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

The magnetic field is
`B = 0.0764 \ T`

Step-by-step 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 = (\mu_k * m * g)/(I * d )`

substituting values

=>
`B = (0.200 * 0.210 * 9.8 )/( 11 * 0.490 )`

=>
`B = 0.0764 \ T`