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Question 1, chap 130, sect 1. part 1 of 1 10 points A wire in which there is a current of 6.45 A is to be formed into a circular loop of one turn. The permeability of free space is 1.25664 × 10−6 T · m/A . If the required value of the magnetic field at the center of the loop is 5.38 µT, what is the required radius? Correct answer: 75.3281 cm (tolerance ± 1 %). Explanation:

wire #1 a

II

45◦ C

O

45◦

y x

O III

a

I IV

wire #2 a D The direction of the magnetic field at C is 1. in the negative y direction. 2. out of the plane.

Let : B = 5.38 µT = 5.38 × 10−6 T , I = 6.45 A , and µ0 = 1.25664 × 10−6 T · m/A . The magnetic field at the center of circular loop is µ0 I 2R µ0 I R= 2B (1.25664 × 10−6 T · m/A) (6.45 A) = 2 (5.38 µT) B=

3. in quadrant I. 4. in the positive y direction. correct 5. in quadrant II. 6. in quadrant IV. 7. in quadrant III. 8. into the plane. 9. in the negative x direction. 10. in the positive x direction.

= 75.3281 cm . Explanation: wire #1 Question 2, chap 130, sect 4. part 1 of 4 10 points Consider two long parallel wires which are perpendicular to the plane of the paper (i.e., the x-y plane). Both wires carry the same current, I. Wire #1 intersects the plane a distance a above point O and wire #2 intersects the plane a distance a below point O. Point C is equidistant from both wires and is a distance a from point O.

B1

B2 C

O wire #2

From this figure, we can see by symmetry that the y components of the magnetic fields cancel, leaving only the component in the

2 positive x direction. Question 3, chap 130, sect 4. part 2 of 4 10 points What is the magnitude of the magnetic field at C ? 1 µ0 I 1. BC = √ 2 2π a 1 µ0 I 2. BC = 4π a 1 µ0 I 3. BC = π a 1 µ0 I 4. BC = √ 2π a 1 µ0 I 5. BC = 4 a

Question 4, chap 130, sect 4. part 3 of 4 10 points The direction of the magnetic field at D is 1. in the negative y direction. 2. in the positive y direction. 3. in quadrant II. 4. in the positive x direction. 5. in quadrant III. 6. in the negative x direction. correct 7. in quadrant I.

6. BC = 0

8. into the plane.

1 µ0 I 7. BC = correct 2π a 1 µ0 I 8. BC = 2 a 1 µ0 I 9. BC = √ 2 2 a 1 µ0 I 10. BC = √ 2 a Explanation: Note: The distance r from each wire to C is √ a r= = 2a. sin 45◦ The magnitude of the x component of the magnetic field due to each wire at C is given by µ0 I √ cos 45◦ 2π 2 a 1 µ0 I √ √ = 2π 2 a 2 µ0 I = 4πa

B1 = B2 =

Hence the contribution from both wires at C is just twice this value. BC = 2 B1 =

µ0 I . 2πa

9. out of the plane. 10. in quadrant IV. Explanation: wire #1

C

O wire #2

B2

B1 D The magnetic field due to wire 1 points to the left at D, while the magnetic field due to wire 2 points to the left at D. Because the magnetic field decreases as the distance increases, the magnetic field due to wire 2 is greater in magnitude than that due to wire 1. The resultant magnetic field vector points to the left (i.e., in the negative x direction).

3 Question 5, chap 130, sect 4. part 4 of 4 10 points What is the magnitude of the magnetic field at D? 1. BD = 2. BD = 3. BD = 4. BD = 5. BD = 6. BD = 7. BD = 8. BD = 9. BD =

5 µ0 I 6 πa 1 µ0 I 4 πa 1 µ0 I 6 πa 2 µ0 I correct 3 πa 1 µ0 I 3 πa 1 µ0 I 2 πa µ0 I πa 1 µ0 I 12 π a 3 µ0 I 4 πa

10. BD = 0 Explanation: The magnitude of the magnetic field at D is   1 µ0 I 1 ~ + kBD k = 2π a 3a 2 µ0 I = . 3 πa Note: Fourth of four versions. Question 6, chap 130, sect 2. part 1 of 1 10 points A long, thin conductor carries a current of 12.9 A. At what distance from the conductor is the magnitude of the resulting magnetic field 5.38 × 10−5 T? Correct answer: 4.79554 cm (tolerance ± 1 %). Explanation:

Let : µ0 = 1.25664 × 10−6 N/A2 , B = 5.38 × 10−5 T , and I = 12.9 A . By the Biot-Savart Law, B= r= = = =

µ0 I 2πr µ0 I 2πB (1.25664 × 10−6 N/A2) (12.9 A) 2 π (5.38 × 10−5 T) 0.0479554 m 4.79554 cm .

Question 7, chap 130, sect 3. part 1 of 1 10 points Two long straight wires carry currents perpendicular to the xy plane. One wire carries a current of 52 A and passes through the point x = 12 cm on the x axis. The second wire carries a current of 120 A and passes through the point y = 3.2 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin? Correct answer: 754.991 µT (tolerance ± 1 %). Explanation: Let : x = 12 cm = 0.12 m, I1 = 52 A, y = 3.2 cm = 0.032 m, I2 = 120 A, and µ0 = 1.25664 × 10−6 T · m/A. Basic Concepts: Magnetic Field due to a Long Straight Wire B =

µ0 I . 2πa

Solution: The magnetic field from the first wire will have magnitude B1 =

µ0 I1 2πx

4 and will point along the y axis at the origin. The magnetic field from the second wire will have magnitude B2 =

µ0 I2 , 2πy

and will point along the x axis at the origin. Since these contributions are perpendicular to each other, the magnitude of the total magnetic field at the origin will be q B = B12 + B22 s µ0 I1 µ0 I2 + = 2πx 2πy s   I1 I2 µ0 · = + y 2π x  1.25664 × 10−6 T · m/A = 2π  1/2 52 A 120 A × + 0.12 m 0.032 m = 0.000754991 T = 754.991 µT . Question 8, chap 130, sect 5. part 1 of 1 10 points An air-core solenoid has a length of 41.4 cm, a cross-sectional area of 0.08 m2, and contains 443 turns. The current through the solenoid increases by 8.5 A. The permeability of free space is 1.25664 × 10−6 N/A2. By how much does the magnetic flux through the solenoid change? Correct answer: 914.37 µWb (tolerance ± 1 %). Explanation: Let : ℓ = 41.4 cm = 0.414 m, A = 0.08 m2 , N = 443 turns, I = 8.5 A, and µ0 = 1.25664 × 10−6 N/A2.

The magnetic field inside the solenoid is B = µ0 n I = (1.25664 × 10−6 N/A2)   443 turns × (8.5 A) 0.414 m = 0.0114296 T .

∆ΦB = B A = (0.0114296 T) (0.08 m2) = 0.00091437 Wb = 914.37 µWb .

Question 9, chap 130, sect 5. part 1 of 1 10 points What current is required in the windings of a long solenoid that has 475 turns uniformly distributed over a length of 0.356 m in order to produce inside the solenoid a magnetic field of magnitude 8.61 × 10−5 T? The permeablity of free space is 1.25664 × 10−6 T m/A. Correct answer: 51.3511 mA (tolerance ± 1 %). Explanation:

Let : N = 475 , ℓ = 0.356 m , and B = 8.61 × 10−5 T . Magnetic field inside the solenoid is µ0 N I , then ℓ Bℓ I= µ0 N (8.61 × 10−5 T) (0.356 m) = µ0 (475)

B=

= 51.3511 mA .

Question 10, chap 130, sect 6. part 1 of 1 10 points

5 A solenoid 78.2 cm long has 840 turns and a radius of 2.28 cm. If it carries a current of 1.82 A, find the magnetic field along the axis at its center. Correct answer: 0.00245671 T (tolerance ± 1 %). Explanation: Let : l = 78.2 cm = 0.782 m , N = 840 turns , I = 1.82 A , and µ0 = 1.25664 × 10−6 N/A2 . Since l >> R, the magnetic field at the center is approximately B ≈ µ0 n I N = µ0 I l = (1.25664 × 10−6 N/A2) 840 turns (1.82 A) × 0.782 m = 0.00245671 T .

6. toward the wire Explanation: Use the right-hand rule to determine the direction of the magnetic field surrounding a long, straight wire carring a current (thumb). We find that the magnetic field points into the page at point R (fingers). Question 12, chap 130, sect 2. part 1 of 5 10 points A long, straight wire is in the plane of the page and carries a 16 A current. Point P is also in the plane of the page and is a perpendicular distance 0.7 m from the wire, as shown below. The permeability of free space is 4 π × −7 10 T · m/A. Gravitational effects are negligible. P z

0.7 m 16 A

y x

Question 11, chap 130, sect 2. part 1 of 1 10 points Consider a straight wire carrying current I , as shown. R I

x is upward from the paper With reference to the coordinate system in the figure above, what is the direction of the magnetic field at point P due to the current in the wire? 1. in the negative x direction

What is the direction of the magnetic field at point R caused by the current I in the wire?

2. in the negative z direction 3. in the positive x direction correct

1. out of the page

4. in the positive z direction

2. into the page correct

5. in the positive y direction

3. to the left

6. in the negative y direction

4. away from the wire 5. to the right

Explanation: Using the right hand rule, the magnetic field at point P due to the current in the wire

6 points out of the page, or along the positive bP = +b x-axis; i.e., B x. Question 13, chap 130, sect 2. part 2 of 5 10 points

A 2.324 × 10−26 kg mass particle with positive 1.12 × 10−18 C charge is initially moving parallel to the wire with a 130 m/s speed when it is at point P , as shown below.

Question 14, chap 130, sect 2. part 3 of 5 10 points Determine the magnitude of the magnetic force acting on the positive particle at point P. Correct answer: 6.656 × 10−22 N (tolerance ± 1 %). Explanation:

−26

2.324 × 10 kg 1.12 × 10−18 C z 0.7 m 16 A

130 m/s

y

x x is upward from the paper With reference to the coordinate system established above, what is the direction of the magnetic force acting on the positive particle at point P ?

Let : q = 1.12 × 10−18 C , µ0 = 1.25664 × 10−6 T · m/A , I = 16 A , d = 0.7 m , and v0 = 130 m/s . The magnitude of the magnetic field at point P is

1. in the negative x-axis

µ0 I 2πd (1.25664 × 10−6 T · m/A) (16 A) = 2 π (0.7 m)

2. in the negative z-axis

= 4.57143 × 10−6 T .

3. in the positive x-axis 4. in the negative y-axis 5. in the positive y-axis 6. in the positive z-axis correct Explanation: The direction of the magnetic force is F~B = q ~v0 × ~BP , so bP FbB = b v0 × B = (−b y ) × (+b x) = +b z .

Using the right hand rule, the direction of the magnetic force acting on the positive particle at point P is along the positive z-axis, or upward.

BP =

Therefore, the magnitude of the magnetic force on the positive particle at point P is FB = q v0 BP = |(1.12 × 10−18 C)| × (130 m/s) (4.57143 × 10−6 T) = 6.656 × 10−22 N . Question 15, chap 130, sect 2. part 4 of 5 10 points An electric field is applied that causes the net force on the positive particle to be zero at point P . With reference to the coordinate system established above, what is the direction of the electric field at point P that could accomplish this? 1. in the negative x-axis

7 ℓ

2. in the negative z-axis correct 3. in the negative y-axis

P

4. in the positive x-axis 5. in the positive z-axis

I

6. in the positive y-axis

What is the magnitude of the magnetic field at point P (at the center of the square loop) due to the current in the wire?

Explanation: To make the net force on the positive particle zero, the electric force on the particle must exactly cancel the magnetic force on the particle, so the electric force must point downward, or along the negative z-axis. Since the charge of the particle is positive, it means that the electric field must point along the negative z-axis. Question 16, chap 130, sect 2. part 5 of 5 10 points Determine the magnitude of the electric field. Correct answer: 0.000594286 N/C (tolerance ± 1 %). Explanation:

~ = 1. kBk ~ = 2. kBk ~ = 3. kBk ~k= 4. kB ~ = 5. kBk ~ = 6. kBk ~ = 7. kBk ~ = 8. kBk ~k= 9. kB

q E = FB FB E= q 6.656 × 10−22 N = 1.12 × 10−18 C = 0.000594286 N/C .

~ = 10. k Bk

2 µ0 I √ 2 3πℓ √ µ0 I 3 πℓ 2 µ0 I √ 3 πℓ 4 µ0 I √ 2 3πℓ µ I √0 2πℓ 2 µ0 I √ 2 correct πℓ 2 µ0 I πℓ µ0 I √ 2 πℓ 4 µ0 I πℓ µ0 I πℓ

Explanation: By the Biot-Savart law,

dB = Question 17, chap 130, sect 1. part 1 of 2 10 points A conductor in the shape of a square of edge length ℓ carries a counter-clockwise current I as shown in the figure below.

µ0 I d~s × ˆr . 4π r2

Consider a thin, straight wire carring a constant current I along the x-axis with the y axis pointing towards the center of the square, as in the following figure.

8 Thus, we have reduced the expression to one involving only the variable θ. We can now obtain the total field at P by integrating Eq. 4 over all elements subtending angles ranging from θ1 to θ2 as defined in the figure. This gives the magnetic field as

y P r

a θ

ˆr ds

x I Let us calculate the total magnetic field at the point P located at a distance a from the wire. An element d~s is at a distance r from P . The direction of the field at P due to this element is out of the paper, since d~s × ˆr is out of the paper. In fact, all elements give a contribution directly out of the paper at P . Therefore, we have only to determine the magnitude of the field at P . In fact, taking the origin at O and letting P be along the positive y axis, with ˆk being a unit vector pointing out of the paper, we see that x

O

d~s × ˆr = kˆ |d~s × ˆr| = kˆ (dx sin θ) . Substituting into Biot-Savart law gives ~ = kˆ dB, with dB dB =

µ0 I dx sin θ . 4π r2

(1)

In order to integrate this expression, we must relate the variables θ, x, and r. One approach is to express x and r in terms of θ. From the geometry in the figure and some simple differentiation, we obtain the following relationship a = a csc θ . (2) sin θ a Since tan θ = − from the right triangle in x the figure, r=

x = −a cot θ and dx = a csc2 θ dθ .

(3)

Substituting Eqs. 2 and 3 into Eq. 1 gives µ0 I a csc2 θ sin θ dθ 4π a2 csc2 θ µ0 I sin θ dθ . = 4πa

dB =

(4)

Z θ2 µ0 I sin θ dθ 4 π a θ1 µ0 I (cos θ1 − cos θ2) . = 4πa

B=

We can apply this result to this problem. For a segment of wire, as set up above, the magnetic field at a point P is B=

µ0 I (cos θ1 − cos θ2) . 4πa

For a square wire loop consider the bottom segment. Using the above general formula for this case gives   π 3π µ0 I   cos − cos Bone = 4 ℓ 4 4π 2 √ µ0 I 2, = 2πℓ where the total magnetic field is B = 4 Bone =

2 µ0 I √ 2 , πℓ

since all four sides contribute. The direction and magnitude of the field is the same for each side. Question 18, chap 130, sect 1. part 2 of 2 10 points What is the direction of the magnetic field b BP at point P due to the downward current in the left-hand side of the square wire? b is to the right. 1. B

b is to the left. 2. B

9 b is zero. 3. B

b is down the page. 4. B

b is out of the page. correct 5. B

b is in to the page. 6. B

b is up the page. 7. B

Explanation: The direction of the magnetic field due to a current element is determined by the cross product in the definition of the magnetic field Z µ0 I d~s × ˆr ~ B= . 4π r2 For the present case the right hand rule gives the direction of the magnetic field as out of the page or out for short....