**D. Magnetism
Part I**

**19.1-19.3**

**19.1 Magnets, Magnetic
Poles, Magnetic Field Direction**

**2, 4, 5 (ask about 4 and 5
in class if you don’t understand)**

2: South pole of a compass is the north pole of
a magnet

**19.2 Magnetic Field strength and magnetic force**

**6 – 12, 15, 16, 18, 19**

7: Point fingers (of right hand) in direction
of + charge and thumb in direction of deflection. Curl your fingers to determine B.

10: a)
charges of 1 and 3 are equal in magnitude but opposite in sign. You decide which one is + and which is
-. Obviously, charge #2 is
neutral. B) Heavy masses can’t be deflected as well)

11: draw it with Ď (B field into the page) and ť (B field out of the page)

12: F = qvBsinq

15: mass of proton = 1.67 x 10^{-27}kg. Charge on proton = 1.6 x 10^{-19}C. Assume the particle is ^ to the field so q = 90)

16: electrons are deflected in the opposite
direction of protons. If it were a
proton, your thumb would point in the direction of deflection (force). But, since this is an electron, your thumb
should be oriented in the direction OPPOSITE of the deflection (eg, it should
point up along the +y axis, not down along the –y axis)

18: EASY!!!!! But, realize there are two
answers.

19: Point your fingers in the direction of the
proton’s velocity and curl them up to 37^{o}. Where your thumb points is the direction of the deflection. Don’t forget your old friend F=ma to
determine the acceleration.

**19.3 Electromagnetism – the source of magnetic
fields**

**20-23 (draw these
out!!!!) 28, 29, 34, 37, 38, 39**

28: a.
The fields are equal in magnitude but opposite in direction. Therefore,
midway between them there will be no net field. (ask yourself this, however, would there be a field between them
at a point which was not mid-way? The
answer is YES, do you understand why?).
b. the distance will be 0.25m
since that is one half the distance between the wires which are separated by a
distance of 0.5m.

29: Both will be the same so you need only solve
for one. Let us solve for the point B
which is in the left side of the figure.
At that point, the field from the left wire is coming out of the page
0.15 m from that left wire. Also, at
that point, the field created by the right wire is coming out of the page but
it is 0.20 m + 0.15 m = 0.35 m away from the right wire. Solve for B_{left} and B_{right}
and add ‘em up!).

34: Earth’s magnetic field ad the equator is 1 x
10-5T. Notice how small of a current it
takes to beat the Earth’s magnetic field.

37: B=m_{o}nI where n = #of
turns/meter)

38: solve for B_{inner} and B_{outer}. Realize that their fields are in opposite
direction. So, you need to subtract
them.

39: Notice B_{2} will go straight down
through the yellow plane. Notice that B_{1}
will go straight OUT along the yellow plane.
Therefore, B_{1}^B_{2} and the resultant
vector can then be found by Pythagorean theorem.

Even
Answers:

2: b

4: ask in class

6: no

8: into the page

10: particle 1 is negative, 2 is neutral, 3 is
positive

12: 0.40T

16: 1.0 x 10^{-6}T in –z direction (-z
is into the page. +z is out of the
page)

18: 30^{o} and 150^{o}

20: a

22: b

28: a) 0 (only at the midpoint!!!!!) b)
6.4 x 10^{-6}T

34: 1.6 A

38: 8.8 x10^{-2}T