Torque on a Current Loop, 2

There is a force on sides 2 & 4 since they are perpendicular to the field

The magnitude of the magnetic force on these sides will be: F2 = F4 = I a B

The direction of F2 is out of the page

The direction of F4 is into the page

Torque on a Current Loop, 3

The forces are equal and in opposite directions, but not along the same line of action

The forces produce a torque around point O

Torque on a Current Loop, Equation

The maximum torque is found by:

The area enclosed by the loop is ab, so τmax = IAB This maximum value occurs only when the field is

parallel to the plane of the loop

2 42 2 2 2max (I ) (I )

I

b b b bτ F F aB aB

abB

Torque on a Current Loop, General

Assume the magnetic field makes an angle of

< 90o with a line perpendicular to the plane of the loop

The net torque about point O will be τ = IAB sin

Use the active figure to vary the initial settings and observe the resulting motion

PLAYACTIVE FIGURE

Torque on a Current Loop, Summary

The torque has a maximum value when the field is perpendicular to the normal to the plane of the loop

The torque is zero when the field is parallel to the normal to the plane of the loop

where is perpendicular to the plane of the loop and has a magnitude equal to the area of the loop

I A B

A

Direction

The right-hand rule can be used to determine the direction of

Curl your fingers in the direction of the current in the loop

Your thumb points in the direction of

A

A

Magnetic Dipole Moment

The product I is defined as the magnetic dipole moment, , of the loop Often called the magnetic moment

SI units: A · m2

Torque in terms of magnetic moment:

Analogous to for electric dipole

A

B

p E

Chapter 30

Sources of the Magnetic Field

Biot-Savart Law – Introduction

Biot and Savart conducted experiments on the force exerted by an electric current on a nearby magnet

They arrived at a mathematical expression that gives the magnetic field at some point in space due to a current

Biot-Savart Law – Set-Up

The magnetic field is at some point P

The length element is

The wire is carrying a

steady current of I

Please replace with fig. 30.1

dB

ds

Biot-Savart Law – Observations

The vector is perpendicular to both and to the unit vector directed from toward P

The magnitude of is inversely proportional to r2, where r is the distance from to P

dBr̂

dB

ds

ds

ds

What does this tell you about the magnetic field, ?

1 2 3

33% 33%33%dB

ds

r̂ds

r̂

dB

1. It goes like the scalar dot product of and

2. It goes like X

3. is usually zero

0 of 30

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

Biot-Savart Law – Observations, cont

The magnitude of is proportional to the current and to the magnitude ds of the length element

The magnitude of is proportional to sin where is the angle between the vectors and

ds

r̂ds

dB

dB

The observations are summarized in the mathematical equation called the Biot-Savart law:

The magnetic field described by the law is the field due to the current-carrying conductor Don’t confuse this field with a field external to the

conductor

Biot-Savart Law – Equation

24oμ d

dπ r

s rB

ˆI

Permeability of Free Space

The constant o is called the permeability of free space

o = 4 x 10-7 T. m / A

Total Magnetic Field

is the field created by the current in the length segment ds

To find the total field, sum up the contributions from all the current elements I

The integral is over the entire current distribution

dB

24oμ d

π r

s rB

ˆI

ds

Biot-Savart Law – Final Notes

The law is also valid for a current consisting of charges flowing through space

represents the length of a small segment of space in which the charges flow For example, this could apply to the electron

beam in a TV set

ds

Compared to

Distance The magnitude of the magnetic field varies as the

inverse square of the distance from the source The electric field due to a point charge also varies

as the inverse square of the distance from the charge

B

E

Compared to , 2

Direction The electric field created by a point charge is

radial in direction The magnetic field created by a current element is

perpendicular to both the length element and the unit vector r̂

ds

B

E

Compared to , 3

Source An electric field is established by an isolated

electric charge The current element that produces a magnetic

field must be part of an extended current distribution Therefore you must integrate over the entire current

distribution

B

E

Which variable can be pulled out of the integral?

1 2 3 4

25% 25%25%25%1. ds

2. sinθ

3. r2

4. None of them

0 of 30

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

How are θ and Φ related?

1 2 3 4

25% 25%25%25%

1. Φ = θ – π/2

2. Φ = θ

3. Φ = π/2 – θ

4. Φ = θ + π/2

0 of 30

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

for a Long, Straight Conductor, Special Case

The field becomes

2

IoμB

πa

B

for a Long, Straight Conductor, Direction The magnetic field lines are

circles concentric with the wire

The field lines lie in planes perpendicular to to wire

The magnitude of the field is constant on any circle of radius a

The right-hand rule for determining the direction of the field is shown

B

for a Curved Wire Segment

Find the field at point O due to the wire segment

I and R are constants

will be in radians4

IoμB θ

πR

B

What about the contribution from the wires coming in and going out?

1 2 3

33% 33%33%

ds

r̂

1. They are distant enough to neglect their contribution

2. X = 0

3. The two currents cancel each other

0 of 30

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

for a Curved Wire Segment

Find the field at point O due to the wire segment

I and R are constants

will be in radians4

IoμB θ

πR

B

for a Circular Loop of Wire

Consider the previous result, with a full circle = 2

This is the field at the center of the loop

24 4 2

o o oμ μ μB θ π

πa πa a

I I I

B

for a Circular Current Loop

The loop has a radius of R and carries a steady current of I

Find the field at point P

B

What can we pull out of the integral this time?

1 2 3 4

25% 25%25%25%1. r2

2. Sin θ

3. ds

4. nothing

0 of 30

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

for a Circular Current Loop

The loop has a radius of R and carries a steady current of I

Find the field at point P

2

32 2 22

ox

μ aB

a x

I

B

Comparison of Loops

Consider the field at the center of the current loop

At this special point, x = 0 Then,

This is exactly the same result as from the curved wire

2

32 2 2 22

o ox

μ a μB

aa x

I I