Magnetic Fields and Equilibrium

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Homework Help Overview

The problem involves a wire bent into a square connected to a 9 Volt battery, requiring the determination of the magnetic field needed to maintain equilibrium. The wire's properties, including its material (copper), dimensions, and the battery's characteristics, are provided, along with relevant equations related to magnetic fields and forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the system, including weight and magnetic force, and question the necessity of considering electric force and the battery's mass. There are attempts to calculate current and magnetic field strength using different approaches, with some participants suggesting corrections to the length used in calculations.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the calculations of current and the magnetic field. Some participants provide guidance on the application of equations and the significance of the battery's mass in the context of equilibrium. There is no explicit consensus on the final values or methods, but productive dialogue is ongoing.

Contextual Notes

Participants note the importance of using the entire length of the wire for resistance calculations and the specific sections of the wire that contribute to the magnetic force. There is also mention of the need to clarify the role of the battery's mass in the overall force balance.

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Homework Statement



A wire is bent into a square connected to the terminals of a 9 Volt battery. The wire is constructed of copper with diameter of 1 mm and is bent into a square with side length of 30 cm. If the battery has a mass of 50 g and
internal resistance 0.1 Ω, what magnetic field (magnitude and direction) is needed in the shaded region to keep the system in equilibrium? The resistivity of copper is 1.7 x 10^(-8) Ω m with a density of 8920 kg/cubic meter.

3013jwy.jpg


Homework Equations



B = (μI)/(2πL)
Force in a wire = LI x B = LIB
V = IR
r = internal resistance = 0.1 Ω
Resistivity ρ = (RA)/L
d = 0.3 m

The Attempt at a Solution



Well...I know that net force = 0 if the system is in equilibrium. The forces that I have identified are the weight and the magnetic force. Should I consider electric force as a possible force? Why do I even need the mass of the battery?

Resistivity ρ = (RA/L)
r = (ρL)/A = (1.7 x 10^(-8) Ω m)(0.3 m)/(π x 0.0005² m²) = 0.0065 Ω
I = V/R = 9V/(0.1 Ω + 0.0065 Ω) = 84.5 A
current flows clockwise from the positive terminal to the negative terminal.
B = (μI)/(2πL) = [(4π x 10^-7 Tm/A)(84.5 A)]/(2π x 0.3 m) = 5.63 x 10^-5 T
length L = 0.3 m
μ = 4π x 10^-7 Tm/A
Mass of the top part of the wire = (density)(volume) = (8920 kg/cubic meter)(π x 0.0005² m²)(0.3 m) = 0.0021 kg
weight = mg = (0.0021 kg)
Net force = magnetic force - weight = 0
LIB - mg = 0
LIB = mg
B = (mg)/(LI) = (0.0021 kg x 9.8 m/s²)/(0.3 m x 84.5 A) = 8.12 x 10^-4 T

In order to find the direction of the magnetic field, I tried using the right hand rule and found that the magnetic field points into the page. Does this seem to make any sense? I also think that the two parallel current carrying wires in the shaded region have currents in opposite directions so their forces should also cause them to repel.

After solving it two different ways, I got two different answers of B = 8.12 x 10^-4 T and B = 5.63 x 10^-5 T. I'm also having trouble telling if the magnetic field points inward or outward; I think it points inward after using the right hand rule. Please help if you can. Thank you so much.
 
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One thing I see is that when you are trying to find the current in the wire, you should take L to be the length of the entire loop, not just the one side. So it should be 1.2 m instead of 0.3 m. The same thing goes for finding the total mass of the wire.

I don't think B = (μI)/(2πL) applies to this problem, as you are trying to find the magnitude of an external magnetic field that will provide the upwards force. This equation is for the magnetic field due to current in a wire.

If the current is clockwise, then the magnetic field will point into the page for an upwards force.
 
Thanks for your help, so now I get...
Net force = Magnetic Force - weight = 0
LIB - mg = 0
LIB = mg
Mass of the wire = (density)(volume) = (8920 kg/cubic meter)(π x 0.0005² m²)(1.2 m) = 0.0084 kg
B = (mg)/(LI) = (0.0084 kg x 9.8 m/s²)/(1.2 m x 84.5 A) = 8.125 x 10^-4 T

Does this look better? Thank you so much for your help. I really appreciate it, but I still don't understand why the problem states the mass of the battery.
 
but I still don't understand why the problem states the mass of the battery.

The mass of the battery is needed because it is providing a downwards force, so it must be included so the magnetic field is strong enough to hold it up (I am assuming the battery is not touching the ground, that is the point of the problem).

I don't think your current value is right (it didn't appear to change anyway), you must include the entire length of the wire when you are calculating the resistance of the wire (since the current goes through the entire wire).

B = (mg)/(LI) = (0.0084 kg x 9.8 m/s²)/(1.2 m x 84.5 A) = 8.125 x 10^-4 T

Now, when you get to this last step of calculating B, you want to use only 0.3 m for L, since it is only the top section of the wire that is interacting with the magnetic field (the vertical parts of the loop don't contribute to the upwards force).

Your approach looks OK you just have to work out a few of the little details. Go through it very carefully and check that everything makes sense!
 
Oh, ok. That makes sense. Wow, I am so incompetent.
Resistivity ρ = (rA/L)
r = (ρL)/A = (1.7 x 10^(-8) Ω m)(1.2 m)/(π x 0.0005² m²) = 0.0260 Ω
I = V/R = 9V/(0.1 Ω + 0.0260 Ω) = 71.4 A
Net force = Magnetic Force - weight of wire - weight of battery = 0
Magnetic Force = weight of wire + weight of battery
LIB = (mass of wire)g + (mass of battery)g
Mass of the wire = (density)(volume) = (8920 kg/cubic meter)(π x 0.0005² m²)(1.2 m) = 0.0084 kg
B = (mg + mg)/(LI) = (0.0084 kg x 9.8 m/s² + 0.05 kg x 9.8 m/s²)/(0.3 m x 71.4 A) = 0.0267 T
The magnetic force points upward so the magnetic field must point inward via the right hand rule.

Wow, my answer is now totally different from my previous answer. I hope this is right. Thank you so much for all your help! :)
 
You're welcome. I hope it's right too! I think it should be.
 

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