A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diame

Question

A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don’t concern us).
What is the magnitude of the force on this wire assuming the solenoid’s field points due east?

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Thiên Ân 4 years 2021-08-08T12:13:39+00:00 1 Answers 49 views 0

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    2021-08-08T12:15:00+00:00

    Complete Question

    A 590-turn solenoid is 12 cm long. The  current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don’t concern us).

    What is the magnitude of the force on this wire assuming the solenoid’s field points due east?

    Answer:

    The force is  F =  0.1602 \ N

    Explanation:

    From the question we are told that

       The number of turns is  N =  590 \ turns

       The  length of the solenoid is  L  = 12 \ cm  =  0.12 \ m

       The current is  I  =  36 \ A

       The  diameter is  D =  4.5 \ cm  =  0.045 \ m

       The  current carried by the wire is  I  = 27 \ A

        The  length of the wire is  l   =  2 cm  =  0.02 \ m

    Generally the magnitude of the force  on this wire assuming the solenoid’s field points due east is mathematically represented as

               F =  B  *  I  *  l

    Here  B  is the magnetic field which is mathematically represented as

              B  =  \frac{\mu_o  *  N  *  I }{L}

    Here   \mu _o is permeability of free space with value  \mu_ o  =  4\pi *10^{-7} \  N/A^2

    substituting values

             B  =  \frac{4 \pi *10^{-7}  *   590  *   36 }{ 0.12}

               B  =   0.2225 \ T

    So

          F =  0.2225   *  36  *  0.02

          F =  0.1602 \ N

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