from the floor to
a uniform magnetic field pointing straight up,
y—direction be
be upward, the xadirection be eastward, and the
A room is filled with
the ceiling. Let the 2-direction
northward.
(a) A positively charge
magnetic force on it, if any?
(b) Now
165
Chapter 39
(d)
culates clockwise or counterclockwise. To
answer this, apply the righthand rule to the cross oint your right fingers in the
direction of v. Initially, that's the ans-direction. Then, curl your fingertips so that they point in the
you fi
Chapter 39 167
° rt J ' : ~Q and mass or gets shot rightward at initial speed so between the two
plates of a parallel~plate capacitor. The plates, fixed a distance 5 apart,
I have length l. The particle enters the capacitor midway between the
plates, a
Chapter 39 1 71
Before moving on to part (c), let me make a theoretical point. As demonstrated in this
problem, when a particle feels both a magnetic field and an electric field, the total force is
FaFelec+Fmg =qE+qva=q(E+va).
Your textbook calls this the
Chapter 39
(b)
169
N
moo S
= S—-———
Q30 '
where in the last step, I used our expression for r from subproblem 1. _
lntuitiveiy, the particle doesn’t deflect if the electric force cancels the magnetic force. The plates
must generate an electric ﬁeld that p
Chapter 39 175
where I denotes the length and A denotes the cross-sectional area. Intuitively, a wider tunnel allows
more “car-current” to flow through, and therefore puts up less resistance. Also, if we assume a fixed
height difference between the two en
Chapter 39 17‘s
OK, here goes. The infinitesimal force on a tiny piece of the rod containing charge dq is
dFmag = dqv x B
qugxﬁ [sincev=dl/dt]
=dqglth
=Idle. [sincev=dl/dtl
You’ll use this crucial formula throughout the course. Even if the magnetic field
Chapter 39 17?“
Let’s begin by ﬁnding 1'2, the torque produced by F2. In this
diagram, I’ll drawn the relevant radius vector, from the center of
mass to the contact point where F: acts. I’ve also broken F2 into its
components parallel and perpendicular to
Chapter 39 i 79
(b)
To speed up an object, you must apply a force. But not just any force. At least a component
of the force must be directed along the object's direction of motion. For instance, consider a hockey
puck sliding rightward on frictionless ic
.
ems : t ’f"
Ampere’s Law and Blot—Saved
(Before attempting this problem, please work through your textbook’s sample problems that use the Biot—
Saoart law to calculate a magneticﬁeld. This problem is too hard to serve as your ﬁrst exposure.) _
Take
Chapter 39 I 181
These two conﬂicting intuitions "cancel out.” Our part (b) answer, B = 21rﬁn/ q, does not
depend on R. In words, the magnetic field needed to keep the particle circling the cyclotron with fre-
quency f does not depend on the cyclotron’s r
Chapter 40 185
Speciﬁcally, you can choose an arbitrary piece of the hexagon, point your fingers in the
direction of current flow (d1), and then curl your ﬁngertips in the direction of it,
towards the center. Your thumb points into the page. And this is t
Chapter 40 187
(c)
undeflected, along an imaginary axis through the center of the hexagon. Let me call that axis the
"central axis.”
Look at your textbook’s drawing of the magnetic ﬁeld produced by a symmetric current
loop. By symmetry, the ﬁeld at an
Chapter 40 191
depends only on where the objects are located, not on how fast they’re moving. Therefore, in parts
(b) and (c), the particle feels the same electric force it felt in part (a). If the particle feels a magnetic
force, it’s in addition to that
Chapter 40 189
So, the particle feels an electrostatic repulsion, and nothing else. Since Page = ngdue ,0 rod, the
meat of this problem is ﬁnding the rod’s electric ﬁeld. As you know, Gauss’ law enables you to cal-
culate the field produced by an “infinit
Chapter 40 1 93
and hence
B _, #_01 : ﬂoquo
2m 22ml. '
where in the last step I used my part (a) expression for the current. By the Way, we just derived the
textbook equation for the field generated by an infinite wire, B :
Now that we know the magne
Chapter 40 195
Fortunately, the superposition principle applies to all kinds of ﬁelds, not just electric ones.
If 31 denotes the field created by semicircle 1, 132 denotes the field generated by wire segment 2, and
so on, then the total field at point P
197
Chapter 40
Two Very long cylindrical wires of length L, each of radius R, run
right next to each other, as pictured here. In the end View, the left wire carries
current I“ into the page, While the right wire carries equal current out of the
page. In
Chapter 40 199
I’ll now rederive this same expression by thinking in terms of current density, the current per
cross—sectional area. Since the current density is uniform, I = Io/Awire z 10/ (1:12)." The Amperian loop
encloses area Am}. Therefore, the loop
Chapter 40 203
derived a special case of the textbook formula for the magnetic force between two parallel
We just 2
I L .
or antiparallel currents. it's P = #37; , where 1' denotes the center-to-center distance between the
two wires. Here, 1' = 2R.
Co
Chapter 40 201
Fortunately, they do agree. Substituting 1' = R into Bin and BM, you get
I
13in =Bout=gjzRﬂ atr=R.
By the reasoning of part (b), both wires generate a down-the—page field at point Y. The total
field has strength
BY = Bdue to left wire 'l’
Chapter 40
(b)
205
§B-ds = LB-ds + IZB'dS + LB-ds + LB-ds.
Let me save side 1 for last, and start With side 2. As mentioned above, the magnetic ﬁeld
inside the solenoid points parallel to its central axis. But side 2 runs perpendicular to the central axis
Chapter 40 207
{c} By What fraction (or percentage) did the magnetic field decrezise when we removed the 250th coil?
Explain intuitively why this fraction works out to be greater than 1/ 500, despite the fact that only
one of the 500 coils is missing.
“
An equilateral triangle is constructed of three thin wires, each with length 5
and resistance R. The triangle sits inside a uniform magnetic field of strength BU,
pointing into the page. At time t1, the magnetic field starts decreasing in strength at
a st
Chapter 40 209
(c)
ally inward, 511 and f are perpendiCular. Therefore, the magnitude of the cross product f reduces to
a regular product. And since the unit vector f has length 1, we get Idl X fl = disin 90" = d1.
Let’s assume the current flows clockwise