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MASSACHUSETTS INSTITUTE OF TECHNOLOGYDepartment of Physics: 8.02

Problem Solving 6: Amperes Law and Faradays Law Solution

Part One: Amperes Law

Summary: Strategy for Applying Amperes Law (Section 9.10.2, 8.02 Course Notes)

Amperes law states that the line integral of d!B s!

!

around any closed loop is proportional

to the total steady current passing through any surface that is bounded by the closed loop:

!B ! d

!s"" =0

!J ! n dA""

To apply Amperes law to calculate the magnetic field, we use the following procedure:

Step 1: Identify the symmetry properties of the current distribution.

Step 2: Determine the direction of the magnetic field

Step 3: Decide how many different spatial regions the current distribution determines

For each region of space

Step 4: Choose an Amperian loop along each part of which the magnetic field iseither constant or zero

Step 5: Calculate the enclosed current through the Amperian Loop, Ienc

=

!

J ! n dA""

Step 6:Calculate the line integral!B !d

!s"" around the closed loop.

Step 7: Equate!B !d

!s"" with 0 e ncI and solve for B

!

.

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Problem 1: Magnetic Field of a Cylindrical Shell

A long cylindrical cable consists of aconducting cylindrical shell of innerradius a and outer radius b. The current

density J!

in the shell is out of the page(see sketch) and varies with radius as

J(r) = !r for a r b< < and is zero

outside of that range, where ! is a

positive constant with units A !m-3

.

Question 1: Problem Solving Strategy

Step 1: Identify Symmetry of Current DistributionEither circular or rectangular which is it?

Circular

Step 2: Determine Direction of magnetic field

Clockwise or counterclockwise?

counterclockwise

Step 3: How many regions?

Three: r< a ; a < r< b ; r> b

For each region, you will repeat the following four steps:

Step 4: Draw Amperian Loop, choose an integration direction.

Step 5: Calculate current enclosed by Amperian Loop. Be careful about signs!

Step 6: Apply Amperes Law to calculate magnitude of the magnetic field.

Step 7: Determine direction of magnetic field based on results of step 6.

The next step is to calculate the current enclosed by this Amperian loop. Because the

current is non-uniform, you will need to set up and solve an integral relation for thecurrent enclosed in each of your Amperian loops for the different regions.

Question 2: Draw a new figure showing the cross section of the cable and draw

Amperian Loop on your figure for the region a < r< b , indicating your choice ofintegration direction.

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Question 3: Calculate current enclosed by Amperian Loop. Be careful about signs!

Remember that!

J is non-uniform so you will need to compute the integral

Ienc

=

!

J ! n dA"" .

Answer. The enclosed current is

Ienc

=

!

J ! d!

A

open surface

"" = # $r2% $r d $rar

" = 2%# $r2 d $r

a

r

" =2%#

3(r3 &a3)

Question 4: Your answer above should be zero when r = a and when r = b , the totalcurrent is Iin the wire. Find an expression for I.

Answer. When r = b , the Amperian loop encloses all the current, so Ienc

=I, and

I = ! "r2# "r d "ra

b

$ = 2#! "r2d "r

a

b

$ =2#!

3(b3 % a3) .

Question 5: What is!B !d

!s"" ? (That is, evaluate the integral, the left hand side of

Amperes law)

Answer.!B !d

!s"" =B(2#r).

Question 6: Now apply Amperes Law. What do you get for the magnitude of themagnetic field in the region a < r< b ?

Answer. Applying Amperes law, we have

!B !d

!s"" = 0Ienc # B(2$r) =

02$%

3(r3 & a3)

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Therefore!

B =B = 0

"

3(r3 # a3)

1

r a < r< b .

Question 7: Repeat the steps above to find the magnitude of the magnetic field in theregion r< a .

Answer. In the region r< a , Ienc

= 0 , and therefore B = 0 .

Question 8: Repeat the steps above to find the magnitude of the magnetic field in the

region r> b .

Answer. In the region r b> , the Amperian loop is shown in the figure below. We use thesame integration element as in part b).

Then the enclosed current is

Ienc

=

!

J ! d!

A

open surface

"" = # $r2% $r drab

" =2%#

3$r 2 d $r

a

b

" =2%#

3(b3 &a3)

Applying Amperes law, we have

!B!d

!s"" = 0Ienc # B(2$r) =

02$%

3(b3 & a3)

or

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!

B =B = 0

"

3(b3 # a3)

1

r

Question 9: Plot B vs. ron the graph below.

Answer:A plot of the magnitude of B vs. ris shown below.

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Problem 2: Magnetic Field of a Slab of Current

We want to find the magnetic field B!

due to an infinite slab of current,using Ampere's

Law. The figure shows a slab of current with current density!

J = (2Jey / d) z , where J

e

is positive constant with units of amps per square meter. The slab of current is infinite inthex andzdirections, and has thickness din they-direction.

Question 1: What is the magnetic field at y = 0 , where y = 0 is the exact center of theslab? Explain your reasoning.

Answer. By symmetry, the magnetic field at y = 0 is zero. Above y = 0 the field points

in the neagtice x-direction. Below y = 0 the field points in the positive x-direction.

Therefore by symmetry, the magnetic field at y = 0 is zero.

Problem Solving Strategy

(1) Identify Symmetry

Either circular or rectangular. Which is it?

Rectangular

(2) How many regions?

Three for this problem: in the slab and above the slab and below the slab.

(3) Determine Direction

Make sure you determine the direction of the magnetic field!

B in all regions. Sketch on

figure above.

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For each region, you will repeat the following four steps.

Step 4: Draw Amperian Loop, choose a integration direction.

Step 5: Calculate current enclosed by Amperian Loop. Be careful about signs!

Step 6: Apply Amperes Law to calculate magnitude of the magnetic field.

Step 7: Determine direction of magnetic field based on results of step 6.

We want to find the magnetic field fory > d/2, and we have from the answer to Question1 for the magnetic field aty =0. Therefore.

Question 2: What Amperian loop do you take to find the magnetic field for y > d/2?

Make a new figure below and draw the Amperian loop on that figure indicating any

variables that you may need.

The next step is to calculate the current enclosed by this Amperian loop. Hint: thecurrent enclosed is the integral of the current density over the enclosed area.

Question 3: What is the total current enclosed by your Amperian loop?

Answer. Consider the loop shown in the figure above. We have to integrate the current

density to get the enclosed current:

Ienc

=

2Jey

ddA!! =

2Je!

dy dy

0

d 2

! =2J

e!

d

y2

20

d 2

=

Je!d

4

Question 4:What is

!B

!d!s

"" ?

Answer. The loop has four segments. Along two of those,!

B is perpendicular to d !

s so!

B !d!

s = 0 . Along the center line!

B =!

0 . On the last side!

B is parallel. Thus,

!B !d

!s"" =B#+0+ 0+ 0 =B# .

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Question 5:Now apply Amperes Law to find the magnitude of the magnetic field in theregiony > d/2?

Answer.

!B !d

!s

"" =B#=

o

Je

#d 4 # !B =$

oJ

ed

4i; y% d/ 2

We now want to find the magnetic field in the region 0< y < d/2.

(4) Draw Amperian Loop:

We want to find the magnetic field for 0

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Question 8:What is!B !d

!s"" ?

Answer. The loop has four segments. Along two of those,!

B is perpendicular to d !

s so!

B !d!

s = 0 . Along the centerline!

B =!

0 . On the last side!

B is parallel. Thus, as in the

previous case

!B !d

!s"" =B#+0+ 0+ 0 =B# .

(7) Solve for B:

Question 9:Now apply Amperes Law to find the magnitude of the magnetic field in theregion 0

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Part Two: Problem Solving Faradays Law

OBJECTIVES

1. To explore a particular situation that can lead to a changing magnetic flux through the

open surface bounded by an electric circuit.

2. To calculate the rate of change of magnetic flux through the open surface bounded bythat circuit in this situation.

3. To determine the sense of the induced current in the circuit from Lenzs Law.

4. To look at the forces on the current carrying wires in our circuit and determine theeffects of these forces on the dynamics of the circuit.

REFERENCE: Sections 10.1 10.4, 8.02 Course Notes.

Problem-Solving Strategy for Faradays Law

In Chapter 10 of the 8.02 Course Notes, we have seen that a changing magnetic fluxinduces an emf:

Bd

dt!

"=#

according to Faradays law of induction. For a conductor that forms a closed loop, theemf sets up an induced current | | /I R!= , where R is the resistance of the loop. To

compute the induced current and its direction, we follow the procedure below:

(1) For the closed loop of area A , define an area vector A!

and choose a positive

direction. For convenience of applying the right-hand rule, let A!

point in the direction ofyour thumb. Compute the magnetic flux through the loop using

( is uniform)

( is non-uniform)B

d

! "#$ = %

"#&''

B A B

B A B

!! !

!! !

(2) Evaluate the rate of change of magnetic flux /B

d dt! . Keep in mind that the change

could be caused by

(i) changing the magnetic field / 0dB dt ! ,

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Answer: !B

=Bwz(t)

Problem Solving Strategy Step (2): Compute /B

d dt!

Question 3:What is /B

d dt! ? Is this positive or negative at time t? Be careful here,

your answer should involve vz(t) (not ( )z t ), and remember that v

z(t) < 0 .

Answer:d

dt!

B =

d

dt(Bwz(t)) =Bwv

z(t)

Because vz(t) < 0 , the change in magnetic flux is negative.

Problem Solving Strategy Step (3): Determine the sign of the induced emf (the same

as the direction of the induced current)

Question 4: If / 0B

d dt! < then your induced emf (and current) will be right-handed

with respect to A!

, and vice versa. What is the direction of your induced current givenyour answer to Question 3, clockwise or counterclockwise?

Answer:Counterclockwise.

Question 5: Lenzs Law says that the induced current should be such as to create a self-magnetic field (due to the induced current alone) which tries to keep things from

changing. What is the direction of the self-magnetic field due to your induced current

inside the circuit loop, into the page or out of the page? Is it in a direction so as to keepthe flux through the loop from changing?

Answer:The magnetic field generated by the induced current is out of the page, hencepositive magnetic flux since we choose out of the page as positive. This opposes the

decrease in external magnetic flux as more of the loop enters the region of zero externalmagnetic field.

Question 6: What is the magnitude of the current flowing in the circuit at the time shown(use | | /I R!= )?

Answer:The magnitude of the induced current is Iinduced

=

Bwvz(t)

R,

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Question 7: Besides gravity, what other force acts on the loop in the k-direction? Give

the magnitude and direction of this force in terms of the quantities given (hint:

d!

Fmag

=I d!

s !!

B ).

Answer:There is a magnetic force directed upwards (opposing the motion) on the top legof the loop given by

!

Fmag

= I d!

s! !

Btop leg

" =Iinduced!

L ! !

B =Bwv

z(t)

R(#wj) ! (Bi) =

B2w2

Rv

z(t) k.

The force on the two side legs cancel and there is no force on the bottom leg.

Question 8: Assume that the loop has reached "terminal velocity"--that is, that it is no

longer accelerating. What is the magnitude of that terminal velocity in terms of given

quantities?

Answer:When the loop has reached terminal speed, the sum of the forces on the loop are

zero, hence

!

Fmag

+

!

Fgrav

=B

2w

2

Rvz,term

k! mgk=!

0 .

So the terminal speed is

vz,term

=

mgR

B2

w2

Question 9: Show that atterminal velocity, the rate at which gravity is doing work onthe loop is equal to the rate at which energy is being dissipated in the loop through Joule

heating (the rate at which a force does work is!

F ! !

v )?

Answer:Because the loops I now moving at terminal speed, the rate that gravity doeswork is given by

!

Fgrav

!

!

v = "mgk( ) ! vy ,termk ="mgvy,term =m

2g

2R

B2w

2

Recall that vy,term

=!

mgR

B2w

2. Also recall that I

ind,term =

Bwvz,term

R=

mg

Bw. Therefore

!

Fgrav

!

!

v =m2g2R

B2w2 =I

ind,term

2 R .