to the resistor is exactly
imvo2.
Problem 7.8 A square loop of wire (side a) lies on a table, a distance s from a
wire, which carries a current I, as shown in Fig. 7.17.
Physics 9
NAME:
Spring 2012
(a) Find the flux of B through the loop.
TA:
(b) If someone now pulls the loop directly away from the wire, at spee
generated? In what direction (clockwise or counterclockwise) does the curr
Quiz 8 - Solutions
(c) What if the loop is pulled to the right at speed v, instead of away?
Make sure your name is on your quiz, and please box your final answer. Because we will
be giving partial credit, be sure to attempt all the problems, even if you don’t finish them!
a
1. A square loop of wire (side a) lies on a
table, a distance s from a very long straight
wire, which carries current I, as shown in
the figure to the right.
s
I
I
0I
(a) Recalling that the magnetic field of an infinite wire is B = µ2πr
, find the flux of
Figure
7.17
~ through the loop. Hint: notice that B changes with distance from the wire!
B
Also, take dA to be a strip of length a and height dr.
(b) If someone now pulls the loop directly away from the wire, at speed v, what
emf, E, is generated? In what direction (clockwise or counterclockwise) does the
Problem
7.9 vAn
infinite number of different surfaces can be fit to a given b
current flow? (Hint:
note that
= ṡ.)
in defining
the magnetic
throughofaaway?
loop, <I> = B .da, I never spec
(c) What if the loop yet,
is pulled
to the right
at speedflux
v, instead
J
surface to be used. Justify this apparent oversight.
————————————————————————————————————
Solution
Problem 7.10
A square loop (side a) is mounted on a vertical shaft and
µ0 I
(Fig.
A uniform
magnetic
points
velocity
1. (a) Since the field is B
= 2πr , w
then
the7.18).
flux through
the loop
(out offield
the B
page)
is to the right. Fin
alternating currentRgenerator.
~
~
ΦM =
B · dA
R
µ0 I s+a 1
= 2π s r adr
out of a thick sheet of aluminum. It is then
Problem 7.11 A =
square
is cut
µ0 Ialoops+a
.
ln
2π
s
top portion is in a uniform magnetic field B, and allowed to fall under grav
(b) The induced emf,the
E, is
E = dtd Φshading
M . So, indicates the field region; B points into the page.) If
diagram,
µ0 Ia d laboratory
s+a
is 1 T (a pretty
field), find the terminal velocity of the l
E =standard
ln
2π dt
s
1
µ0 Ia
1
as aṡ function
of time. How long does it take (in seco
the velocity of=the loop
− s ṡ
2π
s+a
µ0 Ia2
90% of the terminal
velocity?
= − 2πs(s+a)
v, What would happen if you cut a tiny slit in
the circuit? [Note: The dimensions of the loop cancel out; determine the a
where v = ṡ is the velocity of the loop. Because the square is pulled away, the
the units indicated.]
magnetic flux is decreasing.
By Lenz’s law, we need to increase the flux and so
we need a counterclockwise current in the loop. (We can also see this by the fact
that the induced emf is negative.)
(c) If the loop is pulled to the right, then the flux is constant and so E = 0.
1