Question

A brass rod with a mass of 0.300 kg slides on parallel horizontal iron rails, 0.440 m apart, and carries a current of 15.0 A. The coefficient of friction between the rod and rails is 0.300. What vertical, uniform magnetic field is needed to keep the rod moving at a constant speed

Answer 1

The magnitude of the magnetic field is
`B = 0.0890 \ T`

Step-by-step explanation:

From the question we are told that

The mass of the rod is
`m =0.300 \ kg`

The distance of separation is
`d = 0.440 \ m`

The current is
`I = 15.0 \ A`

The coefficient of friction is
`\mu = 0.300`

Generally for the rod the rod to continue moving at a constant speed

The frictional force must equal to the magnetic field force so

`F_m = F_f`

Where
`F_m = B* I * d`

and
`F_f = \mu * m * g`

`B*I *d = \mu * m * g`

=>
`B = (\mu * m * g )/(I * d )`

substituting values

`B = (0.2 * 0.300 * 9.8 )/( 15 * 0.440 )`

`B = 0.0890 \ T`