A circular coil of wire 8.40 cm in diameter has 17.0 turns and carries a current of 3.20 A . The coil is in a region where the magnetic fiel

Question

A circular coil of wire 8.40 cm in diameter has 17.0 turns and carries a current of 3.20 A . The coil is in a region where the magnetic field is 0.610 T.Required:a. What orientation of the coil gives the maximum torque on the coil ?b. What is this maximum torque in part (A) ?c. For what orientation of the coil is the magnitude of the torque 71.0 % of the maximum found in part (B)?

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Thanh Thu 5 months 2021-08-11T11:26:10+00:00 1 Answers 5 views 0

Answers ( )

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    2021-08-11T11:27:15+00:00

    Answer:

    a) for the torque to be maximum, sin should be maximum

    i.e (sinФ)maximum = 1

    b) therefore the Maximum torque is

    Tmax = 0.1838 × 1 = 0.1838  N.m

    c) Given the torque is 71.0% of its maximum value; Ф  = 45.24⁰ ≈ 45⁰

    Explanation:

    Given that; Diameter is 8.40 cm,

    Radius (R) = D/2 = 8.40/2 = 4.20 cm = 0.042 m

    Number of turns (N) = 17

    Current in the loop (I) = 3.20 A

    Magnetic field (B) = 0.610 T

    Let the angle between the loop’s area vector A and the magnetic field B be

    Now. the area of the loop is;

    A = πR²

    A = 3.14 ( 0.042 )²

    A =  0.005539 m²

    Torque on the loop (t) = NIABsinФ

    t = 17 × 3.20 ×0.005539 × 0.610 × sinФ

    t = 0.1838sinФ N.m

    for the torque to be maximum, sin should be maximum

    i.e (sinФ)maximum = 1

    therefore the Maximum torque is

    Tmax = 0.1838 × 1 = 0.1838  N.m

    Given the torque is 71.0% of its maximum value

    t = 0.71 × tmax

    t = 0.71 × 0.1838

    t = 0.1305

    Now

    0.1305 N.m =  0.1838 sinФ N.m

    sinФ = 0.1305 / 0.1838

    sinФ = 0.71001

    Ф = sin⁻¹ 0.71001

    Ф  = 45.24⁰ ≈ 45⁰

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