Question

A wire with a linear mass density of 1.17 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic friction between the wire and the surface is 0.250. If the wire carries a current of 1.24 A westward and moves horizontally to the south, determine the magnitude and direction of the smallest magnetic field that can accomplish this.

Answer 1

The value is
`B = 0.2312 \ T`

The direction is into the surface

Step-by-step explanation:

From the question we are told that

The mass density is
`\mu =(m)/(L) = 1.17 \ g/cm =0.117 kg/m`

The coefficient of kinetic friction is
`\mu_k = 0.250`

The current the wire carries is
`I = 1.24 \ A`

Generally the magnetic force acting on the wire is mathematically represented as

`F_F = F_B`

Here
`F_F` is the frictional force which is mathematically represented as

`F_F = \mu_k * m * g`

While
`F_B` is the magnetic force which is mathematically represented as

`F_B = BILsin(\theta )`

Here
`\theta =90^o` is the angle between the direction of the force and that of the current

So

`F_B = BIL`

So

`BIL = \mu_k * m * g`

=>
`B = \mu_k * (m)/(L) * [(g)/(I) ]`

=>
`B = 0.25 * 0.117 * [(9.8)/(1.24) ]`

=>
`B = 0.2312 \ T`

Apply the right hand curling rule , the thumb pointing towards that direction of the current we see that the direction of the magnetic field is into the surface as shown on the first uploaded image