Question

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.

Answer 1

I = 2.99 A

Step-by-step 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 =
`(7.742)/(2.592)`

= 2.9869

I = 2.99 A

The required minimum current is 2.99 A.