physics 301: intermediate electromagnetism fall...

PHYSICS 301: INTERMEDIATE ELECTROMAGNETISMFALL 2010, PROBLEM SET 10

Name:_______________________________PROBLEM SET Due Date: F 11/12 4pm

Topics/Chapters: Griffiths’ §7.1, 7.2: electrodynamics, electromotive force, Faraday’s law, induction.Relevant Class Meetings: 11/03 W + 11/05 F + 11/08 M

Recommended Reading: Griffiths §7.1, §7.2Required Reading: Griffiths §7.3.1-7.3.3

1. Griffiths 7.07 – changing flux

2. Griffiths 7.09 – flux surface

3. Helmholtz coils with AC current – magnetic and electric fields

4. Griffiths 7.17 – solenoid and a loop

5. Griffiths 7.20 – mutual inductance for two loops

______________________________________

1. Griffiths 7.07. Changing flux for a rectangular loop.

Collaborators:_________________________________

Two key concepts that you used:_________________________________

Useful pages in book and/or dates from class:_____________________________

2. Griffiths 7.09. Flux surface in integral vector calculus.

Collaborators:_________________________________



3. Helmholtz coils with AC current. Magnetic and electric fields.

Collaborators:_________________________________



-q

q

3q !"

#"

$"

%"

&%"

%"

CHAPTER 3. SPECIAL TECHNIQUES 73

But q! = !R

aq (Eq. 3.15), so

V (r, !) "=1

4"#0

!q2r

a2cos ! ! R

aq2

a

R2

r2cos !

"=

1

4"#0

#2q

a2

$#r ! R3

r2

$cos !.

Set E0 = ! 1

4"#0

2q

a2(field in the vicinity of the sphere produced by ±q):

V (r, !) = !E0

#r ! R3

r2

$cos ! (agrees with Eq. 3.76).

Problem 3.47The boundary conditions are

(i) V = 0 when y = 0,(ii) V = V0 when y = a,(iii) V = 0 when x = b,(iv) V = 0 when x = !b.

%&&'

&&(

Go back to Eq. 3.26 and examine the case k = 0: d2X/dx2 = d2Y/dy2 = 0, so X(x) = Ax+B, Y (y) = Cy+D.But this configuration is symmetric in x, so A = 0, and hence the k = 0 solution is V (x, y) = Cy +D. PickD = 0, C = V0/a, and subtract o! this part:

V (x, y) = V0y

a+ V̄ (x, y).

The remainder (V̄ (x, y)) satisfies boundary conditions similar to Ex. 3.4:

(i) V̄ = 0 when y = 0,(ii) V̄ = 0 when y = a,(iii) V̄ = !V0(y/a) when x = b,(iv) V̄ = !V0(y/a) when x = !b.

%&&'

&&(

(The point of peeling o! V0(y/a) was to recover (ii), on which the constraint k = n"/a depends.)The solution (following Ex. 3.4) is

V̄ (x, y) =")

n=1

Cn cosh(n"x/a) sin(n"y/a),

and it remains to fit condition (iii):

V̄ (b, y) =)

Cn cosh(n"b/a) sin(n"y/a) = !V0(y/a).

Invoke Fourier’s trick:

)Cn cosh(n"b/a)

* a

0sin(n"y/a) sin(n!"y/a) dy = !V0

a

* a

0y sin(n!"y/a) dy,

a

2Cn cosh(n"b/a) = !V0

a

* a

0y sin(n"y/a) dy.

c#2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.

Adapted from Griffiths, Introduction to Electrodynamics, Fig. 5.60. Adapted from Purcell, Electricity and Magnetism, Fig. 7.16.

We recently learned about the configuration called the "Helmholtz coilds", shown above, in which cur-rent flows in the !̂-direction through two coils separated by a distance d = R. As indicated in theright-hand figure, the field will be nearly uniform in the middle of the coils.

For the following calculations, a time-dependent current flows through the coils:I(t) = I0 sin("t) with a maximum value of 100 Amps and a frequency of f=100/2# Hz. Assume theresulting field is uniform in the regions of interest, and is parallel (or antiparallel) to the z-axis.

(a). Write a analytical expression for the maximum $B-field at the origin, as a function of time. Thisshouldn’t be arduous: you may use the results for B(z) that we previously calculated. What is thenumerical value of the maximum field at the origin?

(b). Sketch a plot of the flux !(t) through loop C, which has radius a = 10cm and is centered at theorigin in the xy-plane. What is the maximum value of flux?

(c). Sketch a plot of the electromotive force E(t) for the loop C. Indicate the maximum value of emf.

(d). There are several options for finding the electric field along the loop C for this configuration.Without doing intensive calculation, sketch the 3D vector field on the curve C at a single momentin time (what time did you pick?) and indicate the direction that current flows in the loop (if any).As part of your justification, verify that your field satisfies Faraday’s Law:

$!" $E = #% $B

%t.

4. Griffiths 7.17. Solenoid and a loop.

Collaborators:_________________________________



5. Griffiths 7.20. Mutual inductance for two loops.

Collaborators:_________________________________



physics 301: intermediate electromagnetism fall...

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