Question

A 2.5 cm-long horizontal wire segment has a mass of 0.15 g and carries a current sufficiently large that it levitates when the segment is in a region of uniform B field of magnitude 1.5 T. Assume that the B field is horizontal and perpendicular to the wire segment. The current is passed to the segment by way of massless electrical leads. What is the magnitude of the current in the wire

Answer 1

The magnitude of the current in the wire is 3.92x10⁻²A

Step-by-step explanation:

Given:

L = 2.5 cm = 0.025 m

m = 0.15 g = 1.5x10⁻⁴kg

B = 1.5 T (horizontal and perperdicular to the wire segment)

g = gravity = 9.8 m/s²

Question: What is the magnitude of the current in the wire, I = ?

The intensity of the current in the wire is given by

`I=(mg)/(BL)`

Here,

m is the mass, g is the gravity, B is the magnetic field, and L is the length

Substituting values:

`I=(1.5x10^(-4)*9.8 )/(1.5*0.025) =3.92x10^(-2) A`