Suppose a straight 1.50 mm -diameter copper wire could just "float" horizontally in air because of the force due to the Earth's magnetic field B, which is horizontal, perpendicular to the wire, and of magnitude 3.0×10−5 T . What current would the wire carry?
To determine the current the wire would carry, we need to use the formula that represents the force experienced by a current-carrying wire in a magnetic field:
F = BIL
Where:
F is the force experienced by the wire
B is the magnitude of the magnetic field
I is the current flowing through the wire
L is the length of the wire within the magnetic field
In this case, since the wire is floating horizontally, we can assume that the force experienced by the weight of the wire is exactly equal to the force experienced due to the magnetic field. Therefore, the force due to the magnetic field can be calculated as:
F = m * g
Where:
m is the mass of the wire (assuming uniform density, m = (πr²Lρ), where r is the radius of the wire, and ρ is the density of copper, which is 8,960 kg/m³)
g is the acceleration due to gravity (approximately 9.8 m/s²)
Now, since the magnetic force is given by F = BIL and the gravitational force by F = m * g, we can equate the two expressions:
BIL = m * g
Next, let's solve for the current I:
I = (m * g) / (B * L)
Now we can plug in the given values and calculate the current.
Given:
Diameter of the wire = 1.50 mm
Radius of the wire (r) = diameter / 2 = 1.5 mm / 2 = 0.75 mm = 0.75 × 10^-3 m
Magnetic field (B) = 3.0 × 10^-5 T
Density of copper (ρ) = 8,960 kg/m³
Gravity (g) = 9.8 m/s²
To find the length (L) of the wire required, additional information is needed. Could you provide the length of the wire within the magnetic field?