and the direction of the field would be coming out of the page if it means the force is up in the field of the page.
In what direction is the current relative to the page?
 
  • #11
The current is to the east. I wasn't sure if they meant up as in a 3-D view of the ground up or just up in the plane of the page. If they mean that the force is up in the plane of the page, the field would be coming out of the page. If they mean that the force is coming out of the page as if it were a 3-D view of the ground, then the field would be going up in the plane of the page.
 
  • #12
Suppose you hold the page so that its plane is parallel to the ground. Draw a line right to left and turn the page about a vertical axis so that left-to-right is west to east. Gravity acts perpendicular to the page and down. You want the magnetic force to be up. In what direction should the magnetic field point?
 
  • #13
The magnetic force on what? The magnetic force is q VxB, so it depends on the velocity of the charge as well as the magnetic field. In what direction is the charge moving?
 
  • #14
phyzguy said:
The magnetic force on what? The magnetic force is q VxB, so it depends on the velocity of the charge as well as the magnetic field. In what direction is the charge moving?
Please read the statement of the problem carefully to answer this question for yourself.
 
  • #15
kuruman said:
Please read the statement of the problem carefully to answer this question for yourself.

Sorry, I missed the original statement. Please ignore my post.
 
  • #16
So, the magnetic field should point North.
 
  • #17
Correct.
 
  • #18
Thanks for your time kuruman!
 

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