Question

A wire with a mass of 1.50 g/cm is placed on a horizontal surface with a coefficient of friction of 0.200. The wire carries a current of 1.52 A eastward and moves horizontally to the north. What are the magnitude and the direction of the smallest magnetic field that enables the wire to move in this fashion?

Answer 1

0.193 T

Inward direction

Step-by-step explanation:

`(m)/(L)` = Mass per unit length of wire = 1.50 g/cm =
`(1.50* 10^(-3)kg)/(0.01 m)` = 0.15 kg/m

`i` = magnitude of current = 1.52 A

`B` = magnitude of magnetic field = ?

`m` = mass of the wire

`L` = length of the wire

μ = Coefficient of friction = 0.200

For the wire to move,

magnetic force = frictional force

`i`
`B`
`L` = μ
`m`
`g`

(1.52)
`B`
`L` = (0.200) (9.8)
`m`

(1.52)
`B` = (0.200) (9.8)
`(m)/(L)`

(1.52)
`B` = (0.200) (9.8) (0.15)

`B` = 0.193 T

Direction of magnetic force is towards north and current is directed towards east. hence using right hand rule, the direction of magnetic field comes out to be in inward direction.