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

The magnetic field is
`B = 8.20 *10^(-3) \ T`

Step-by-step 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 = (F_k)/(L * I )`

=>
`B = (0.212)/( 6.3 * 4.1 )`

=>
`B = 8.20 *10^(-3) \ T`