# Homework Assignment #4

Due: Tuesday 30 June 1998

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.

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 = Kf - Ki)
Q 29-11) Yes, orient the plane of the loop perpendicular to the magnetic field.
Q 29-21)
1. The magnetic field is into the picture. (Remember that electrons have negative charge.)
2. The magnetic field would point out of the picture and the electron beam would be deflected up.

Q 30-3)
1. 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.
2. 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)
1. the magnetic field is halved.
2. 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)
1. 5.41 x 10-3 A.m2
2. 4.33 x 10-3 N.m

P 29-26)
1. 1.18 x 10-4 N.m
2. -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) rd = ralpha = rp sqrt(2)

P 29-40) 2.44 x 105 V/m
P 30-3) 0.314 m
P 30-8A) B = mu0I/(2 pi R) + mu0I/(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 = mu0I1/(2 pi R1) toward I2 + mu0I2/(2 pi R2) away from I1
P 30-21) 1.30 x 10-5 T down the page (negative y direction)
P 30-24) 31.8 mA
P 30-40)
1. pi R2 B cos(theta)
2. same as above since the net flux through the CLOSED surface S1 + S2 is zero.

P 30-43)
1. 1.13 x 1010 V.m/s
2. 0.10 A (same as the current in the wire)

P 30-44)
1. 7.2 x 1011 V/(m.s) (that is, V/m per second)
2. 2 x 10-7 T

P 30-Review Problem)
1. upper: V/3R in the direction shown by the arrows
lower: V/R opposite the direction shown by the arrow
2. zero
3. [sqrt(2)-1]mu0V2/(18 pi R2) repulsive
4. [sqrt(2)-1]mu0V2/(6 pi R2) attractive
5. 3d/4 above the bottom side, d/4 below the top side

Please report any corrections to Professor Scalise.