     # Sources of magnetic field - Northeastern ITS Sources of magnetic field Magnetic Field created by a single

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• Lots of details so far on how Magnetic Fields exert forces. But, where did those Magnetic Fields come from?

B-fields are created by moving charges (currents).

This seems quite opposite to what we might expect!

Sources of magnetic field

Magnetic Field created by a single moving charged particle at a position

2 0 ˆ

4 r

rvq B

×= v

v

π µ

Constant µ0 = 1.257 x 10-6 Tesla meter/Amp.

is a vector in the direction of , but with magnitude = 1 (often referred to as a unit vector). direction is from charge to where you are evaluating B !

Field created by a moving charge

r r

r̂ r̂

r r

• Field created by a moving charge A new Right Hand Rule…

Thumb in direction of the motion, and fingers then curl in direction of the B-field.

2 0 ˆ

4 r

rvq B

×= v

v

π µ

2 0 ˆ

4 r

rvq B

×= v

v

π µClicker Question

What is the direction of the B- field at point P indicated?

A)Up and to the Left B)Down and to the Left C)Out of the Page D)Into the Page E)None of the Above

Point P

r

• Example: Forces between two moving protons

z r

ev

r

rve B ˆ

4

ˆ

4 2 0

2 0

π µ

π µ =×=

v

v

y r

e FE ˆ4

1 2

2

0πε =

r

( ) y r

ve zx

r

ev ve

z r

ev xve

BveFB

ˆ 4

ˆˆ 4

)(

ˆ 4

)ˆ(

)(

2

22 0

2 0

2 0

π µ

π µ

π µ

=×−

=⎟ ⎠

⎞ ⎜ ⎝

⎛×−

=×−= r

r

r

Now the B-field created by many moving charges (i.e. current flowing in a wire).

I

2 0

2 0

2 0

ˆ

4

ˆ )(

4

ˆ

4

r

rldI

r

r

rvdQ Bd

d

d

×

=×=

r

r

r

v

π µ π

µ π

µ

dl

X r

dB

Biot-Savart first discovered this experimentally.

Magnetic field of a current element

• I

2 0 ˆ

4 r

rldI Bd

×= r

v

π µ

Magnetic field of a current element

2 0 ˆ

4 r

rLdI Bd

×= v

v

π µ

∫∫ ×== 2

0 ˆ

4 r

rLdI BdBtot

v

vv

π µ

This can be a very difficult integral to evaluate.

Magnetic field of a current-carrying wire

• 2 0 ˆ

4 r

rLdI Bd

×= v

v

π µ

page) theinto( sin 4 2

0 θ π

µ r

dyI Bd = v

222 xyr += 22//sin yxxrx +==θ

Now a little geometry

Magnetic field from an infinite straight wire

( ) 2/322 0

4 yx

xdyI Bd

+ =

π µv

∫∫ +∞

∞− + ==

2/322 0

)(4 yx

xdyI BdB

π µvv

page) the(into 2

0

x

I B

π µ=

v

( ) 2/322 0

4 yx

xdyI Bd

+ =

π µv

Magnetic field from an infinite straight wire (cont.)

• 2

|| 0 R

I B

π µ=

v

This is a key result!

B-field a perpendicular distance x away from an infinite (or very long) wire.

Magnetic field from an infinite straight wire (cont.)

Clicker Question

i

A long wire has a current moving as shown. What is the direction of the B-field created by the wire just below the wire?

B=?

A) Into the Page B)Out of the Page C)To the right D)Down E)None of the Above

2 0 ˆ

4 r

rLdi Bd

×= v

v

π µ

L

r

• Clicker Question

i

A long wire has a current moving as shown. What is the direction of the B-field created by the wire just above the wire?

B=?

A) Into the Page B)Out of the Page C)To the right D)Down E)None of the Above

2 0 ˆ

4 r

rLdi Bd

×= v

v

π µ

Lr

Power line has 500 Amps going through it.

What is the B-field strength 15 meters below on the ground?

T m

AATm

R

I B 6

-6 0 107.6

)15(2

)500)(/10(1.26

2 −×=×==

ππ µ

Example with numbers

• I I’ • Think of the Red Wire as creating a B-field.

• Then think of that B-field creating a force on the moving charges (current) in the Blue Wire.

Interaction between two current carrying wires

The two wires may exert forces on each other through Magnetic Interactions.

Interaction between two current carrying wires

BLIF vvv

×= '

r r

I B ˆ

2 0

π µ=

r

r

LII LBIF

π µ

2

' ' 0==

• Clicker Question

I I What is the direction of the Force

acting on the Blue Wire?

A)Up B)Right C)Left D)Into the Page E)Out of the Page

BLIF vvv

×=

i1 i2

The B-field from the Red Wire at the location of the Blue Wire is into the page.

X

left)( 2

10 2122 R

i LiBLiF

π µ=×=

vvv

Then the Force on the Blue Wire is to the left.

page) theinto( 2

10 1 R

i B

π µ=

v

R

• Clicker Question

i i What is the direction of the Force

acting on the Red Wire?

A)Up B)Right C)Left D)Into the Page E)Out of the Page

BLIF vvv

×=

i1 i2

The B-field from the Blue Wire at the location of the Red Wire is out of the page.

.

right)( 2

20 1211 R

i LiBLiF

π µ=×=

vvv

Then the Force on the Red Wire is to the right.

page) theofout ( 2

20 2 R

i B

π µ=

v

R

• i i

Wires with parallel currents attract each other.

i

Wires with antiparallel currents repel each other. Try following the procedure we just outlined to confirm this for yourselves.i

What happens if we flip the direction of one current?

Interaction between current-carrying wires

What about the B-field at the center of a circular loop of wire of radius a and current I?

)(44 22 0

2 0

ax

dlI

r

dlI dB

+ ==

π µ

π µ

2/322 0

)(4 cos

ax

== π

µθ

Field through a circular loop

∫∫∫ + =

+ == dl

ax

Ia

ax

0 2/322

0

)(4

)(4 π µ

π µ

2/322

2 0

)(2 ax

Ia Bx +

= µ

• Field through a circular loop

Which point A or B has the larger magnitude Magnetic Field?

A B C : The B-field is the same at A and B.

A

B

I I

Answer: Case B has the larger magnetic field. Use the Biot-Savart Law to get the directions of the B-field due to the two semi-circular portions of the loop. In A the two fields oppose each other; in B they add.

Clicker Question

2 0 ˆ

4 r

rldI Bd

×= r

v

π µ

• Clicker Question

2/322

2 0

)(2 ax

Ia B

+ = µIn the limit as x >> a, the expression

for the B-field becomes?

A)

B)

C)

3

2 0

2 x

Ia B

µ=

a

I B

2 0µ=

)(2 22

2 0

ax

Ia B

+ = µ

a

I

x

B=?

x x

Ia B ˆ

2 3

2 0µ=

v

x x

IA x

x

aI B ˆ

2 ˆ

2 3 0

3

2 0

π µπ

π µ ==

a

I

x

B=?

* A = area of loop

AI v

v =µ Magnetic Dipole Moment

3 0

2 x B

µ π

µ vv =

Magnetic Dipole moment

• AI v

v =µ

3 0

2 x B

µ π

µ vv =

Any current loop looks like a Magnetic Dipole far away.

The B-field drops as the distance^3 And depends on the Magnetic Dipole Moment.

Magnetic Dipole moment

A square loop of side length a of wire carrying current I is in a uniform magnetic field B. The loop is perpendicular to B (B out of the page). What is the magnitude of the net force on the wire?

I

B A: IaB B: 4IaB C: 2IaB D: 0 E: None of these

Clicker Question

BLIF vvv

×=

• Clicker Question

r B Bz= ⋅\$ , where B = B(y) = A y

D

B

C

A

B stronger

B weaker E: net force is zero

The same loop is now in a non-uniform field.

where A is a constant. The direction of the net force is?

y

x

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