Exercise 28 - notendur.hi.isisa13/Efni/Einn/S.hl/Edlisfraedi II... · The figure is a sectional...

Heimadæmi 7Due: 11:00pm on Thursday, March 3, 2016

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Exercise 28.11

A long, straight wire lies along the axis and carries a 3.80 current in the . Find the magnetic field(magnitude and direction) produced at the following points by a 0.600 segment of the wire centered at the origin.

Part A

,

Enter your answers numerically separated by commas.

ANSWER:

All attempts used; correct answer displayed

Part B

, ,


ANSWER:

Correct

Part C

, ,


ANSWER:

Correct

Part D

, ,


ANSWER:

z − − A +z − direction− mm

x = 2.00 m, y = 0 z = 0

= 0,5.70×10−11,0 , ,Bx By Bz T

x = 0 y = 2.00 m z = 0

= −5.70×10−11,0,0 , ,Bx By Bz T

x = 2.00 m y = 2.00 m z = 0

= −2.02×10−11,2.02×10−11,0 , ,Bx By Bz T

x = 0 y = 0 z = 2.00 m

Correct

Exercise 28.18

Certain fish, such as the Nile fish (Gnathonemus), concentrate charges in their head and tail, thereby producing an electricfield in the water around them. This field creates a potential difference of a few volts between the head and tail, which inturn causes current to flow in the conducting seawater. As the fish swims, it passes near objects that have resistivitiesdifferent from that of seawater, which in turn causes the current to vary. Cells in the skin of the fish are sensitive to thiscurrent and can detect changes in it. The changes in the current allow the fish to navigate.Since the electric field is weakfar from the fish, we shall consider only the field running directly from the head to the tail. We can model the seawaterthrough which that field passes as a conducting tube of area and having a potential difference across its ends. These fishnavigate by responding to changes in the current in seawater. This current is due to a potential difference of around 3.00 generated by the fish and is about 12.0 within a centimeter or so from the fish. Receptor cells in the fish are sensitiveto the current. Since the current is at some distance from the fish, their sensitivity suggests that these cells might beresponding to the magnetic field created by the current. To get some estimate of how sensitive the cells are, we can modelthe current as that of a long, straight wire with the receptor cells 2.00 away.

Part A

What is the strength of the magnetic field at the receptor cells?

ANSWER:

Correct

Exercise 28.23

Two long, straight, parallel wires, 10.0 apart carry equal 4.00 currents in the same direction, as shown in the figure .

= 0,0,0 , ,Bx By Bz T

VmA

cm

= 0.12 B μT

cm A

Part A

Find the magnitude of the magnetic field at point , midway between the wires.

ANSWER:

Correct

Part B

What is its direction?

ANSWER:

Correct

Part C

Find the magnitude of the magnetic field at point , 25.0 to the right of .

ANSWER:

Correct

P1

= 0 B T

to the left

to the right

upward

downward

no field

P2 cm P1

= 6.67×10−6 B T

Part D


ANSWER:

Correct

Part E

Find the magnitude of the magnetic field at point , 20.0 directly above .

ANSWER:

Correct

Part F


ANSWER:

Correct

Problem 28.67

The figure is a sectional view of two circular coils with radius , each wound with turns of wire carrying a current ,circulating in the same direction in both coils. The coils are separated by a distance equal to their radii. In thisconfiguration the coils are called Helmholtz coils; they produce a very uniform magnetic field in the region between them.

to the left

to the right

upward

downward

no field

P3 cm P1

= 7.53×10−6 B T

to the left

to the right

upward

downward

no field

a N Ia

Part A

Derive the expression for the magnitude of the magnetic field at a point on the axis a distance to the right of point , which is midway between the coils.

ANSWER:

Correct

Part B

From part (a), obtain an expression for the magnitude of the magnetic field at point .

Express your answer in terms of the variables , , , and appropriate constant ( ).

ANSWER:

Correct

Part C

Calculate the magnitude of the magnetic field at if 310 turns, = 7.00 , and = 7.00 .

B xP

B = ( + )N Iμ0 a2

21

((x+a/2 −)2 a2)3/21

((x−a/2 −)2 a2)3/2

B = ( − )N Iμ0 a2

21

((x+a/2 −)2 a2)3/21

((x−a/2 +)2 a2)3/2

B = ( + )N Iμ0 a2

21

((x+a/2 +)2 a2)3/21

((x−a/2 +)2 a2)3/2

B = ( − )N Iμ0 a2

21

((x+a/2 +)2 a2)3/21

((x−a/2 +)2 a2)3/2

P

N I a μ0

= BNIμ0

a 54

32

P I A a cm

ANSWER:

Correct

Part D

Calculate at .

ANSWER:

Correct

Part E

Calculate at .

ANSWER:

Correct

Part F

Discuss how your results in parts d) and e) show that the field is very uniform in the vicinity of .

ANSWER:

Graded, see 'My Answers' for details

Force between an Infinitely Long Wire and a Square Loop

A square loop of wire with side length carries a current . The center of the loop is located a distance from an infinitewire carrying a current . The infinite wire and loop are in the same plane; two sides of the square loop are parallel to thewire and two are perpendicular as shown.

= 2.79×10−2 B T

dB/dx P(x = 0)

= 0dBdx

∣∣x=0

B/dd2 x2 P(x = 0)

= 0Bd2

dx2∣∣x=0

P

3754 Character(s) remaining

Breytingin er engin við agnarsmáa breytingu dx

a I1 dI2

Part A

What is the magnitude, , of the net force on the loop?

Express the force in terms of , , , , and .

Hint 1. How to approach the problem

You need to find the total force as the sum of the forces on each straight segment of the wire loop. You'll savesome work if you think ahead of time about which forces might cancel.

Hint 2. Determine the direction of force

Which of the following diagrams correctly indicates the direction of the force on each individual line segment?

Hint 1. Direction of the magnetic field

In the region of the loop, the magnetic field points into the plane of the paper (by the righthand rule).

Hint 2. Formula for the force on a currentcarrying conductor

The magnetic force on a straight wire segment of length , carrying a current with a uniform magneticfield along its length, is

,

where is a vector along the wire in the direction of the current.

F

I1 I2 a d μ0

l IB

F = I ×l B

l

ANSWER:

Hint 3. Determine the magnitude of force

Which of the following diagrams correctly indicates the relative magnitudes of the forces on the parallel wiresegments?

Hint 1. Find the magnetic field due to the wire

What is the magnitude, , of the wire's magnetic field as a function of perpendicular distance from thewire, .

Express the magnetic field magnitude in terms of , , and .

Hint 1. Ampère's law

Use Ampère's law to obtain the magnetic field. Ampère's law states that

,where the line integral can be done around any closed loop.

ANSWER:

ANSWER:

a

b

c

d

Br

I2 r μ0

∮ ⋅ =B dl→

μ0 Iencl

= Bμ0I2

2πr

a

b

c

d

Hint 4. Find the force on the section of the loop closest to the wire

What is the magnitude of the force on the section of the loop closest to the wire, that is, a distance from it?

Express your answer in terms of , , , , and .

Hint 1. Formula for the force on a currentcarrying conductor

The magnetic force on a straight wire segment of length , carrying a current with a uniform magneticfield along its length, is

,

where is a vector along the wire in the direction of the current.


What is the magnitude, , of the wire's magnetic field as a function of perpendicular distance from thewire, .





ANSWER:

ANSWER:


What is the magnitude, , of the wire's magnetic field as a function of perpendicular distance from the wire, .





ANSWER:

F1 d − a/2

μ0 I1 I2 a d

l IB

F = I ×l B

l

Br

I2 r μ0

∮ ⋅ =B dl→

μ0 Iencl

= Bμ0I2

2πr

= F1aI2

μ0I1

2π(d− )a

2

B r

I2 r μ0

∮ ⋅ =B dl→

μ0 Iencl

ANSWER:

Correct

Part B

The magnetic moment of a current loop is defined as the vector whose magnitude equals the area of the loop timesthe magnitude of the current flowing in it ( ), and whose direction is perpendicular to the plane in which thecurrent flows. Find the magnitude, , of the force on the loop from Part A in terms of the magnitude of its magneticmoment.

Express in terms of , , , , and .

ANSWER:

Correct

The direction of the net force would be reversed if the direction of the current in either the wire or the loop werereversed. The general result is that "like currents" (i.e., currents in the same direction) attract each other (or, morecorrectly, cause the wires to attract each other), whereas oppositely directed currents repel. Here, since the likecurrents were closer to each other than the unlike ones, the net force was attractive. The corresponding situationfor an electric dipole is shown in the figure below.

Magnetic Field due to a Wire Conceptual Question

The same amount of current is flowing through two wires, labeled 1 and 2 in the figure, in the directions indicated by thearrows. In this problem you will determine the direction of the net magnetic field at each of the indicated points (A C).

= Bμ0I2

2πr

= F ( − )aμ0I1I2

2π1

d− a

2

1d+ a

2

m m = IA

F

F m I2 a d μ0

= F ( − )mμ0I2

2πa1

d− a

2

1d+ a

2

I

B net

Part A

What is the direction of the magnetic field at point A? Recall that the currents in the two wires have equalmagnitudes.

Hint 1. The magnitude of the magnetic field due to a long, straight currentcarrying wire

The magnitude of a magnetic field is directly proportional to the amount of current flowing in the wire andinversely proportional to the distance from the wire:

.

Hint 2. The direction of the magnetic field due to a long, straight currentcarrying wire

The magnetic field surrounding a long, straight wire encircles the wire, as shown in the figure:

The direction of the field is determined by a righthand rule: Grasp the wire with the thumb of your right hand inthe direction of the current flow. The direction in which your fingers encircle the wire is the direction in which themagnetic field encircles the wire.

Hint 3. How to approach the problem

To determine the direction of the magnetic field at point A, you must determine the contribution to the field fromboth of the wires. The field at point A is the vector sum of these two contributions.Because point A is in the same plane as the wires, the contribution to the net magnetic field at point A from wire1 or wire 2 will either point into or out of the screen. Keep in mind that if the magnetic fields are in opposite

B net

B B Ir

B = Iμ0

2πr

directions, the larger field will decide the direction of the net magnetic field. If they are the same size, the netmagnetic field will be zero.

Hint 4. Find the direction of the magnetic field at point A due to wire 1

Is the magnetic field from wire 1 directed into or out of the screen at point A? Be sure to point your thumb in thedirection of the current, in this case to the right.

ANSWER:

Hint 5. Find the direction of the magnetic field at point A due to wire 2

Is the magnetic field from wire 2 directed into or out of the screen at point A? Be sure to point your thumb in thedirection of the current, in this case downward.

ANSWER:

ANSWER:

Correct

Part B

What is the direction of the magnetic field at point B?

Hint 1. Find the direction of the magnetic field at point B due to wire 1

Is the magnetic field from wire 1 directed into or out of the screen at point B?

ANSWER:

Hint 2. Find the direction of the magnetic field at point B due to wire 2

Is the magnetic field from wire 2 directed into or out of the screen at point B?

ANSWER:

in

out

in

out

points out of the screen at A.

points into the screen at A.

points neither out of nor into the screen and at A.

at A.

B net

B net

B net ≠ 0B

net

= 0B net

B net

in

out

ANSWER:

Correct

Part C

What is the direction of the magnetic field at point C?

Hint 1. Find the direction of the magnetic field at point C due to wire 1

Is the magnetic field from wire 1 directed into or out of the screen at point C?

ANSWER:

Hint 2. Find the direction of the magnetic field at point C due to wire 2

Is the magnetic field from wire 2 directed into or out of the screen at point C?

ANSWER:

ANSWER:

Correct

Score Summary:

in

out

points out of the screen at B.

points into the screen at B.

points neither out of nor into the screen and at B

at B.

B net

B net

B net ≠ 0B

net

= 0B net

B net

in

out

in

out

points out of the screen at C.

points into the screen at C.

points neither out of nor into the screen and at C.

at C.

B net

B net

B net ≠ 0B

net

= 0B net

Your score on this assignment is 94.9%.You received 6.65 out of a possible total of 7 points.

Exercise 28 - notendur.hi.isisa13/Efni/Einn/S.hl/Edlisfraedi II... · The figure is a sectional...

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Transcript of Exercise 28 - notendur.hi.isisa13/Efni/Einn/S.hl/Edlisfraedi II... · The figure is a sectional...