Consider a straight piece of copper wire of length 1.5 m and diameter 1 mm that

Question

Consider a straight piece of copper wire of length 1.5 m and diameter 1 mm that carries current I = 4.0 A. A magnetic field of magnitude B is 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 9,000 kg/m3.

FB = ILB sin

Fgrav = L(d2/4)g

Using these relations for FB and Fgrav, what is the minimum value of B needed to just levitate the wire?

Explanation / Answer

here FB=FGRAV

ILB sin = L(d2/4)g

B = 9000*3.14*10^-6*9.8/4*4

B = 0.017 T

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