Download Physics 2049 Exam 3 Solutions Wednesday, October

Physics 2049 Exam 3 Solutions
Wednesday, October 24, 2001
~ as shown in the figure (both ~v and B
~ are in
1. An electron moves with a velocity ~v in a magnetic field B
the plane of the paper). What is the direction of the force on the electron?
B
v
Answer: into the page
~ Using the right hand rule, we see
Solution: The magnetic force on a charged particle is F~M = q~v × B.
~ is out of the page; however, an electron is negatively charged, so the magnetic force is into
that ~v × B
the page.
2. A straight horizontal copper wire of length 20 m carries a current of 10 A. The mass of the wire is
1 kg. What is the magnitude of the smallest magnetic field needed to suspend the wire (i.e., balance
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the force of gravity)? Recall that g = 9.8 m/s .
Answer: 4.9 × 10−2 T
Solution: The magnetic field needs to be horizontal, directed so that the magnetic force points up.
The magnitude of the magnetic force in this case is ilB, with i the current, l the length of the wire,
and B the magnitude of the magnetic field. In equilibrium this force must balance the weight of the
wire, which is mg; equating these two forces, we have ilB = mg, so that B = mg/(il). Substituting
the numbers, we obtain 4.9 × 10−2 T.
3. A proton, moving in the x-direction with a velocity ~v = v0 î, enters a region of space in which there are
~ = E0 ĵ,
crossed electric and magnetic fields. The electric field is in the y-direction and is given by E
~ = B0 k̂, with
with E0 = 10 V/m, and the magnetic field is in the z-direction and is given by B
B0 = 0.2 T. For what value of v0 does the particle continue with the same speed and direction?
Answer: 50 m/s
Solution: The magnetic force on the proton is F~M = q~v × B = −qv0 B0 ĵ; the electric force is F~E =
~ = qE0 ĵ, so the net force is F~ = q(E0 − v0 B0 )ĵ. Since we want the proton to continue on its original
qE
path, the net force should be zero, so that v0 = E0 /B0 = 50 m/s.
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4. A long hairpin is formed by bending an infinitely long wire as shown. If R = 4.5 cm and the wire
carries a current of i = 4 A, what is the magnetic field (magnitude and direction) at the point P ?
i
R
P
Answer: 4.6 × 10−5 T, out of the page
Solution: The magnetic field at the point P is the superposition of the fields due to the semi-infinite
segment on the top, the semicircle of radius R, and the semi-infinite segment on the bottom. Using
the right hand rule, we see that all of these contributions point out of the page at the point P , so we’ll
add them together. The semi-infinite segments each contribute a field Bstraight = µ0 i/(4πR), and the
semicircle contributes Bsemicircle = µ0 iπ/(4πR), so the magnitude of the total field is
B=2
µ0 iπ
µ0 i
+
= 4.6 × 10−5 T.
4πR
4πR
(1)
5. Three long, straight current-carrying wires which are parallel, coplanar, and a distance of d = 10 cm
from each other are shown in the figure below (the wires are perpendicular to the page). The left and
right wires carry a current of 1.0 A out of the page. What current i0 (magnitude and direction) is
needed in the center wire so that the system is in equilibrium?
d
d
?
i
i’
i
Answer: 0.5 A into the page
Solution: Let’s focus on the wire on the right. It is attracted by the wire on the left, so the only way
for it to be in equilibrium is if it is repelled by the center wire; therefore, the current i0 must be into
the page. Assuming that all wires have a length l, The force on the right wire is
F =−
µ0 i2 l µ0 i|i0 |l
+
.
2d
d
In equilibrium F = 0, so we find that |i0 | = i/2 = 0.5 A (i.e., 0.5 A into the page).
2
(2)
6. The figure below shows a cross-section of a collection of current-carrying wires, each
of which carries
H
~ · d~s for the path
a current of 3.0 A either into the page or out of the page. What is the value of B
marked “A”?
A
Answer: 7.5 × 10−6 T · m
H
~ · d~s = µ0 ienc , where ienc is the current which passes through
Solution: Ampére’s Law tells us that B
the area inside the path A. The net current through this path is 2 × (3 A) = 6 A, so
I
~ · d~s = (4π × 10−7 T · m/A)(6 A) = 7.5 × 10−6 T · m.
(3)
B
7. A cross section of a hollow conducting cylinder, with inner radius a and outer radius b, is shown below.
The cylinder carries a current out of the page, with the current density given by J = cr2 , with r the
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distance from the axis of the cylinder and c = 3.0 × 106 A/m . If a = 1.0 cm and b = 2.0 cm, what is
the magnitude of the magnetic field at r = b?
b
a
Answer: 7.1 × 10−6 T
Solution: Ampére’s Law applied to the a point on the outer surface of the cylinder tells us that
I
~ · d~s = (2πb)B = µ0 ienc ,
B
(4)
so that B = µ0 ienc /(2πb). Here ienc is the total current passing through the cylinder; this was deterR
~ = 0.71 A. Substituting the values of
mined in Exam 2, Problem 6 (see the solutions); ienc = J~ · dA
b and µ0 , we obtain 7.1 × 10−6 T.
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8. The Colliding Beam Fusion Reactor (being developed by Prof. Monkhorst and collaborators) will have
a cylindrical fusion chamber, whose inside dimensions will be 2 m in diameter and 12 m in length.
External coils will create a uniform magnetic field of magnitude 10 T, oriented along the axis of the
cylinder. How much magnetic field energy will be stored in this chamber?
Answer: 1.5 × 109 J
Solution: The total energy stored in the magnetic field is B 2 /(2µ0 ) times the volume inside the chamber;
letting l = 12 m be the length and r = 1.0 m its radius, we have
B2
(πr2 l) = 1.5 × 109 J.
2µ0
UB =
(5)
9. A rigid circular loop consisting of 10 turns of wire has a radius of 2 cm. Its plane is perpendicular to
a magnetic field whose time dependence is given by B(t) = 0.1 + 0.2t − 0.5t2 , where B is in tesla and t
is in seconds. If the resistance of the loop is 10 Ω, what is the induced current in the loop at t = 1 s?
Answer: 1.0 mA
Solution: The flux through the loop is ΦB = (πr2 )B(t), so the induced emf is
E
= −N
dΦB
dt
dB
dt
= −πr2 N (0.2 − 1.0t).
= −(πr2 )N
(6)
Evaluating this at t = 1 s, with r = 0.02 m and N = 10, we obtain an emf of E = 10.1 mV. If the
resistance of the loop is 10 Ω, then the induced current is i = 10.1 mV/10 Ω = 1.0 mA.
10. A rectangular loop of wire, with sides of length a and b shown in the figure below, is coplanar with a
long wire carrying a current i. The distance between the wire and the left side of the loop is r. The
loop is pulled to the right as indicated. What is the direction of the induced current and the directions
of the magnetic forces on the left and right sides of the loop?
r
b
i
a
Answer: Current direction: clockwise; force on left side: to the left; force on right side: to the right
Solution: The magnetic field from the long wire is into the page in the region to the right of the wire.
By pulling the loop to the right, we decrease the flux into the page; therefore, Lenz’s Law tells us
the the induced current will produce a magnetic field which is into the page inside the loop. This is
accomplished with a clockwise current. On the left side this current is parallel to the current on the
wire, so that segment is attracted to the wire and the force is to the left; on the right side this current
is antiparallel to the current in the wire and the force is to the right. The force on the left side is
larger, so the net force on the loop will be to the left.
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