Suppose a straight 1.50 -diameter copper wire could just "float" horizontally in air because of the force due to the Earth's magnetic field , which is horizontal, perpendicular to the wire, and of magnitude 4.1×10−5 . Find the current the wire would carry.
a) let the current needed is I
Now,density of copper, p = 8960 Kg/m^3
Now, for the wire to levitate
lamda * g = B * I
8960 * pi * (0.0015/2)^2 = 5.5 *10^-5 * I
solving for I
I = 287 A
the current needed is 287 A
b) No
c) as this current would causes too high heat losses
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Let’s say you want to “float” a section of copper wire, which is
7.65 cm long and 2.34 mm in diameter within the magnetic field of
Earth near the Earth’s surface. Assume the magnetic field of Earth
makes a 31° with respect to the horizontal and has a magnitude of
3.5.×10-5 T. The current is flowing from west to east. See the
diagram given below.
Let's say you want to "float" a section of copper wire, which is 7.65 cm...
Consider a straight piece of copper wire of length 8 m and diameter 8 mm that carries a current I = 4.5 A. There is a magnetic field of magnitude B directed perpendicular to the wire, and the magnetic force on the wire is just strong enough to “levitate” the wire (i.e., the magnetic force on the wire is equal to its weight). Find B. Hint: The density of copper is 9000 kg/m3 .
Let's say you want to "float" a section of copper wire, which is 7.65 cm long and 2.34 mm in diameter within the magnetic field of Earth near the Earth's surface. Assume the magnetic field of Earth makes a 31° with respect to the horizontal and has a magnitude of 3.5.x10-5 T. The current is flowing from west to east. See the diagram given below. 5.Draw a Free Body Diagram of the wire. 22-degrees B 6. What is the value...