Physics 2B Final
Name:
Summer Session I 2009
PID:
Version 0
Quiz ID:
READ FIRST: Make sure to write/fill in your scantron form your name,
quiz ID and the test version.
e ≈ 1.6×10−19 C , k ≈ 8.9×109 N m2 /C2 , 0 ≈ 8.9×10−12 C2 /N m2 , µ0 = 4π×10−7 N/A2 = 4π×10−7 H/m .
Z
Z
dx
dx
x
= ln |x| ,
= √
2
2
3/2
x
(x + a )
a2 a2 + x2
All coordinate system are right-handed.
1. In the following setup, two solenoids connected to a battery and a resistor is threaded
with an iron rod. The circuit on the right is moved in the direction of the arrow. What
is the direction of the induced current through the resistor?
Solution: According to the direction of the battery and the winding of the left
solenoid, it creates a magnetic field pointing from right to left. As the second solenoid
moves away, the magnetic flux decreases. Lenz’s rule dictates that the second solenoid
will create a magnetic field in the same direction to compensate for the decreasing
magnetic flux. In order to create such a magnetic field, the current flows upward on
the visible side of the solenoid and from left to right across the resistor.
A. Left to right.
B. Right to left.
C. Zero.
D. Cannot tell.
2. A closed loop conductor with radius 2.0 m is located in a uniform magnetic field of 0.1
T. At what angular rate should one rotate the loop such that the peak emf induced in
the loop is 5.0 V?
Physics 2B Final
Summer Session I 2009
Version 0
Solution: The area of the loop is A = πr2 ≈ 12.56m2 . The induced EMF is given
by the rate of change of the magnetic flux:
E =−
dΦB
d
d
= − BA cos θ = − BA cos ωt = BAω sin ωt
dt
dt
dt
The peak EMF is therefore Ep = BAω. Solving for angular rate ω gives:
ω=
A. 4.0 s−1 .
B. 0.4 s−1 .
Ep
5.0 V
=
≈ 4.0 s−1
2
BA
0.1 T × 12.56 m
C. 1.0 s−1 .
D. 12.5 s−1 .
3. A very long rectangular conductor loop of resistance 10 Ω and mass 20 g is falling into a
region of uniform, perpendicular 3 T magnetic field. Its width w is 20 cm. As it enters,
it is falling with speed 2 m/s. What is current on the conductor as it enters the magnetic
field?
w
g
v
Solution: The motional EMF is, as shown in class:
E = Bwv = 3 T × 0.2 m × 2 m/s = 1.2 V
The current, using Ohm’s law is:
I = E/R = 1.2 V/10Ω = 0.12 A
A. 0.12 A
B. 2.0 A
C. 0.03 A
D. 3 A
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Physics 2B Final
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4. As the conductor loop from the last problem falls into the region. It will eventually—if
it is long enough—reach a terminal velocity when the net force acting on it is zero. Find
the terminal velocity.
Solution: Suppose the terminal velocity if vT . Same as last problem, the current in
B w vT
the loop is I =
. The magnetic force on the left and right segment of the loop
R
cancels. Only the magnetic force of the bottom segment contributes: F = wIB =
B 2 w 2 vT
. Terminal velocity is reached when this force balances out the gravitational
R
force:
B 2 w2 vT
mg =
R
mgR
0.2 N × 10Ω
vT = 2 2 =
≈ 5.56 m/s
B w
9 T2 × 0.04 m2
A. 5.56 m/s
B. 50.0 m/s
C. 3.33 m/s
D. 15.6 m/s
5. Two identical conductor loops—each with 1000 turns, 2–cm diameter and negligible
length—are placed 2–m apart in parallel on the same axis. Find the mutual inductance
between them.
Solution: The separation between the two loops is much bigger than their radius, it
suffices to assume that the magnetic field in the vicinity of one loop due to the other
is roughly constant and is given by the on-axis field we calculated in class (Example
30-1 in textbook). When one of the ring has current I, the magnetic flux in the other
loop is approximately:
ΦB = N BA ' N ·
µ0 N I r 2
· πr2
2d3
The mutual induction is:
M=
ΦB
µ0 N 2 πr4
4π × 10−7 N/A2 · (1000)2 · π(1 cm)4
=
≈ 2.47 × 10−9 H
=
I
2d3
2 · (2 m)3
A. 2.47 nH
B. 39.5 nH
C. 4.94 nH
D. 0.79 nH
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Physics 2B Final
Summer Session I 2009
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6. A small conductor loop of radius 5 cm is placed concentrically and in the same plane as
a big conductor loop of radius 1 m. What is the mutual inductance between them?
Solution: When the big conductor carries current I, recall from class (or formula
sheet) that the magnetic field at the center is:
B=
µ0 I
2R
Since the other conductor is very small, we can assume that the magnetic field
through it is constant. This gives magnetic flux:
ΦB = BA =
µ0 I
· πr2
2R
The mutual inductance is therefore:
M=
A. 4.93 nH
B. 628 nH
µ0 πr2
= 4.93 × 10−9 H
2R
C. 1.57 nH
D. 27.6 nH
7. A 7.0-µF capacitor is charged and has a 9.0–V potential difference across its plates. If
the capacitor is discharged through a 140.0–kΩ resistor, how long does it take for the
potential difference across the plates of the capacitor to drop to 1% of its initial voltage.
Solution: The RC constant is τ = R C = 7.0 µF · 140 kΩ = 980 ms. The voltage
across a discharging capacitor as function of time is: V = V0 e−t/τ . The time for the
required voltage is:
t = −τ ln
A. 4.5 s
B. 45 s
C. 45 ms
V
= −(980 ln 0.01)ms ≈ 4510ms
V0
D. 450 ms
8. A 180 Ω resistor, a 2.5 mH inductor, and a 12 V battery are connected in series. How
long after the connection is made does the current reach 99% of its saturation value?
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Physics 2B Final
Summer Session I 2009
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Solution: After long time, the inductor acts as a short and the saturation current
is I0 = 12 V/180 Ω = 66 mA. 99% of that corresponds to I = 66 × 0.99 = 65.34 mA.
This, however, is not needed. Just solve for time from the formula for current as
function of time:
I = I0 (1 − e−t/τL ) t = −τL ln(1 − I/I0 ) = −
A. 64 ms
B. 14 µs
C. 32 ms
L
ln 0.01 ≈ 64.0 × 10−6 s
R
D. 6.1 ms
9. A toroid has 2–cm side square cross section. Its inner radius is 10–cm and it has 1000
turns. What is its self-inductance?
Solution: When there is a current I, the magnetic field of a toroid is:
B=
µ0 N I
,
2πr
where r is the distance to the axis of revolution. The magnetic flux through the
toroid can be computed by integrating the magnetic field over the cross-section:
Zb
Z
ΦB = N
Bda = N h
µ0 N I
dr
2πr
a
2
L=
A. 0.73 mH
ΦB
µ0 N h b
=
ln
I
2π
a
4π × 10−7 N/m · (1000)2 · 0.02 m 12
=
ln
≈ 0.00073 H
2π
10
B. 6.4 mH
C. 8.0 mH
D. 4.6 mH
10. You are to wind a long wire into a solenoid 10-cm long and 2-cm diameter cross section.
How many turns do you have to wind to have an inductance of 100 mH. Assume uniform
winding and ignore fringe field.
Solution: The magnetic field inside the solenoid is uniform and is B = µ0 nI. n is
the number of turn per length. The magnetic flux through the solenoid is:
µ0 N I
µ0 N 2 A
ΦB = N BA = N
A=
·I
l
l
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Physics 2B Final
Summer Session I 2009
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The inductance is simply the coefficient before I:
s
r
Ll
0.1 × 0.1
=
N=
≈ 5033
µ0 A
12.6 × 10−7 × π · 0.012
A. 5032
B. 2.53 × 106
C. 159
D. 4460
11. A sinusoidal AC source is connected in series to a 2 µF capacitor and a 100 Ω resistor.
What is the frequency of the AC source if the peak voltage across the resistor is half
that of the source?
Solution: The voltage across the resistor is VR = IZR , where the current can be
calculated using Ohm’s law: Vs = I(ZC + ZR ). Therefore:
ZR
VR = Vs ×
ZR + ZC
1 VR R 1
R
√
⇒ω= √
≈ 2886 s−1
= =
1 =
2
Vs
R + iωC
R2 + ω −2 C −2
3RC
The frequency is therefore f =
A. 460 Hz
B. 397 Hz
ω
≈ 459 Hz
2π
C. 795 Hz
D. 562 Hz
12. For an RLC circuit with resistance of 150 Ω, a capacitance of 9.0 µF, and an inductance
of 37.0 mH, what frequency is needed to minimize the impedance?
Solution: The total impedance is:
s
Z=
R2
+
2
1
+ωL
ωC
Impedance is minimized at resonance frequency:
f0 =
A. 275 Hz
ω0
1
1
√
= √
=
≈ 275 Hz
2π
2π · 37 × 10−3 H · 9 × 10−6 F
2π L C
B. 118 Hz
C. 1.732 kHz
D. 645 Hz
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Physics 2B Final
Summer Session I 2009
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13. A device with impedance Z = 400 + i300 Ω is being driven by a sinusoidal AC current
Is (t) = 0.3 sin(20t) A. What is the voltage drop V (t) across the device?
Solution: The current is also sinusoidal. The peak current is related to the peak
voltage by:
Vp = Ip |Z| = 0.3 A · 500Ω = 150 V
30
The phase of the impedance in radian is φz = arctan
≈ 0.644. This is the amount
40
in phase for which the voltage is ahead of the current. Therefore the current is:
V (t) = IP |Z| sin(ωt + φz ) = 150 sin(20t + 0.644) V
A. 150 sin(20t + 0.64) V.
D. 120 sin(20t − 0.64) V.
B. 150 sin(20t − 0.64) V.
C. 120 sin(20t + 0.64) V.
14. In the following circuit, both batteries are 9 V and all five resistors are 50 Ω. Which of
the following is not a correct equation obtained by applying Kirchoff’s rules?
i1
i2
i3
A. 150i3 − 50i2 = 9
B. i1 − i2 + i3 = 0
C. −50i1 + 150i3 = 0
D. 50(i1 + i2 ) = 9
15. Two identical, 2-kg small spheres carrying the same charge is hung from the same point
off the ceiling by a 10-cm long string. At equilibrium, their distance is 2 cm. How much
charge is on each sphere? Hint: draw a picture.
Solution: Three forces need to balance here: gravity, electrostatic, and tension.
Work out the forces in the horizontal and vertical direction separately on one of the
sphere:
kq q 2
d/2
d2
d/2
Fx = 2 − T
=0
⇒ q2 =
·T
d
l
kq
l
p
l2 − (d/2)2
l
Fy = T
− mg = 0 ⇒ T = mg p
2
l
l − (d/2)2
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Physics 2B Final
Summer Session I 2009
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Solving the two equations for q gives:
q2 =
d2
l
d
0.022
1
mg √
=
≈ 8.76 × 10−14 C2
× 2 × 9.8 × √
9
2
2
2
2
kq
8.9 × 10
l +d l
10 + 1
Therefore q ≈ 2.96 × 10−7 C.
A. 296 nC
B. 938 nC
C. 294 nC
D. 806 nC
16. On the x–axis there are two point charges: a 4–nC charge at the x = −1 and a −9–nC
charge at x = 1. On what point of the x-axis is the electric field zero?
Solution: To the left of 4–nC charges and to the right of −9–nC charge, the direction
of the electric field due to the two charges are in opposite direction. So we expect
the answer to be either x < −1 or x > 1. Use the principle of superposition:
4
−9
E(x) ∝
+
=0
(x + 1)2 (x − 1)2
9(x + 1)2 = 4(x − 1)2
3(x + 1) = ±2(x − 1)
This give x = −5 with the upper sign and x = −1/5 with the lower sign. The former
is the correct answer.
A. x = −5
B. x = −1/5
C. x = −13/5
D. x = −5/13
17. A long solid rod 2 cm in radius carries a uniform charge. If the electric field strength
at the surface of the rod (not near either end) is 4 kN/C. What is the volume charge
density?
Solution: For a uniformly charged rod of radius a, the relationship between volume
charge density and linear charge density is:
λ = ρπa2
The electric field of on the surface of the long rod is, as shown in class:
E=
λ
ρa
=
.
2π0 a
20
From this one can solve for volume charge density ρ:
ρ=
2ε0 E
≈ 3.54 × 10−6 C/m3
a
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Physics 2B Final
A. 3.54 µC/m3
Summer Session I 2009
B. 7.08 µC/m3
C. 0.35, µC/m3
Version 0
D. 11.1 µC/m3
7.1
>
18. A circular plate of 10–cm
radius carries a uniform charge of 5–µC. Approximately how
many times stronger is the electric field 1 mm above the center of plate than the electric
field 10 m above the center plate? Hint: what does the plate look like from these
distances?
Solution: 1 mm above the plate, the electric field is essentially that of an infinite
plate:
σ
Q
5 × 10−6 C
Enear =
=
=
.
20
20 πr2
20 π(0.1 m)2
10 m above the plate, the electric field is essentially that of a point charge:
Efar =
The ratio is
A. 40000
5 × 10−6 C
Q
=
.
4π0 d2
4π0 (10 m)2
Enear
4d2
2 × 102
= 2 =
≈ 40000.
Efar
2r
0.0712
B. 10000
C. 62500
D. 1000
19. How much work is done to assemble eight 20–µC charges into eight corner of a 10–cm
side cube?
5
1
8
4
6
2
7
3
Solution: Move the charge in one by one in any order:
1. 0.
2. U2 = kq 2 /a.
√
3. U3 = kq 2 (1/a + 1/ 2a).
√
4. U4 = kq 2 (2/a + 1/ 2a).
√
√
5. U5 = kq 2 (1/a + 2/ 2a + 1/ 3a).
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Physics 2B Final
Summer Session I 2009
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√
√
6. U6 = kq 2 (2/a + 2/ 2a + 1/ 3a).
√
√
7. U7 = kq 2 (2/a + 3/ 2a + 1/ 3a).
√
√
8. U8 = kq 2 (3/a + 3/ 2a + 1/ 3a).
Add them up:
U = kq
2
12
4
12
+√ +√
a
2a
3a
−6 2
9
= (8.9 × 10 )(20 × 10 )
12
12
4
√ +
√ ≈ 820.6 (J)
+
0.1 0.1 2 0.1 3
√
Alternative: Each √
charge sees three charges at distance 10 cm, three charges 2·10
cm, and one charge 3 · 10 cm away. The total energy is the sum of in-place energy
for each charge divide by two to account for double-counting:
8
1 X X ke qj
U=
qi
2 i=1 j6=i rij
3
3
1
1
9
−6 2
√ +
√ ≈ 820.6 (J)
+
= · 8 · (8.9 × 10 ) · (20 × 10 ) ×
2
0.1 0.1 2 0.1 3
A. 821 J
B. 1641 J
C. 205 J
D. 82 J
20. How much work is done to assemble six 20–nC charges into the configuration that each
occupy a center of the six faces of a 2–mm side cube?
Solution: Use the alternative
solution of the last problem. Each charge sees four
√
charges at distance 2 mm (red line) and one charge 2 mm (blue line) away. Sum
over the in-place energy for each charge and divide by two to account for double
Page 10
Physics 2B Final
Summer Session I 2009
Version 0
counting:
U=
1 X X ke qi qj
2 i j6=i rij
4
1
8.9 × 109 · (20 × 10−9 )2
×6× √
+
≈ 0.036 (J)
=
2
2 × 10−3 2 × 10−3
Alternative: Count the number of bonds. There are 3 bonds (blues) of length a
and 12 bonds (red) of length a/sqrt(2). Add up the energy of each bond:
U =3×
kq 2
kq 2
+ 12 × √
a
a/ 2
12
3
+√
≈ 0.036 (J)
= 8.9 × 10 · (20 × 10 ) ×
2 × 10−3
2 × 10−3
−9 2
9
A. 36 mJ
B. 72 mJ
C. 6 mJ
D. 41 mJ
21. At each of the eight corners of a cube is an proton. The sides of the cube are 0.5 nm.
What is electric potential at the center of the cube?
√
Solution: The center is the same distance away from each charges: 0.5 nm 3/2 ≈
0.433 nm. Calculate the potential using principle of superposition:
J·m
9 × 109 2 · 1.6 × 10−19 C
kq
C
φ=8
≈8×
= 26.6 V
d
0.433 × 10−9 m
A. 26.6 V
B. 3.33 V
C. 0 V
D. 6.65 V
22. A 2–nC charge is located at the center of a cube with 20–cm sides. What is the electric
flux through each surface of the cube?
Solution: According to Gauss’s Law, the electric flux through any closed surface is
the total charged enclosed divided by the permittivity of free space.
ΦE =
2 × 10−9 C
N m2
Qenclosed
=
≈
226
0
C
8.85 × 10−12 C2 /N m2
N m2
By symmetry, each face shares 1/6 of that: 37.6
.
C
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Physics 2B Final
Summer Session I 2009
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N m2
N m2
N m2
N m2
B. 226
C. 18.0
D. 3.0
C
C
C
C
23. A long co-axial cable consists of a cylindrical inner shell of radius ra = 1.0 cm and a
concentric outer shell of radius rb = 30.0 cm. They are each carrying 10 A of current
in the opposite direction. How much energy is stored in the magnetic field between the
inner and outer shells in a 1–m segment of the cable?
A. 37.6
Solution: The magnetic field between the core and the shell can be found using
Ampère’s law. It is:
µ0 I
B=
2πr
This gives magnetic energy density:
uB =
B2
µ0 I 2
= 2 2.
2µ0
8π r
Integrate over the cross section area gives the energy per length:
Zrb
Z
U=
uB dV =
µ0 I 2 l
uB 2πrdr l =
4π
ra
A. 34 µJ
B. 10 µJ
Zrb
dr
µ0 I 2 l rb
≈ 34 × 10−6 J
=
ln
r
4π
ra
ra
C. 8 µJ
D. 70 µJ
24. Two capacitors, C1 =3.0 µF and C2 =9.0 µF, are charged separately to 5 V. After the
battery is removed, the positively charged plate of each capacitor is connected to the
negatively charged plate of the other capacitor. What is the final charge of C1 ?
Solution: C1 is initially charged to Q01 = C1 V0 = 15 µC and C2 is initially charged
to Q02 = C2 V0 = 45 µC.
Suppose the final charges are Q1 and Q2 respectively. Conservation of charge gives:
Q1 + Q2 = Q01 − Q02 = 30 µC .
When the capacitors are connected, they have the same voltage:
Q1 /C1 = V1 = V2 = Q2 /C2 .
Solving these two equations gives
Q1 =
C1
(Q0 − Q02 ) = 7.5 µC .
C1 + C2 1
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Physics 2B Final
A. 7.5 µC
Summer Session I 2009
B. 22.5 µC
C. 15 µC
Version 0
D. 45 µC
25. Mercury has resistivity 9.84 × 10−7 Ω·m. A long tube of mercury of 2.0–mm diameter
has a uniform 25.0-V/m electric field inside. What is the magnetic field at the surface
of the tube?
Solution: Relating conductivity to resistivity, the current density is:
J = σE =
25.0 V/m
E
=
≈ 2.54 × 107 A/m2
−7
ρ
9.84 × 10 Ω · m
The current is then just the product of current density and the cross-sectional area:
I = JA = 2.54 × 107 A/m2 · π(1.0 × 10−3 m)2 ≈ 80 A
The magnetic field at the surface is:
B=
A. 16 mT
B. 8 mT
µ0 I
4π × 10−7 N/A2 · 80 A
=
≈ 160 × 10−4 T
2πr
2π · 1 × 10−3 m
C. 32 mT
D. 64 mT
26. What is the equivalent resistance of the following network of four identical resistors, each
with resistance R?
R2
R1
R4
R3
Solution: Going from left to right:
1. First two resistors are in series: R1−2 = 2R.
2. The third one is in parallel with the first two: R1−3 =
R1−2 R
2R
=
.
R1−2 + R
3
3. The fourth resistor is in series with the first three: R1−4 =
Page 13
5R
.
3
Physics 2B Final
A.
5R
3
B.
3R
5
Summer Session I 2009
C.
R
4
Version 0
D. 4R
27. A 10–F parallel plate capacitor is charged to 3 V. After it is disconnected, it is then
submerged in a non-conducting liquid with electric permittivity 4 times that of free
space and the distance between the two plates is halved. What is the voltage across the
capacitor now?
A
Solution: The formula for parallel plate capacitance is C =
. Since the capacid
tance is proportional to and inversely proportional to distance, the new capacitance
is 8 times that of the original: C = 8C0 . The voltage is reduced by a factor of 8
given the same charge.
A. 3/8 V
B. 3 V
C. 24 V
D. 3/2 V
28. Remember that magnetic monopole do not exist. Which of the following vector fields
cannot be a magnetic field?
Solution: Mathematically the statement that magnetic monopole do not exist trans~ = 0. Out of the choices,
~ ·B
lates to that the magnetic field has zero divergence: ∇
only x x̂ + y ŷ has non-zero divergence:
div(x x̂ + y ŷ) =
A. x x̂ + y ŷ
B.
y x̂ − x ŷ
x2 + y 2
∂x ∂y
+
=2
∂x ∂y
C. sin(x − y) x̂ + sin(x − y) ŷ
D. 3x2 y 3 x̂ − 2x2 y 3 ŷ
29. Two protons (mass 1.67 × 10−27 kg), moving in a plane perpendicular to a uniform
magnetic field of 0.5 T, undergo an elastic head-on collision and flies apart. How much
time elapses before they collide again? Ignore the electrostatic repulsion between them.
Solution: A charge particle in a magnetic field undergoes circular cyclotron motion.
The frequency of the motion does not depend on the velocity of the particle or the
radius of the trajectory. The two proton will come back to the original point of
collision—regardless of their speed after the collision—after a period of cyclotron
motion:
1
2πmp
2π · 1.67 × 10−27 kg
T =
=
≈
≈ 13.1 × 10−8 s
fc
eB
1.6 × 10−19 C · 0.5 T
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Physics 2B Final
A. 131 ns.
Summer Session I 2009
B. 762 ns.
C. 262 ns.
Version 0
D. Never.
30. An electric dipole centered at the origin has electric dipole moment 3.0 µC · m is oriented
at an angle of π/6 with respect to the x-axis. The positive charge is in the first quadrant.
An electric field parallel to the y-axis exert a torque of 14 ẑ N·m. What is the magnitude
and direction of the electric field?
Solution: The electric dipole moment can be written in vector form as
π π
~p = 3.0 cos x̂ + sin ŷ µC · m .
6
6
Notice that both the x and y components are negative because the dipole moment
points from the first quadrant toward the third quadrant.
The torque of an electric field on a dipole is:
√
~
~τ = ~p × E
14 ẑ = 3.0
√
3 3
·E
14 =
2
A. 5.38 ŷ N/C
B. -5.38 ŷ N/C
1
3
x̂ + ŷ
2
2
!
× E ŷ
28
E = √ ≈ 5.38
3 3
C. -9.3 ŷ N/C
D. 9.3 ŷ N/C
31. Two long wires are both carrying a current of 0.3 A. One wire runs along the x-axis with
the current going from x < 0 to the x > 0 direction. The other runs along the y-axis
with the current going from y < 0 to the y > 0 direction. What is the magnetic field on
the z-axis, at z = 1.5 m? Drawing a picture would help.
Solution: By the right hand rule, the x-axis running wire creates a magnetic field
in the −ŷ direction on the z-axis and the y-axis running wire creates a magnetic field
in the x̂ direction.
The point is 1.5 m away from both wires. The strength of the magnetic field by each
wire is:
2 × 10−7 TAm × 0.3 A
µ0 I
B=
=
= 4 × 10−8 T .
2π z
1.5 m
Using the principle of super position, the total magnetic field is
~ = 4 × 10−8 (−x̂ + ŷ) T .
B
A. 40(x̂ − ŷ) nT.
B. 40(−x̂ + ŷ) nT.
C. 80(x̂ − ŷ) nT.
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D. 80(−x̂ + ŷ) nT.