A circular loop of wire with area A lies in the xy-plane. As viewed along the z-axis looking in the −z-direction toward the origin, a curren

Question

A circular loop of wire with area A lies in the xy-plane. As viewed along the z-axis looking in the −z-direction toward the origin, a current I is circulating clockwise around the loop. The torque produced by an external magnetic field B⃗ is given by τ⃗ =D(5i^−5j^), where D is a positive constant, and for this orientation of the loop the magnetic potential energy U=−μ⃗ ⋅B⃗ is negative. The magnitude of the magnetic field is B0=14D/IA.

A) Determine the vector magnetic moment of the current loop.

Express your answer in terms of the variables I, A, i^, j^, and k^.

B) Determine the component Bx of B⃗ .

Express your answer using one significant figure.

C) Determine the component By of B⃗ .

Express your answer using one significant figure.

D) Determine the component Bz of B⃗ .

Express your answer using two significant figures.

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Orla Orla 5 years 2021-08-13T21:12:07+00:00 2 Answers 119 views 0

Answers ( )

  1. trucchi
    0
    2021-08-13T21:13:18+00:00
    Reply

    Answer:

    A) Vector μ = -IA•k^

    B) Bx = 5D/IA

    C) By = 5D/IA

    D) Bz = -12D/IA

    Explanation:

    I have attached the explanation for ease of understanding

  2. giabao
    0
    2021-08-13T21:13:28+00:00
    Reply

    Answer:

    a. μ~ = – IA •k

    b. Bx = 5D/IA

    c. By = 5D/IA

    d. Bz = -12.1D/IA

    Explanation:

    Given that,

    Torque, τ = D(5i^ − 5j^) Nm

    Potential energy is gvwn as

    U=− μ•B

    Magnitude of magnetic field is

    Bo=14D/IA

    a. The vector magnetic moment of the current loop is given as

    μ~ = – μ•k

    μ~ = – IA •k

    b. Now to find the component of the magnetic field B.

    Assume B = Bx •i + By •j + Bz •k

    Then, torque is given as

    τ = μ~ ×B

    τ = – IA •k × (Bx •i + By •j + Bz •k)

    Note that

    i×i=j×j×k×k=0

    i×j=k. j×i=-k

    j×k=i. k×j=-i

    k×i=j. i×k=-j

    Then,

    τ = – IA •k × (Bx •i + By •j + Bz •k)

    τ= -IABx•(k×i) – IABy•(k×j) – IABz•(k×k)

    τ= -IABx•j + IABy•i

    τ= IABy•i – IABx•j

    The given torque is τ = D(5i^ − 5j^)

    Comparing coefficient

    Then,

    IABy=5D

    Then, By= 5D/IA

    c. Also,

    -IABx=-5D

    Bx=-5D/-IA

    Bx=5D/IA

    d. To get Bz, let use the magnitude of magnetic field Bo

    Bo²=Bx²+By²+Bz²

    (14D/IA)²=(5D/IA)²+(5D/IA)² + Bz²

    Bz²=(14D/IA)²- (5D/IA)²-(5D/IA)²

    Bz²=196D²/I²A²-25D²/I²A²-25D²/I²A²

    Bz²=(196D²-25D²-25D²)/I²A²

    Bz²=146D²/I²A²

    Bz=√(146D²/I²A²)

    Bz=± 12.1 D/IA

    So we want to determine if Bz is positive or negative

    From the electric potential,

    U=− μ•B

    U= – – IA k•(Bx i+By j+Bz k)

    Note, -×- =+, i.i=j.j=k.k=1

    i.j=j.k=k.i=0

    Then,

    U= IA k•(Bx i+By j+Bz k)

    U=IABz

    Since we are told that U is negative, then this implies that Bz is negative

    Then, Bz= -12.1D/IA

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