# Physics 1304/1404 - Spring 1999

# Homework Assignment #4

**Due:** Thursday 4 March 1999

### READING

Chapters 29 and 30.
### QUESTIONS

Chapter 29 - 3, 4, 8, 11, 21.

Chapter 30 - 3, 10, 12, 16, 19, 27.
### PROBLEMS

Chapter 29 - 10, 14, 22, 26, 33, 40.

Chapter 30 - 3, 8A, 10, 19, 21A, 21, 24, 40, 43, 44, Review Problem.
### ANSWERS

These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to
copy these and turn them in as homework. You must show your work.
**Q 29-3)** No, there might be a magnetic field parallel or antiparallel
to the velocity vector of the charged particle.

**Q 29-4)** They move in circles with the opposite orientation (one
clockwise and one counterclockwise). The radius of the proton's orbit
is larger than the radius of the electron's orbit, because the proton
is more massive than the electron. R=mv/qB.

**Q 29-8)** The magnetic force is always perpendicular to the charged
particle's velocity, therefore the magnetic force can only change the
direction of the particle and not its speed. In other words, magnetic
forces do no work. (Remember the work-energy theorem:
W = K_{f} - K_{i})

**Q 29-11)** Yes, orient the plane of the loop perpendicular to the
magnetic field.

**Q 29-21)**
- The magnetic field is into the picture. (Remember that electrons
have negative charge.)
- The magnetic field would point out of the picture and the electron
beam would be deflected up.

**Q 30-3)**
- The magnetic fields due to the two wires add constructively
between the wires; the resultant magnetic field is stronger than
it would be if just one wire were present.
- The magnetic fields due to the two wires partially cancel each other
outside the wires; the resultant magnetic field is weaker than
it would be if just one wire were present.

**Q 30-10)** B=0 inside the tube because no current is enclosed in an
Amperian loop inside the tube.

B is nonzero outside the tube. An Amperian loop outside the tube will
enclose current.

**Q 30-12)**
- the magnetic field is halved.
- the magnetic field is doubled.

**Q 30-16)** The end of a bar magnet will stick to the center of a bar of
metal, but the end of a bar of metal will not stick to the center of a
bar magnet.

**Q 30-19)** The North Pole of the Earth is the south pole of a giant
magnet.

**Q 30-27)** All substances are diamagnetic, some to a very small degree.
Only substances with unpaired outer electrons (which have some net electron
spin, and therefore also a permanent magnetic moment) can be paramagnetic
or ferromagnetic. These last two are really the same phenomenon --
ferromagnetism is just very strong paramagnetism.
Diamagnetic materials are repelled by a magnetic field (north or south).
Para- and ferromagnetic materials are attracted by a magnetic field (north
or south). Usually, the order of strengths of the interactions is:
diamagnetism < paramagnetism < ferromagnetism.

**P 29-10)** 2.09 x 10^{-2} T in the negative y direction

**P 29-14)** 0.109 A to the right

**P 29-22)**
- 5.41 x 10
^{-3} A.m^{2}
- 4.33 x 10
^{-3} N.m

**P 29-26)**
- 1.18 x 10
^{-4} N.m
- -1.18x 10
^{-4} J < U < +1.18x 10^{-4} J

-1.18x 10^{-4} J, aligned parallel to B

0 J, perpendicular to B

+1.18x 10^{-4} J, anti-aligned with B

**P 29-33)** r_{d} = r_{alpha} = r_{p} sqrt(2)

**P 29-40)** 2.44 x 10^{5} V/m

**P 30-3)** 0.314 m

**P 30-8A)** B = mu_{0}I/(2 pi R) + mu_{0}I/(2R) into page

**P 30-10)** 3 x 10^{-5} T into the page

**P 30-19)** 2.70 x 10^{-5} N toward the infinite wire

**P 30-21A)** B =
mu_{0}I_{1}/(2 pi R_{1}) toward I_{2} +
mu_{0}I_{2}/(2 pi R_{2}) away from I_{1}

**P 30-21)** 1.30 x 10^{-5} T down the page (negative y direction)

**P 30-24)** 31.8 mA

**P 30-40)**
- pi R
^{2} B cos(theta)
- same as above since the net flux through the CLOSED surface
S
_{1} + S_{2} is zero.

**P 30-43)**
- 1.13 x 10
^{10} V.m/s
- 0.10 A (same as the current in the wire)

**P 30-44)**
- 7.2 x 10
^{11} V/(m.s) (that is, V/m per second)
- 2 x 10
^{-7} T

**P 30-Review Problem)**
- upper: V/3R in the direction shown by the arrows

lower: V/R opposite the direction shown by the arrow
- zero
- [sqrt(2)-1]mu
_{0}V^{2}/(18 pi R^{2}) repulsive
- [sqrt(2)-1]mu
_{0}V^{2}/(6 pi R^{2}) attractive
- 3d/4 above the bottom side, d/4 below the top side

Please report any corrections to
Professor Scalise.

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