Chapter 29 Magnetic Fields Due to Currents

In this chapter we will explore the relationship between an electric current and the magnetic field it generates in the space around it. We will follow a two-prong approach, depending on the symmetry of the problem.

For problems with low symmetry we will use the law of Biot-Savart in combination with the principle of superposition.

For problems with high symmetry we will introduce Ampere’s law.

Both approaches will be used to explore the magnetic field generated by currents in a variety of geometries (straight wire, wire loop, solenoid coil, toroid coil) We will also determine the force between two parallel current carrying conductors. We will then use this force to define the SI unit for electric current (the Ampere)

(29 – 1)

A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratioof their speeds, vproton/valpha, is:

A. 0.5B. 1C. 2D. 4E. 8

A proton (charge e), traveling perpendicular to a magnetic field in a circular orbit, experiences the same force as an alpha particle (charge 2e, mass 4p) which is also traveling perpendicular to the same field. The ratio of their orbital radii rproton/ralpha, is:

A. 0.5 B. 1 C. 2 D. 4 E. 8

A proton (charge e), traveling perpendicular to a magnetic field in a circular orbit, experiences the same force as an alpha particle (charge 2e, mass 4p) which is also traveling perpendicular to the same field. The ratio of their orbital frequency fproton /f alpha is:

This law gives the magnetic field generated by a wire

segment of length that carries a current . Consider

the geometry shown in the figure. Associated with the

element

dB

ds i

ds

The law of Biot - Savart

we define an associated vector that has

magnitude whch is equal to the length . The direction

of is the same as that of the current that flows through

segment .

ds

ds

ds

ds

A

3

The magnetic field generated at point P by the element located at point A

is given by the equation: . Here is the vector that connects 4

point A (location of element ) w

o

dB dS

i ds rdB r

r

ds

7 6

2

ith point P at which we want to determine .

The constant 4 10 T m/A 1.26 10 T m/A and is known as

sin" ". The magnitude of is:

4Here is the angle

o

o

dB

i dsdB dB

r

permeability constant

between and . ds r

34oi ds r

dBr

(29 – 2)

The magnitude of the magnetic field

generated by the wire at point P

located at a distance from the wire

is given by the equation:

2

oiBR

R

Magnetic field generated by a long

straight wire

2

oiBR

The magnetic field lines form circles that have their centers

at the wire. The magnetic field vector is tangent to the

magnetic field lines. The sense for is given by the

. We po

B

B right

hand rule

int the thumb of the right hand in the

direction of the current. The direction along which the

fingers of the right hand curl around the wire gives the

direction of .B

(29 – 3)

2

Consider the wire element of length shown in the figure.

The element generates at point P a magnetic field of

sinmagnitude Vector is pointing

4 the page. The magnetic field

o

ds

i dsdB dB

r

into

20 0

2 2 2 2

generated by the whole

wire is found by integration.

sin2

2

sin sin / /

oi dsB dB dB

r

r s R R r R s R

3/ 2

3/ 2 2 2

2 2 22 2

2 20 0

2 2 2o o oi i

dx x

a x a

iRd

x

s sB

R Rs Rs R

a

2oiBR

(29 – 4)

. .dB

B

R

Magnetic field generated by

a circular wire arc of radius

at its center C

2 2

A wire section of length generates at the center C a magnetic field

sin 90The magnitude The length

4 4

Vector points of the page4

The

o o

o

ds dB

i ids dsdB ds Rd

R Ri

dB d dBR

out

0

net magnettic field =4

The angle must be expressed in radians

For a circular wire 2 . In this case we get:

4

2

o

ocirc

o i

R

iB

iB dB

R

R

d

Note :

(29 – 5)

4oiBR

2 2

The wire element generates a magnetic field

sin 90whose magnitude

4 4

We decompose into two components:

One (

o o

ds dB

i ids dsdB

r r

dB

Magnetic field generated by a circular loop

along the loop axis

d ) along the z-axis.The second component (d )

is in a direction perpendiculat to the z-axis. The sum of

all the d is equal to zero. Thus we sum only the d terms.

B B

B B

2

3/ 22 2

2 22

3/ 23 2 2

3/ 2 3/ 22 2 2 2

cos cos cos

4

4 4

24 4 2

o

o o

o oo

i ds Rr R z dB dB

r riR iRds ds

dB B dBr R z

iR iRB ds R

R z R z

iR

R z

(29 – 6)

The law of Biot-Savart combined with the

principle of superposition can be used to

determine if we know the distribution

of currents. In situations that have high

symmetry we can use inst

B

Ampere's Law.

ead Ampere's

law, because it is simpler to apply.

Ampere's law can be derived from the law of Biot-Savart with which it is

mathematically equivalent. Ampere's law is more suitable for advances

formulations of electronmagnetism. It can be expressed as follows:

o

The line integral of the magnetic field along any closed path

is equal to the total current enclosed inside the path multiplied by

B ds B

The closed path used is known an an " ". In its present form

Ampere's law is not complete. A missing term was added by Clark Maxwell.

The complete form of Ampere's law will be discussed in

Amperian loop

chapter 32.

o encB ds i

(29 – 7)

o encB ds i

1 2

1 1

Determination of . The closed path is

divided into n elements , , ...,

We then from the sum:

cos Here is the

mag

n

n n

i i i i i ii i

B ds

s s s

B s B s B

Implementation of Ampere's law :

1.

1

netic field in the i-th element.

lim as

Calculation of We curl the fingers

of the right hand in the direction in which

the Amperian loop was traversed. We note

the

n

i ii

enc

B ds B s n

i

2.

direction of the thumb.

All currents inside the loop to the thumb are counted as .

All currents inside the loop to the thumb are counted as .

All currents outside the loop are not count

pa

antiparal

rallel positive

lel negative

1 2

ed.

In this exampl e : enci i i (29 – 8)

We already have seen that the magnetic field lines

of the magnetic field generated by a long straight

wire that carries a current have the form of

circles

i

Magnetic field outside a long straight wire

which are cocentric with the wire.

We choose an Amperian loop that reflects the

cylindrical symmetry of the problem. The

loop is also a circle of radius that has its center

on the wire. The magnetic fi

r

eld is tangent to the

loop and has a constant magnitude .

cos0 2

Ampere's law holds true for any closed path.

We choose to use the path that makes the

2

o enc

o

o

B

B ds Bds B ds rB i

iB

r

i

Note :

calculation

of as easy as possible.

B

(29 – 9)

We assume that the distribution of the current

within the cross-section of the wire is uniform.

The wire carries a current and has radius .

We choose an Amper

i R

Magnetic field inside a long straight wire

2 2

2 2

ian loop is a circle of

radius ( ) that has its center

on the wire. The magnetic field is tangent to the

loop and has a constant magnitude .

cos0 2 o enc

enc

r r R

B

B ds Bds B ds rB i

r ri i i

R R

2

2 222 o

o

iB r

R

rrB i

R

R

2oi

R

r

B

O (29 – 10)

The solenoid is a long tightly wound helical wire

coil in which the coil length is much larger than the

coil diameter. Viewing the solenoid as a collection

of single circular loops one ca

The solenoid

n see that the magnetic

field inside is approximately uniform.

The magnetic field inside the solenoid is parallel to the solenoid axis. The sense of

can be determined using the right hand rule. We curl the fingers of the right hand

along the direction of the c

B

urrent in the coil windings. The thumb of the right hand

points along . The magnetic field outside the solenoid is much weaker and can be

taken to be approximately zero.

B

(29 – 11)

We will use Ampere's law to determine

the magnetic field inside a solenoid. We

assume that the magnetic field is uniform

inside the solenoid and zero outside. We

assume that the solenolis has turns per

u

n

nit length

We will use the Amperian loop abcd. It is a rectangle with its long side parallel

to the solenoid axis. One long side (ab) is inside the solenoid, while the other (cd)

is outside. b

a

B ds B ds

cos0 0

The enclosed current

c d a

b c d

b b b c d a

a a a b c d

enc

o en o oc

B ds B ds B ds

B ds Bds B ds Bh B ds B ds B ds

B ds Bh i nhi

B ds i Bh nh B nii

oB ni

(29 – 12)

A toroid has the shape of a doughnut (see figure)

We assume that the toroid carries a current and that

it has windings. The magnetic field lines inside the toroid

form cir

i

N

Magnetic field of a toroid :

cles that are cocentric with the toroid center.

The magnetic field vector is tangent to these lines.

The sense of can be found using the right hand rule.

We curl the fingers of the right hand

along the

B

direction of the current in the coil windings.

The thumb of the right hand points along . The magnetic

field outside the solenoid is approximately zero.

B

We use an Amperian loop that is a circle of radius (orange circle in the figure)

cos0 2 . The enclosed current

Thus: 2

The magnetic field i 2o

enc

o

r

B ds Bds B ds

Ni

rB i N

rB B

i

Nr

i

Note :

nside a toroid is not uniform.

2oNiBr

(29 – 13)

2

2

Consider the magnetic field generated by a wire coil of

radius R which carries a current i. The magnetic field

at a point P on the z-axis is given by:

2

oiRBR

The magnetic field of a magnetic dipole.

3/ 22

2

3

2

3 3 3

Here is the distance between

P and the coil center. For points far from the loop

( ) we can use the approximation: 2

Here is the magnetic2 2 2

dipole m

o

o o o

zz

iRz R B

z

i R iAB

pz z z

3

oment of the loop. In vector form:

( )2

The loop generates a magnetic field that has the same

form as the field generated by a bar magnet.

oB zz

3( )

2oB zz

(29 – 14)

An electron and a proton each travel with equal speeds around circular orbits in the same uniform magnetic field. The field is into the page. Because the electron is less massive than the proton and because the electron is negatively charged and the proton is positively charged:

A. the electron travels clockwise around the smaller circle and the proton travels counter- clockwise around the larger circleB. the electron travels counterclockwise around the smaller circle and the proton travels clockwise around the larger circleC. the electron travels clockwise around the larger circle and the proton travels counterclockwise around the smaller circleD. the electron travels counterclockwise around the larger circle and the proton travels clock- wise around the smaller circleE. the electron travels counterclockwise around the smaller circle and the proton travels counterclockwise around the larger circle

The diagram shows a straight wire carrying current i in a uniform magnetic field. The magnetic force on the wire is indicated by an arrow but the magnetic field is not shown. Of the following possibilities, the direction of the magnetic field is:

A. opposite the direction of the currentB. opposite the direction of ~FC. in the direction of ~FD. into the pageE. out of the page

i

F

The magnitude of the magnetic field at point P, at the center of the semicircle shown, is givenby:

A. 2μ0i/RB. μ0i/RC. μ0i/4πRD. μ0i/2RE. μ0i/4R

Pi