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

The magnetic field is

Explanation:

From the question we are told that

   The  mass of the metal rod is  

    The current on the rod is  

    The distance of separation(equivalent to length of the rod ) is

     The coefficient of kinetic friction is

      The kinetic frictional force is  

     The constant speed is

Generally the magnetic force on the rod is mathematically represented as  

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

=>      

=>      

=>      

v=ir = 3x15 =45 v