Physics 2220 – Module 08 Homework
01.
An electron moves along the z-axis with vz = 2.0 × 107 m/s. As it passes the origin, what are the strength
and direction of the magnetic field at the (x, y, z) positions of (a) (1 cm, 0 cm, 0 cm), (b) (0 cm, 0 cm, 1
cm), and (c) (0 cm, 1 cm, 1 cm)?
Treat this as a current of one electron which creates a magnetic field as it moves. The Biot-Savart Law
then becomes:
B=
μ 0 qv sin (θ)
4 π
r2
Where θ is the angle between the velocity vector and the vector pointing to the point in question.
(a)
(1 cm, 0 cm, 0 cm) - The angle θ is this case equal to 90 degrees, so:
−7
B = (10
(1.60 × 10−19 C) (2.0 × 10 7 m/s) sin (90o )
T m / A)
= 3.2 × 10−15 T
−2
2
(1.0 × 10 m)
By definition the current is going opposite the direction of the electron. Use the right hand rule to
see the direction of the magnetic field is in the negative y-axis.
B = (−3.2 × 10− 15 ̂j) T
(b)
(0 cm, 0 cm, 1 cm) - The angle θ is this case equal to zero, so:
B=0
(c)
(0 cm, 1 cm, 1 cm) - The angle θ is this case equal to 45 degrees, so:
−7
B = (10
(1.60 × 10−19 C) (2.0 × 10 7 m/s) sin (45 o)
T m / A)
= 1.13 × 10−15 T
−2
2
(1.0 × 10 m)
By definition the current is going opposite the direction of the electron. Use the right hand rule to
see the direction of the magnetic field is in the positive x-axis.
B = (1.13 × 10−15 ̂i ) T
02.
The magnetic field at the center of a 1.0-cm-diameter loop is 2.5 mT.
(a)
What is the current in the loop?
Use the equation found for the center of the current loop:
μ0 I
2R
−3
2 (2.5 × 10 T) (0.5 × 10−2 m)
2BR
I = μ0 =
= 20 A
4 π (10−7 T m / A )
B=
(b)
A long straight wire caries the same current you found in part (a). At what distance from the wire
is the magnetic field of 2.5 mT?
Use the equation found for a straight wire:
μ0 I
2π d
−7
μ I
4 π (10 T m / A) (20 A)
d= 0
=
= 1.6 × 10−3 m
−3
2πB
2 π (2.5 × 10 T)
B=
03.
What are the magnetic field strength and direction at points a to c in the figure??
Both currents will contribute to the magnetic field at each point.
Label the top wire as 1 and the bottom wire as 2.
Consider into the paper as negative and out of the paper as positive.
(a)
Using the right hand rule for current, B1 will be out of the paper, and B2 will be into the paper.
μ0 I
1
1
−
2 π 2.0 cm 6.0 cm
T m / A ) (10 A)
1
1
−
2π
2.0 cm 6.0 cm
B A = B1 − B2 =
BA =
4 π (10−7
(
)
(
)
B A = 6.7 × 10−5 T (out of the page)
(b)
Using the right hand rule for current, B1 will be into the paper, and B2 will be into the paper.
μ0 I
1
1
−
−
2π
2.0 cm 2.0 cm
T m / A ) (10 A)
1
1
−
−
2π
2.0 cm 2.0 cm
(
B B = −B1 − B2 =
BB =
4 π (10−7
)
(
)
BB = −2.0 × 10− 4 T (into the page)
(c)
Using the right hand rule for current, B1 will be into the paper, and B2 will be out of the paper.
μ0 I
1
1
−
2 π 2.0 cm 6.0 cm
T m / A) (10 A)
1
1
−
2π
2.0 cm 6.0 cm
BC = B 2 − B 1 =
BC =
4 π (10−7
(
)
(
BC = 6.7 × 10−5 T (out of the page)
)
04.
The Earth's magnetic dipole moment is 8.0 × 1022 A m2.
(a)
What is the magnetic field strength on the surface of the Earth at the Earth's north magnetic
pole? How does this compare to the value in Table 32.1? You can assume that the current loop
is deep inside the Earth.
Use the current loop model and assume the size of the loop is much much smaller than the
radius of the Earth (6.38 × 106 m). The magnetic field then becomes:
B=
μ0 μ
2 π z3
4 π (10−7 T m / A) (8.0 × 1022 A m2 )
B=
= 6.2 × 10−5 T
6
3
2 π (6.38 × 10 m)
Not a terrible approximation.
(b)
Astronauts discover an earth-sized planet without a magnetic field. To create a magnetic field
with the same strength as Earth's, the propose running a current though a wire around the
equator. What size current would be needed?
Use the magnetic dipole moment.
μ = IA
→
I=
μ
2
π z
8.0 × 10 22 A m2
I=
= 6.3 × 10 8 A
6
2
π (6.38 × 10 m)
05.
What is the line integral of B between points i and f in the figure?
Line integral will be:
f
∮ ⃗B ⋅d ⃗s
i
Note that the direction of the magnetic field and and
direction of the integration path is always 90
degrees.
f
f
∮ ⃗B ⋅d ⃗s = ∮ B ds cos (90) = 0
i
06.
i
The value of the line integral of B around the closed path in the figure is 1.38 × 10 -5 T m. What are the
direction (in or out of the page) and magnitude of I 3?
Recall Ampere's Law for an enclosed
current:
∮ ⃗B ⋅d ⃗s = μ0 I enc
The enclosed current in this case are wires two and three, therefore:
∮ ⃗B ⋅d ⃗s = μ0 (I 2 + I 3 ) = 1.38 × 10−5 T m
Solve for the current in the third wire:
I3 =
I3 =
1.38 × 10−5 T m
− I2
μ0
1.38 × 10−5 T m
− (12 A) = −1.0 A
4 π (10−7 T m / A )
Since the current is negative, and positive current was assumed for wire two going into the page, the
direction of the current in the third wire is out of the page.
07.
The two 10-cm-long parallel wires in the figure are separated by 5.0 mm. For what value of the resistor
R will the force between the two wires be 5.4 × 10-5 N?
The currents for the two sections of wire separated by 5.0 mm will be the same direction. The force per
unit length between the two wires is:
F μo I 1 I 2
=
L
2π a
Both circuits will obey Ohm's Law respectively:
I1 =
Δ V1
R1
I2 =
Δ V2
R2
Combine these equations and solve for R2.
F μo Δ V 1 Δ V 2
=
L
2 π a R1 R2
μ Δ V1 Δ V2 L
R2 = o
2 π a R1 F
−7
−2
4 π (10 T m / A ) (9 V) (9 V ) (10 × 10 m)
R2 =
= 3.0 Ω
2 π (5.0 × 10−3 m ) (2 Ω) (5.4 × 10−5 N )