Question

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?

Answer 1

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

$BIL=mg$

$I=\dfrac{mg}{BL}$

$I=\dfrac{0.016\times 9.8}{1.2\times 0.12}$

I = 1.08 A

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