Suppose a straight 1.25 mm diameter copper wire could just "float" horizontally in air because of the force of the Earth's magnetic field B, which is horizontal, perpendicular to the wire, and of magnitude 5.00 multiplied by 10-5 T. What current would the wire carry?
I want to use the formula F=IlBsin(theta); but am confused as to what F would be, what does just "float" horizontally imply?
I won't be able to answer exactly, but we know that the net force on the wire would be zero. So it means that Force(gravity) = Force(magnetic)
Hey, I thought about it a little more and I have your answer here.
B = mu(naught) * I / (2pi * r)
You can solve for I in that equation. I hope it looks familiar.
mu(naught) = vacuum permeability
I = current
r = radius of wire
In this scenario, "float" horizontally implies that the gravitational force acting on the wire is balanced by the magnetic force exerted on it. Therefore, we can set the magnitudes of these two forces equal to each other:
Fgravitational = Fmagnetic
The gravitational force acting on the wire can be calculated using the formula:
Fgravitational = mg
where m is the mass of the wire and g is the acceleration due to gravity (approximately 9.8 m/s²).
The magnetic force exerted on the wire can be calculated using the formula:
Fmagnetic = IlBsin(theta)
where I is the current flowing through the wire, l is the length of the wire, B is the magnitude of the Earth's magnetic field, and theta is the angle between the wire and the magnetic field (which is 90 degrees in this case).
Since the wire is floating, the gravitational force and the magnetic force must be equal:
mg = IlBsin(theta)
Simplifying the equation, we can solve for the current I:
I = (mg) / (lBsin(theta))
Given that the diameter of the wire is 1.25 mm, we can calculate the radius as 0.625 mm or 0.000625 m. Assuming the wire is a cylinder, its length l can be taken as the radius.
Using a typical value for the density of copper (8.96 g/cm³), we can find the mass m of the wire:
m = (density) x (volume)
= (8.96 g/cm³) x (πr²l)
= (8.96 g/cm³) x (π(0.000625 m)²(0.000625 m))
= (8.96 g/cm³) x (π x 0.00000000039 m³)
= 0.000000012 g
Converting the mass to kilograms:
m = 0.000000012 kg
Plugging in the known values into the equation for the current, we have:
I = ((0.000000012 kg) x (9.8 m/s²)) / ((0.000000625 m) x (5.00 x 10^-5 T) x (sin(90°)))
I = 0.196 A
Therefore, the wire would carry approximately 0.196 Amps of current in order to just "float" horizontally in the Earth's magnetic field.
In this scenario, the term "float" horizontally implies that the magnetic force acting on the wire balances out the weight of the wire, causing it to remain suspended in midair. In other words, the magnetic force upwards is equal to the gravitational force downwards.
To calculate the current that the wire would carry, you can start by analyzing the forces acting on the wire. In this case, we have the magnetic force (F) and the gravitational force acting on the wire.
The magnetic force is given by the formula F = IlBsin(theta), where:
F = magnetic force
I = current flowing through the wire
l = length of the wire
B = magnitude of the Earth's magnetic field
(theta) = angle between the wire and the magnetic field (which is perpendicular in this case)
Since the wire is floating horizontally, the angle (theta) between the wire and the magnetic field is 90 degrees, which means sin(theta) = 1. Therefore, the formula becomes F = IlB.
Now, we know that the magnetic force balances out the weight of the wire, so we have:
IlB = mg
Where:
m = mass of the wire
g = acceleration due to gravity
The mass of the wire can be calculated using its length and radius. Since the diameter is given as 1.25 mm, the radius is half that value (0.625 mm = 0.625 x 10^-3 m). The length of the wire is not provided in the question, so let's assume it is 1 meter.
Substituting the equation IlB = mg and solving for I, we get:
I = mg / (lB)
Substituting the known values:
m = πr^2ρ (where ρ is the density of copper)
g = 9.81 m/s^2
l = 1 m (given in the question)
B = 5.00 x 10^-5 T (given in the question)
r = 0.625 x 10^-3 m (radius of the wire)
ρ = 8.96 x 10^3 kg/m^3 (density of copper)
Now we can plug in these values into the equation I = mg / (lB) to find the current carried by the wire.