Question

A wire of 0.50m length is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.98T. The vurrent in the wire is 2.0A in the direction shown. What is the mass of the wire if the current and the magnetic field are sufficient to remove the tension in the supporting leads?

Answer 1

100 g

Step-by-step explanation:

Answer 2

0.1 kg or 100 g

Step-by-step explanation:

The length of the wire = 0.5 m

the field magnitude = 0.98 T

the current through the wire = 2.0 A

magnetic force due to a wire carrying current is

F =
`IlB`

where

F is the force

`I` is the current = 2 A

`l` is the length of the wire

B is the magnetic field strength

Substituting, we have

F = 2 x 0.5 x 0.98 = 0.98 N

This force balances the weight of the mass

weight = mg

where m is the mass of the wire

g is acceleration due to gravity = 9.81 m/s^2

therefore, weight = m x 9.81 = 9.81m

equating this weight with the force, we have

0.98 = 9.81m

m = 0.98/9.81 = 0.099 kg ≅ 0.1 kg or 100 g