A current of 4.0 A is maintained in a single circular loop having a circumference of 80 cm. An external magnetic field of 2.0 T is directed

Question

A current of 4.0 A is maintained in a single circular loop having a circumference of 80 cm. An external magnetic field of 2.0 T is directed so that the angle between the field and the plane of the loop is 20°. Determine the magnitude of the torque exerted on the loop by the magnetic forces acting upon it. Group of answer choices 0.27 N ⋅ m 0.41 N ⋅ m 0.38 N ⋅ m 0.77 N ⋅ m 0.14 N ⋅ m

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Thanh Thu 3 years 2021-07-19T05:40:40+00:00 1 Answers 50 views 0

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  1. Amity
    0
    2021-07-19T05:42:14+00:00
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    Answer:

    0.14 N.m

    Explanation:

    Applying,

    T = BIAsinФ………….. Equation 1

    Where T = Torque, B = magnetic Field, I = current, A = Area of the circular loop, Ф = angle between the field and the plane loop.

    But,

    A = πr²…………… Equation 2

    And,

    c = 2πr…………….. Equation 3

    Where c = circumference of the circular loop, r = radius of the circular loop

    make r  the subject of the equation in equation 3

    r = c/2π………….. Equation 4

    Give: c = 80 cm = 0.8 m, π = 3.14.

    Substitute into equation 4 to get r

    r = 0.8/(2×3.14)

    r = 0.127 m.

    Substitute into equation 2 to get the area of the circular loop.

    A = 3.14(0.127²)

    A = 0.051 m².

    Also Given: B = 2 T, I = 4 A, Ф = 20°

    Substitute into equation 1

    T = 2(4)(0.051)(sin20°)

    T = 0.408(0.342)

    T = 0.1395 N.m

    T ≈ 0.14 N.m

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