Question

A current-carrying wire passes through a region of space that has a uniform magnetic field of 0.92 T. If the wire has a length of 2.6 m and a mass of 0.60 kg, determine the minimum current needed to levitate the wire. A

Answer 1

Answer:

Current, I = 2.45 T

Explanation:

It is given that,

Magnetic field, B = 0.92 T

Length of wire, l = 2.6 m

Mass, m = 0.6 kg

We need to find the minimum current needed to levitate the wire. It is given by balancing its weight to the magnetic force i.e.

$Ilb=mg$

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

$I=\dfrac{0.6\ kg\times 9.8\ m/s^2}{2.6\ m\times 0.92\ T}$

I = 2.45 A

So, the minimum current to levitate the wire is 2.45 T. Hence, this is the required solution.