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

Answer:

The value is $B = 0.2312 \ T$

The direction is into the surface

Explanation:

From the question we are told that

   The mass density is  $\mu =\frac{m}{L} = 1.17 \ g/cm =0.117 kg/m$

   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

      $BIL = \mu_k * m * g$

=>   $B = \mu_k * \frac{m}{L} * [\frac{g}{I} ]$

=>   $B = 0.25 * 0.117 * [\frac{9.8}{1.24} ]$

=>   $B = 0.2312 \ T$

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