Skc Biot-Savart Law

8/13/2019 Skc Biot-Savart Law

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The Biot-Savart Law

AP Physics C

Biot-Savart sounds like Leo Bazaar


2/28

Biot & Savart produced an equation that gives themagnetic field at some point in space in terms ofthe current that produces the field.

a wire carrying steady current I, themagnetic field dB at some point P hasthe following properties:

The vector dB is to both ds

(direction of I) & to the vector rdirected from the element ds to thepoint P.

The magnetic field wraps in circles

around a wire. The direction is foundusing the right-hand rule.

Thumb of right hand in the directionof the current, fingers curl in thedirection of B. Field at any point is

tangent to curl

dlI


3/28

1. Which drawing below shows the correctdirection of the magnetic field,B, at the pointP?

A. I.B. II.

C. III.

D. IV.

Direction of Magnetic Field

I II III IV V

i i ii

P P P P P

iB

B

Bintopage

Bintopage

Bintopage


4/28

The magnitude of dB is inverselyproportional to r2, where r is the distancefrom the element ds to the point P.

The magnitude of dB is proportional to thecurrent I and to the length ds of theelement.

The magnitude of dB is proportional to sin q,where qis the angle between the vectors ds

and r.

Biot-Savart law:

To determine the total magnetic field B at somepoint due to a conductor of specified size, addup contribution from all elements ds that make

up the conductor (integrate)!

2

o

r4

sdsIdB

qinm0= permeability constant

exactly m/AT104 7


5/28

Magnetic Field of a long straight wire Consider a thin, straight wire

carrying a constant current I alongthe y axis. To determine the totalmagnetic field B at the point P at adistance R from the wire:

Let ds = dx, then dssin q becomesdxsin q.

The contribution to the totalmagnetic field at point P from eachelement of the conductor ds is:

2

o

r

sindx

4

IdB


6/28

Express r in terms of R and x.

Express sin qin terms of R and r.

21

22xR

R

r

Rsin

2

122222

xRrxRr

2

o

2

o

r

sindx

4

I

r

sindx

4

IdB

23

22

o

23

22

o

212222

o

xR

dx

4

RI

xR

dxR

4

IB

xR

R

xR

dx

4

IB


7/28

From the table of integrals:

2

12222

322

xRR

x

xR

dx

R4

I211

R4

I

R4

IB

R4

IB

RRR4IB

xR

x

R4

RI

xRR

x

4

RIB

ooo

21

221

2

o

21

2221

22

o

21

222

o

21

222

o

R2

IB

o


8/28

Find B for a conductor of length l

From the table of integrals:

l

l

l

l

l

l

l

l

l

l

l

l

R 23

22

o

23

22

o

21

2222

o

2

o

2

o

xR

dx

4

RI

x

dxR

4

IB

xR

R

xR

dx

4

IB

r

sindx

4

I

r

sindx

4

IdB

2

12222

322

xRR

x

xR

dx


9/28


10/28

Just add up all of the contributionsds

to thecurrent, but now distance r=Ris constant,and .

Notice that . So the integralbecomes

For a complete loop, f= 2, so

Bat Center of a Circular Arc of Wire

fRdds

f f

m

0 02

0

4ds

R

idBB

R

iRd

R

iB

fmf

m f

44

0

02

0

R

iB

fm

4

0

sdr

R

iB

2

0m

B at center of a full circle


11/28

Find Magnetic Field on the Axis of aCircular Current Loop

Consider a circular loop of wire of radius R inthe yz plane and carrying a steady current I:

Note: Each element of length ds is

to the r from ds to point P.ds90sinds

& the direction of dB

from ds is at an angleqwith the x axis.


12/28

The direction of the net magnetic fieldis along the x axis and directed away

from the circular loop.

cosxR

ds

4

I

cos

xR

ds

4

I

cos

22

o

22

o

dB

dB

dBdBx

21

2222

o

xR

R

xR

ds

4

I

dB

ds

23

22

o

xR4

RIB


13/28

The sum of the elements of length dsaround the closed current loop is the

circumference; s = 2RThe net magnetic field B at point P is :

23

22

2

o

2

3

22

2

o

2322

o

xR2

RIB

xR4

RI2B

R2xR4

RIB


14/28

For large distances along the x axis fromthe current loop, where x is very large in

comparison to R:

3

2

o

6

2

o

3

2

2

o

23

2

2

o

23

22

2

o

x2

RIB

x2

RI

x2

RIB

x2

RI

xR2

RIB


15/28

B field for Circular Loop

R

B

x

x

0

R

iB

2

0m

23

22

2

o

xR2

RIB

3

2

o

x2

RIB


16/28

Remember how to Calculate Electric Field

Either:

Coulomb's Law:

Gauss' Law

What are the analogous equations for the

Magnetic Field?


17/28

Calculation of Magnetic Field

"High symmetry"

I

Two ways to calculate the Magnetic Field:

Biot-Savart Law:

Ampere's Law


18/28

Draw an amperian loop around a systemof currents (like the two wires at right). Theloop can be any shape, but it must be closed.

Add up the component of along the loop,for each element of length dsaround this

closed loop. The value of this integral is proportional to

the current enclosed:

Amperes Law

B

encisdB 0m

i1 i

2

Amperes Law


19/28

Magnetic Field Outside a LongStraight Wire with Current

We already used the Biot-Savart Law to showthat, for this case, .

Lets show it again, using Amperes Law:

First, we are free to draw an Amperian loop of

any shape, but since we know that themagnetic field goes in circles around a wire,lets choose a circular loop (of radius r).

ThenBand dsare parallel, and Bis constanton the loop, so

And solving for B gives our earlier expression.

r

iB

m

2

0

encisdB 0m

Amperes Law

enc

irBsdB0

2

m

r

iB

m

2

0


20/28

Magnetic Field Inside a LongStraight Wire with Current

Calculate B inside the wire. Draw a circular Amperian loop around the

axis, of radius r < R.

The enclosed current is less than the totalcurrent, because some is outside the

Amperian loop. The amount enclosed is

2

2

R

riA

A

Ji en

total

enc

inside a straight wire

2

2

002

R

riirBsdB

enc mm

rR

iB

2

0

2

mrR

~1/r~r

B


21/28

Two cylindrical conductors each carrycurrent Iinto the screen as shown.The conductor on the left is solid andhas radius R=3a. The conductor on

the right has a hole in the middle &carries current only between R=a &R=3a.

(a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a)


What is the relation between the magnetic field at R =

2a for the two cases (L=left, R=right)?

1B

1A

3a

a

3a

I I

2a

What is the relation between the magnetic field at R= 6a for the two cases (L=left, R=right)?


22/28

What is the relationbetween the

magnetic field at R =6a for the two cases(L=left, R=right)?


1A

Amperes Law can be used to find the field in both cases. The Amperian loop in each case is a circle of radius R=6a in the plane

of the screen.

3a

a

3a

I I

2a

The field in each case has cylindrical symmetry, being everywheretangent to the circle. Therefore the field at R=6a depends only on the total currentenclosed!!

In each case, a total currentI

is enclosed.


23/28


What is the relation between

the magnetic field at R = 2afor the two cases (L=left,R=right)?

1B

3a

a

3a

I I

2a

For the LEFT conductor:

Once again, the field depends only on how much current is enclosed.

For the RIGHT conductor:

I9

4I)3(

)2(I2

2L

a

a

I8

3I)3(

)2(I22

22R

aa

aa


24/28

Solenoids We saw earlier that a complete

loop of wire has a magnetic field atits center:

We can make the field stronger bysimply adding more loops. A many

turn coil of wire with current iscalled a solenoid.

R

iB

2

0m

The field near the wires is still circular,but farther away the fields blend into anearly constant field down the axis.

The actual field looks more like this:


25/28

Solenoids

We can use Amperes Law to calculate Binside the solenoid.

Characterize the windings in terms ofnumber of turns per unit length, n. Eachturn carries current i, so total current over

length his inh.

Compare with electric field in a capacitor. Like a capacitor, the field is uniform inside (except near the ends), but the

direction of the field is different.

Approximate that the field is constant inside and zero outside (just likecapacitor).

inhiBhsdBenc 00

mm

only section that has non-zerocontribution

inB0

m ideal solenoid


26/28

Toroids

Notice that the field of the solenoid sticks outboth ends, and spreads apart (weakens) at theends.

We can wrap our coil around like a doughnut, sothat it has no ends. This is called a toroid.

Now the field has no ends, but wraps uniformlyaround in a circle.

What is B inside? We draw an Amperian loopparallel to the field, with radius r. If the coil hasa total ofNturns, then the Amperian loopencloses currentNi.

iNirBsdBenc 00

2 mm

r

iNB

2

0

m inside toroid


27/28

Recall that a wire carrying a current in amagnetic field feels a force.

When there are two parallel wires carryingcurrent, the magnetic field from one causes aforce on the other.

When the currents are parallel, the two wires arepulled together.

When the currents are anti-parallel, the two wiresare forced apart.

Force Between Two Parallel Currents

FF

To calculate the force on bdue to a,abba

BLiF

d

ia

2

0

mR

iB

m2

0

d

LiiF

ba

ba

m

2

0 Force between two parallel currents

BLiFB


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3. Which of the four situations below has thegreatest force to the right on the centralconductor?

A. I.

B. II.

C. III.

D. IV.E. Cannot

determine.

Forces on Parallel Currents

I.

II.

III.

IV.

Fgreatest?

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Skc Biot-Savart Law