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

Answer:

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

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     $F_f = \mu * m * g$

   $B*I *d = \mu * m * g$

=>    $B = \frac{\mu * m * g }{I * d }$

substituting values

       $B = \frac{0.2 * 0.300 * 9.8 }{ 15 * 0.440 }$

       $B = 0.0890 \ T$