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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

1 Answer

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Answer:

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

User Ian Hopkinson
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