Question

Consider a straight piece of copper wire of length 7 m and diameter 3.5 mm that carries a current I = 5.5 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 .

Answer 1

Given

Current I = 5.5 A

Length of wire l = 7m

Diameter of wire d = 3.5mm = 3.5 x 10^-3 m

Radius of wire r = 3.5/2 x 10^-3 m = 1.75 x 10^-3 m

Density of copper = 9000 kg/m³

Given magnetic force on the wire is equal to its weight

That is at particular distance R magnetic force is balanced with the force due to gravity therefore B =

B=μ0I4πR=mg

therefore B can also be calculated by calculating B = mg

where g = 9.8m/s²

we know that

Density = Mass/Volume

so, Mass of copper wire m = Density x Volume = Density x r²l

m = 9000 x (3.14 x (1.75 x 10^-3)² x 7 = 605823.75 x 10^-6 kg = 0.606 kg

so, Magnetic force B = mg = 0.606 x 9.8 = 5.9388 Tesla

B = 5.9388 Tesla