A copper wire 12 cm long and a mass of 16 g is in a magnetic field of 1.2 T oriented perpendicular to the wire. What current would create a force strong enough to levitate the wire?

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aachen aachen · Answer 1 · 2019-09-11T13:42:12+02:00

Answer:

Current, I = 1.08 A

Explanation:

Given that,

Length of the copper wire, l = 12 cm = 0.12 m

Mass of the wire, m = 16 g = 0.016 kg

Magnetic field, B = 1.2 T

The magnetic field is oriented perpendicular to the wire. We need to find the current in the wire.

Magnetic force, F = BIL

Force of gravity, F' = mg

F' = F

[tex]BIL=mg[/tex]

[tex]I=\dfrac{mg}{BL}[/tex]

[tex]I=\dfrac{0.016\times 9.8}{1.2\times 0.12}[/tex]

I = 1.08 A

So, the current through the wire is 1.08 A. Hence, this is the required solution.