Many metals become superconductors (zero resistance, zero loss, zero heating) at low temperatures. If they are warmed above the critical tempareature, the superconductivity is lost. Also, if the metal experiences a magnetic field above a critical value, the superconductivity is lost. Consider an aluminum wire with 1 mm diameter below its critical temperature of 1.2 Kelvin. It's critical magnetic field is 0.01 Tesla. What is the maximum current in Amps that you can send down this wire for it to remain superconducting?
r=d/2 =0.5 mm =0.0005 m
I(max) = B(cr) c r /2 = 0.01•3•10⁸•0.0005 /2 = 750 A
Well, well, well, looks like we have a wire in a "super" situation here! Let's calculate the maximum current that our aluminum wire can handle without losing its superconductivity.
To find the maximum current, we need to know the critical magnetic field value and the diameter of the wire. You've already provided us with those values, so we're good to go!
Now, the formula we need to use is called the critical current density (Jc) formula. It goes like this: Jc = Bc / (μ0 * wire diameter), where Bc is the critical magnetic field and μ0 is the permeability of free space.
First things first, let's find μ0. It's a constant value, and its numerical value is approximately 4π x 10^-7 T m/A.
Now, plugging in the values, we can calculate the critical current density (Jc). Bearing in mind that the diameter of the wire is given in millimeters, we need to convert it to meters.
Once we have Jc, we can calculate the maximum current using the formula I = Jc * wire cross-sectional area (π * (wire diameter / 2)^2).
Let the calculations begin!
To determine the maximum current that can be sent down the superconducting aluminum wire without losing its superconductivity, we need to use the critical magnetic field value and the diameter of the wire.
The critical magnetic field (Bc) is given as 0.01 Tesla, and the diameter of the wire is 1 mm. To find the maximum current (I) we can apply the formula:
I = Bc * (d/2)
Where:
I is the maximum current
Bc is the critical magnetic field
d is the diameter of the wire
Converting the wire diameter to meters:
1 mm = 0.001 meters
Substituting the values into the formula:
I = 0.01 Tesla * (0.001m/2)
I = 0.01 Tesla * 0.0005m
I = 0.000005 Tesla-m/amp
Hence, the maximum current that can be sent down the superconducting aluminum wire for it to remain superconducting is approximately 0.000005 Amps.
To find the maximum current in Amps that you can send down the superconducting aluminum wire, we need to consider the critical magnetic field and the properties of the wire.
First, we need to determine the cross-sectional area of the wire. We know the diameter of the wire is 1 mm, which is equal to 0.001 meters. The area of a circle can be calculated using the formula A = πr^2, where r is the radius of the wire. In this case, the radius is half of the diameter, so it is 0.001/2 = 0.0005 meters.
Now we can calculate the area:
A = π(0.0005)^2
A ≈ 0.000000785 square meters
Next, we consider the critical magnetic field. The critical magnetic field for the aluminum wire is given as 0.01 Tesla.
To calculate the maximum current, we can use the formula for the critical current density (Jc) of a superconductor:
Jc = Bc / μ₀,
where Jc is the critical current density, Bc is the critical magnetic field, and μ₀ is the permeability of free space (approximately 4π × 10^-7 T·m/A).
Plugging in the values:
Jc = 0.01 / (4π × 10^-7)
Jc ≈ 79577.4715 A/m^2
To find the maximum current, we need to multiply the critical current density by the cross-sectional area of the wire:
Max Current = Jc × A
Max Current ≈ 79577.4715 × 0.000000785
Max Current ≈ 0.0624 Amps
Therefore, the maximum current that you can send down the superconducting aluminum wire with a 1 mm diameter, below its critical temperature of 1.2 Kelvin, without losing its superconductivity is approximately 0.0624 Amps.