sources of the magnetic field · magnetic force between two current carrying wires from before we...

Fall 2008Physics 231 Lecture 8-1

Sources of the Magnetic Field


Magnetic Field of a Point Charge

Given a point charge, q, we know that it generates an electric field regardless of whether it is moving or not

If the charge is in fact moving then the charge also generates a magnetic field

The magnetic field generated by the charge however does not behave in the same way the electric field of the charge does


Magnetic Field of a Point ChargeIt is found that the magnetic field is perpendicular to both the velocity of the charge and the unit vector from the charge to the point in question

The magnitude of the fieldis given by

20 sin

4 rvq

Bφ

πµ

=

where φ is the angle betweenthe velocity and unit vector

The magnetic field lines form concentric circles about the velocity vector


Magnetic Field of a Point ChargeThe direction of the magnetic field is given by one of the right-hand rules

Point your thumb, of your right hand, in the direction of the velocity and curl your fingers. The direction of the field is then the direction your fingers curl. The magnetic field for a negative charge will be in the opposite direction.

The vector equation for this is

20 ˆ

4 rrvqB ×

=rr

πµ


Magnetic Field of a Current Element

If instead of a single charge, we have a current what is the magnetic field then given by

We start with the principle of superposition – we add the magnetic fields of the individual charges that make up the current

Given n moving charges per unit volume each carrying a charge q, then the total charge in this element is

dlAqndQ =where A is the cross sectional area


Magnetic Field of a Current Element

We then have for dB

20

20 sin

4sin

4 rdlAvqn

rvdQ

dB dd φπµφ

πµ

==

AvqnI d=But from our work on currents we also have that

20 sin

4 rdlIdB φ

πµ

=Therefore

20 ˆ

4 rrldIBd ×

=r

r

πµ

Or in vector form


Biot-Savart LawFor a complete circuit we then use the integral form of the previous equation

∫×

= 20 ˆ

4 rrldIB

rr

πµ

This can be a very difficult integral to evaluate

The level of difficulty depends upon the variable chosen to integrate over and the presence of any symmetries that can be exploited

We will develop another, much simpler method, shortly


Field of a Straight WireWe set up the problem as shown

The integral becomes

( )∫−

⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

a

a yxdyxIB 2/322

04πµ

The result of this integral is

220 2

4 axxaIB+

=π

µ

In the limit that a >> xxIB

πµ2

0=


Magnetic Force Between Two Current Carrying Wires

We have two wires carrying currents Iand I’ separated by a distance r

The first wire carrying current I sets up a magnetic field at the second wire given by

rIB

πµ2

0=



From before we know that a wire carrying a current that is placed in a magnetic field will experience a force per unit length that is given by r

IILF

πµ

2'

0=

This is in fact what the second wire will experience

BLIFrrr

×= 'And from

This force is directed towards the other wire – the force is attractive



If we had started with the wire carrying the current I’, we would have ended up with the same result

If the wires were carrying currents in opposite directions, thenthe force between the wires would be repulsive


Circular Current LoopAnother circuit for which it is worthwhile to calculate the magnetic field is the circular current loop

We start by utilizing theBiot-Savart law

20 ˆ

4 rrldIBd ×

=r

r

πµ

222 axr += a being the radius of the loopwith


Circular Current LoopA current element at the top will yield the magnetic field as shown

A current element at the bottom will yield a magnetic field of the same magnitude, but directed downward

The vertical components will cancel and all that will be left is a horizontal magnetic field of incremental value

( ) 2/3220

220

4cos

4 ax

dlaIax

dlIdBx+

=+

=π

µθπ

µ


Circular Current LoopThe integral is over just the circumference of the loop as all other variables are constant

( ) ( ) 2/322

20

2/3220

24 ax

aIdlax

aIB x+

=+

= ∫µ

π

µ

If there are N loops, turns, then the result is just N times above results

( ) 2/322

20

2 ax

aINB x+

=µ


Circular Current Loop

An important point for the field of a current loop is at the center of the loop where x = 0

aINBx 2

0µ=

A current loop consisting of N turns is often referred to as aSolenoid

A final caution: These results for a current loop only apply on the axis of the loop


Ampere’s LawJust as we developed an easy way to calculate the electric field in symmetric situations, Gauss’ Law, there is a similar technique we can use for calculating magnetic fields

∫ =⋅ IldB 0µrr

This is known as Ampere’s Law

where the integral is around a closed path and I is the total current passing through the area bounded by this path


Ampere’s LawThere is one big difference to remember between the two methods

Gauss’ Law is an integral over a closed surface

∫ =⋅0ε

QAdErr

Whereas Ampere’s Law is an integral over a closed path

∫ =⋅ IldB 0µrr


Ampere’s LawGiven a path around which the integration is to be done, there is a sign convention for the current

Determine the direction of the normal, as defined by the sense of movement around the integration path

If the current is going in the same direction as the normal, the current is considered to be positive otherwise it is negative


Long Straight WireCurrent in same direction as normal of integration path

rIB

IrB

IdlB

IldB

πµ

µπ

µ

µ

2

2

0

0

0

0

=

=

=

=⋅

∫∫

rr

This is the same as we derived from the Biot-Savart Law


Long Straight WireCurrent in opposite direction to normal of integration path

rIB

IrB

IdlB

IldB

πµ

µπ

µ

µ

2

2

0

0

0

0

=

−=−

−=−

=⋅

∫∫

rr

Again this is the same as we derived from the Biot-Savart Law


Caution

Ampere’s Law gives you only the magnetic field due to any currents that cut through the area bounded by the integration path

For the indicated integration path, the result of Ampere’s Law is zero


A B C

Example

Given three currents and their associated loops

1) For which loop is ∫ B • dl the smallest?→ →

A B C Same

The currents that are used in Ampere’s Law are the currents that are within the loop that is used for integration

Loop A has no current going through it whereas the other two loops do have currents going through them, the integral for Loop A is zero


B C Same

2) Now compare loops B and C. For which loop is ∫ B • dl the greatest?

→ →

A B C

Example

Given three currents and their associated loops

The right hand side of Ampere’s law is the same for both loops, therefore the integral for both loops is the same


Example

(a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(a) > 0

xx

xx

xx

x

x

2II

a

b

x

y

1) What is the magnetic field Bx(a) at point a, just outside the cylinder as shown?

A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius Rcarries current 2I in the -z direction.

The loop used for application of Ampere’s Law will be outside the cylinderThe total current going through this loop is the sum of the two current yielding a current of –I. This gives a magnetic that points in the positive x direction at a

xI

B

BB

B


Example

(a) Bx(b) < 0 (b) Bx(b) = 0 (c) Bx(b) > 0

2) What is the magnetic field Bx(b) at point b, just inside the cylinder as shown?

xx

xx

xx

x

x

2II

a

b

x

y

A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius Rcarries current 2I in the -z direction.

This time, the Ampere loop only encloses current I which is in the +z direction — the loop is inside the cylinder!

The current in the cylindrical shell does not contribute to at point b.


Long Cylindrical Conductor

The problem can be divided into two regions

Outside the conductor

Inside the conductor


Long Cylindrical ConductorOutside the conductor

We start with an integration path around the conductor at an arbitrary distance rfrom the central axis of the conductor

∫ =⋅ IldB 0µrr

Then using Ampere’s Law

IrBdlB 02 µπ ==∫we then have

rIB

πµ2

0=or


Long Cylindrical ConductorInside the conductor

Since we have not been told otherwise, we assume that the current is uniformly distributed across the cross section of the conductor

We define a current density 200

RI

AIJ

π==

We now apply Ampere’s Law around an integration path centered on the cylindrical axis and radius r < R

∫ =⋅ IldB 0µrr with I being the current going through

the area defined by the integration path


Long Cylindrical ConductorInside the conductor

2

2

02

2

02

RrI

RrIrJI ===

πππ

2

2

000 RrII µµ =rBldB π2=⋅∫

rrand

200 2 RrIBπ

µ=So after equating, we then have

The magnetic field increases linearly with the distance from the axis of the cylinder


Inside the wire: (r < a)

Outside the wire: ( r > a )

r

Ba

Long Cylindrical Conductor

200 2 RrIBπ

µ=

rIB

πµ2

0=


Solenoid

We choose a path of integration as shown with one part of the path inside the solenoid and one part of the path outside

We assume that the path on the outside, cd, is sufficiently far from the solenoid, compared to the diameter of the solenoid, that the field is approximately equal to zero


We use Ampere’s LawSolenoid

∫ =⋅ IldB 0µrr

First the Left-Hand-Side

∫∫

∫∫ ∫

⋅+⋅+

⋅+⋅=⋅

a

d

d

c

c

b

b

a

ldBldB

ldBldBldB

rrrr

rrrrrr

a b: Using the right hand rule for currents, the B field is in the same direction as the path and it is constant, we then have

∫ =⋅b

a

BLldBrr


Solenoidb c: Along the part of the path that is inside the solenoid, B is perpendicular to dl, so that contribution is zero, while for the part of the path that is outside of the solenoid, B ~0, so we then have

0=⋅∫c

b

ldBrr

0=⋅∫a

d

ldBrr

0=⋅∫d

c

ldBrr

The same is true for d a

c d: Here B is zero so we again have


SolenoidNow for the Right-Hand-SideThe total current passing through the area defined by the integration path is NI where N is the number of turns and I is the current in the loop, so the right-hand-side is

IN0µ

Now equating the left and right-hand-sides we haveINLB 0µ=

Or solving for BI

LNB 0µ=


SolenoidThe magnetic field of a solenoid looks like the following with the exact shape of the field lines being dependent upon the the number of turns, the length, and the diameter of the solenoid

sources of the magnetic field · magnetic force between two current carrying wires from before we...

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