Question

A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficient of kinetic friction between the rod and rails is 0.18 and the kinetic friction force is 0.212 N , what vertical magnetic field is required to keep the rod moving at a constant speed of 5.1 m/s

Answer 1

Answer:

The magnetic field is $B = 8.20 *10^{-3} \ T$

Explanation:

From the question we are told that

   The  mass of the metal rod is  $m = 0.12 \ kg$

    The current on the rod is  $I = 4.1 \ A$

    The distance of separation(equivalent to length of the rod ) is $L = 6.3 \ m$

     The coefficient of kinetic friction is $\mu_k = 0.18$

      The kinetic frictional force is  $F_k = 0.212 \ N$

     The constant speed is $v = 5.1 \ m/s$

Generally the magnetic force on the rod is mathematically represented as  

      $F = B * I * L$

For  the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

        $F_ k = B* I * L$

=>      $B = \frac{F_k}{L * I }$

=>       $B = \frac{0.212}{ 6.3 * 4.1 }$

=>       $B = 8.20 *10^{-3} \ T$