The Magnetic Dipole Moment of a Coil Problem A rectangular coil of dimensions 5.40 cm ✕ 8.50 cm consists of 25 turns of wire and carries a current of 20.0 mA. A 0.350 T magnetic field is applied parallel to the plane of the loop. A. What is the magnitude of its magnetic dipole moment?B. What is the magnitude of the torque acting on the loop?

Respuesta :

Answer:

a

 [tex]\mu = 0.0023 \ A\cdot m^2[/tex]

b

[tex]\tau = 0.00080 \ N \cdot m[/tex]

Explanation:

From the question we are told that

  The dimensions of the  rectangular coil is 5.40 cm ✕ 8.50 cm  = 0.054 m X 0.085 m

   The  number of turns is  [tex]N = 25 \ turns[/tex]

   The current it  is carrying is  [tex]I = 20 \ mA = 0.02 \ A[/tex]

    The magnetic field is [tex]B = 0.350 \ T[/tex]

Generally the magnitude of the magnetic dipole moment is mathematically represented as

       [tex]\mu = N * I * A[/tex]

Here  A  is the area which is mathematically represented as

     [tex]A = 0.054 * 0.085[/tex]

=>  [tex]A = 0.00459 \ m^2[/tex]

So  

     [tex]\mu = 25 * 0.02 * 0.00459[/tex]

=>  [tex]\mu = 0.0023 \ A\cdot m^2[/tex]

Generally the  magnitude of the torque acting on the loop is mathematically represented as

    [tex]\tau = \mu * B[/tex]

=>[tex]\tau = 0.0023 * 0.350[/tex]

=>[tex]\tau = 0.00080 \ N \cdot m[/tex]

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