A current-carrying wire passes through a region of space that has a uniform magnetic field of 0.96 T. If the wire has a length of 2.7 m and a mass of 0.79 kg, determine the minimum current needed to levitate the wire.

I = 2.99 A

Explanation:

magnetic field, B = 0.96 T, length of wire, l = 2.7 m, mass of wire, m = 0.79 kg.

Since it is expected that the wire would vibrate or move in the magnetic field, from Newton's second law of motion;

F = mg

where F is the force on the wire, m is the mass of the wire and g is the acceleration due to gravity.

But,

F = BIL

⇒ BIL = mg

0.96 x I x 2.7 = 0.79 x 9.8

2.592I = 7.742

I =

 = 2.9869

I = 2.99 A

The required minimum current is 2.99 A.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.