A circular loop of wire with area AA lies in the xyxy-plane. As viewed along the zz-axis looking in the −z−z-direction toward the origin, a current I is circulating clockwise around the loop. The torque produced by an external magnetic field B⃗ B→ is given by τ⃗ =D(2i^−2j^)τ→=D(2i^−2j^), where DD is a positive constant, and for this orientation of the loop the magnetic potential energy U=−μ⃗ ⋅B⃗ U=−μ→⋅B→ is negative. The magnitude of the magnetic field is B0=13D/IAB0=13D/IA.

The remaining part of the question is;

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.

A) μ^ = - IA•k

B) Bx = 2D/IA

C) By = 2D/IA

D) Bz = -13D/(IA)

Step-by-step explanation:

We are given;

Torque; τ = D(2i^ − 2j^) Nm

Potential energy; U =− μ•B

Magnitude of magnetic field;

Bo=13D/IA

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

μ^ = - μ•k

μ^ = - IA •k

b. Now, let's find the component of the magnetic field B.

If we 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(2i^ − 2j^)

Comparing coefficients;

Then,

IABy=2D

Then, By= 2D/IA

c. Also,

-IABx = -2D

Bx = -2D/-IA

Bx = 2D/IA

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

Bo² = Bx² + By² + Bz²

(13D/IA)²=(2D/IA)²+(2D/IA)² + Bz²

Bz²=(13D/IA)²- (2D/IA)²-(2D/IA)²

Bz² = 169D²/(I²A²) - 4D²/(I²A²) - 4D²/(I²A²)

Bz² = (169D² - 4D²- 4D²)/I²A²

Bz² = 161D²/I²A²

Bz = √(161D²/(I²A²))

Bz = ± 13D/(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= -13D/(IA)

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !

c

step-by-step explanation: