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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.

Respuesta :

Kazeemsodikisola

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

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

Ver imagen AFOKE88
Ver imagen AFOKE88

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