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

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

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Gia Bảo 4 years 2021-08-14T09:46:54+00:00 1 Answers 11 views 0

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  1. minhkhang
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    2021-08-14T09:47:55+00:00
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    Answer:

    The magnetic field is B = 8.20 *10^{-3} \ T

    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 = \frac{F_k}{L * I }

    =>       B = \frac{0.212}{ 6.3 * 4.1 }

    =>       B = 8.20 *10^{-3} \ T

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