Question

A 0.210-kg metal rod carrying a current of 11.0 A glides on two horizontal rails 0.490 m apart. If the coefficient of kinetic friction between the rod and rails is 0.200, what vertical magnetic field is required to keep the rod moving at a constant speed?

Answer 1

Answer:

The  magnetic field is  $B = 0.0764 \ T$

Explanation:

From the question we are told that  

    The mass of the metal is  $m = 0.210 \ kg$

     The current is  $I = 11.0 \ A$

      The distance between the rail(length of the rod ) is  $d = 0.490 \ m$

      The coefficient of kinetic friction is  $\mu_k = 0.200$

Generally the magnetic force is mathematically represented as

      $F_b = B * I * d$

Given that the rod is moving at a constant velocity, it

=>    $F_b = F_k$

Where $F_k$ is the kinetic frictional force which is mathematically represented as

       $F_k = \mu_k * m * g$

So

    $B * I * d = \mu_k * m * g$

=>   $B = \frac{\mu_k * m * g}{I * d }$

substituting values

=>   $B = \frac{0.200 * 0.210 * 9.8 }{ 11 * 0.490 }$

=>   $B = 0.0764 \ T$