Lecture18 Magnetic Magnetic Field of a Current Loop; Ampere's Law; Solenoids; Toroids

8/3/2019 Lecture18 Magnetic Magnetic Field of a Current Loop; Ampere's Law; Solenoids; Toroids

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http://www.nearingzero.net(work005.jpg)


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Comment on exam scores

y FS2011 Exam 2 grade distribution (regrades not included)

I need to comment on grading of problems 7 and 10a.


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Announcements

y Homework note: use the Biot-Savart Law if you must

calculate the magnetic field of a ring or coil. Dont start with arandom equation from the text or lecture notes. Example:Special Homework #6. For the loop, start with the Biot-SavartLaw.

y Special Homework #6 is due tomorrow. Download it fromthe Physics 24 web site if you dont pick it up today.


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Todays agenda:

Magnetic Field Due To A Current Loop.You must be able to apply the Biot-Savart Law to calculate the magnetic field of a currentloop.

Amperes Law.You must be able to use Amperes Law to calculate the magnetic field for high-symmetrycurrent configurations.

Solenoids.You must be able to use Amperes Law to calculate the magnetic field of solenoids and

toroids. You must be able to use the magnetic field equations derived with Amperes Lawto make numerical magnetic field calculations for solenoids and toroids.


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Magnetic Field of a Current Loop

P

a

x

dB

x

U

A circular ring of radius a carries a current I as shown.Calculate the magnetic field at a point P along the axis of thering at a distance x from its center.

dl r&r

90-U

I

y

dBy

dBx

U

z

Complicated diagram!

You are supposed tovisualize the ringlying in the yz plane.

dlis in the yz plane. r

is in the xy plane andis perpendicular to dl.*Thus

r

.v&" "d r =d

Also, dB must lie in the xy plane* (perpendicular to dl) and is

perpendicular to r. *Only when dlis centered on the y-axis!


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P

a

x

dB

x

U

dl r&r

90-U

I

y

dBy

dBx

U

z

4

v&

& "02

I d rdB =

r

4

"02

I ddB =

r

4

"02 2

I d

dB = x a

4 4

" "U0 0x 1/22 2 2 2 2 2

I d I d adB = cos =

x a x a x a

4 4

" "U0 0y 1/22 2 2 2 2 2

I d I d xdB = sin =

x a x a x a

By symmetry, By will be 0. Do you see why?


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Px

dBz x

dlr

&r

I

y

dBy

dBxz

When dlis not centered at z=0, there will be a z-componentto the magnetic field, but by symmetry Bz will still be zero.


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x xring

B = dB

4

"0x 3/2

2 2

I a ddB =

x a

I, x, and a are constant as you integrate around the ring!

P

a

x

dB

x

U

dl r&r

90-U

I

y

dBy

dBx

U

z

4 4T

"

0 0

x 3/2 3/22 2 2 2ring

I a I a

B = d =2

ax a x a

2

0x 3/2

2 2

I aB =

2 x aThis is not on your starting equation sheet. I willadd it if we have homework or a test problem whereI judge that you need it.


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At the center of thering, x=0.

P

a

x

dB

x

U

dl r&r

90-U

I

y

dBy

dBx

U

z

2

0x,center 3/2

2

I aB =

2 a

This is not on your starting equation sheet, but I willadd it if we have homework or a test problem whereI judge that you need it. For homework, if aproblem requires this equation, you need toderive it!

20 0x,center 3

I a IB = =2a 2a

For N tightly packed concentric rings (a coil)

0x,center

N IB =

2a


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Magnetic Field at the center of a Current Loop

A circular ring of radius a lies in the xy plane and carries acurrent I as shown. Calculate the magnetic field at the centerof the loop.

a

z

dlI

x

y

One-page derivation, so you

wont have to go through all theabove steps (I will work it at theblackboard)! (reward for students who attend lecture)

The direction of the magneticfield will be different if the planeof the loop is not in the xy plane.

Ill explain the funny orientationof the axes in lecture.Homework hint: use this derivation in

Special Homework #6.


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Quiz time (maybe for points, maybe not!)


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Todays agenda:






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I

B

r

Amperes Law

Just for kicks, lets evaluate the line integralalong the direction of B over a closed circularpath around a current-carrying wire.

ds

& &

B ds =B ds =B 2 r

& &B ds

& &

0

0

IB ds = 2 r = I

2 r

The above calculation is only for the special case of a longstraight wire, but you can show that the result is valid ingeneral.


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& &

0B ds = I Amperes Law

I is the total current that passes through a surface bounded bythe closed (and not necessarily circular) path of integration.

Amperes Law is useful for calculating the magnetic field due tocurrent configurations that have high symmetry.

I

B

r

ds

The current I passing through a loop is positive if the directionof integration is the same as the direction of B from the righthand rule.

I

B

r

dspositive I negative I


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& &

0 enclB ds = IYour text writes

Your starting equation sheet has

I1

ds

If your path includes more than one source of current, add allthe currents (with correct sign).

I2

because the current that you use is the current enclosed bythe closed path over which you integrate.

& &

0 1 2B ds = I -I

*

OI& &

E

0 encl

dB ds = I

dtThe reason for the 2

nd

term on the right will become apparent later. Set it equal to zero fornow.


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Example: a cylindrical wire of radius R carriesa current I that is uniformly distributed overthe wires cross section. Calculate themagnetic field inside and outside the wire.

I

R

Cross-section of the wire:

Rr

direction of I

B

& &

22

0 encl 0 0 0 22

rA enclosed by r rB ds = I = I = I = I

A enclosed by R RR

Choose a path that matches thesymmetry of the magnetic field (sothe dot product and integral areeasy to evaluate); in this case, thefield is tangent to the path.


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Rr

direction of I

B

Over the closed circular path r:

& &

B ds =B ds = 2 rB

Solve for B:

2

0 2

r2 rB = I

R

2

00 02 2 2

Ir rB = I = I = r

2 rR 2 R 2 R

B is linear in r.


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R

r

direction of I

B

Outside the wire:

r

& &

0B ds =B ds = 2 B = I

0 IB =2 r

(as expected).

B

rR

Plot:

A lot easier than usingthe Biot-Savart Law!


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Calculating Electric and Magnetic Fields

Electric Field

in general: Coulombs Law

for high symmetryconfigurations: Gauss Law

Magnetic Field

in general: Biot-Savart Law

for high symmetryconfigurations: Amperes Law

This analogy is rather flawed because Amperes Law is not

really the Gauss Law of magnetism.


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Demo: magnetic field of a current loop.

Demo: magnetic field of a solenoid.

Demo: a solenoid application.

Possible Demos

(depends on time available)


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Todays agenda:






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Magnetic Field of a Solenoid

A solenoid is made of many loops of wire, packed closely*together. Heres the magnetic field from a loop of wire:

Some images in thissection are fromhyperphysics.

*But not so closely that youcan use 0 N I

B =

2a


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Stack many loops to make a solenoid:

Ought to remind you of the magnetic field of a bar magnet.


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You can use Amperes law to calculate the magnetic field of asolenoid.

8 8 8 8 8 8 8 8 8 8 8 8

B

I

& & & & && & & & &

1

2

34

B ds = B ds B ds B ds B ds

l

0 0 0 & &

" 0 enclosedB ds = B = I

" 0B = N IN is the number of loopsenclosed by our surface.


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8 8 8 8 8 8 8 8 8 8 8 8

B

I

l

"

0

NB = I Magnetic field of a solenoid of

length l, N loops, current I.n=N/l(number of turns per

unit length).0B = n I

The magnetic field inside a long solenoid does not depend on the positioninside the solenoid (if end effects are neglected).


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A toroid* is just a solenoid hooked up to itself.

& &

B ds =B ds =B 2 r

& &

0 enclosed 0B ds = I = N I

0B 2 r = N I

0 N IB =2 r

Magnetic fieldinside a toroid of Nloops, current I.

The magnetic field inside a toroid is not subject to end effects, but is notconstant inside (because it depends on r).

*Your text calls this a toroidal solenoid.


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"0

NB = I

-7 400T mB = 4 10 2 AA 0.1 m

B = 0.01 T

Example: a thin 10-cm long solenoid has a total of400 turns ofwire and carries a current of2 A. Calculate the magnetic fieldinside near the center.


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Help! Too many similar starting equations!

0

IB = 2 r long straight wire

0 N IB =2a

center of N loops of radius a

"

0NB = I

0B = n I

solenoid, length l, N turns

solenoid, n turns per unit length

0 N IB =2 r

toroid, N loops

How am I going to know which is which on the next exam?

use Amperes law (or note the lack of N)

probably not a starting equation

field inside a solenoid is constant

field inside a solenoid is constant

field inside a toroid depends on position (r)

Lecture18 Magnetic Magnetic Field of a Current Loop; Ampere's Law; Solenoids; Toroids

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Transcript of Lecture18 Magnetic Magnetic Field of a Current Loop; Ampere's Law; Solenoids; Toroids