HMWK 4
Ch.25: Q5, 9, 14; P 1, 4, 12, 16, 18, 54, 59
Q25.5. Reason: The flux is defined in Equation 25.9: Φm = AB cosθ.
For circular loops A = π r2.
We compute the flux for each case.
2
Φ 1 = (π 1 ) B cos 0° = π B
Φ 2 = (π 12 ) B cos 45° = 0.707 Φ 1
2
Φ3 = (π 2 )B cos 45° = 2.8 3 Φ 1
Φ 4 = (π 2 2 )B cos 90° = 0
So the final ranking is Φ 3 > Φ1 > Φ 2 > Φ4 .
Assess: We see that both the area A and the cosine of the angle matter in the
calculation of the flux, even where B is uniform and constant.
Q25.9. Reason: The original field is into the page within the loop and is changing strength. The induced,
counterclockwise current produces a field out of the page within the loop that is opposing the change. This
implies that the original field must be increasing in strength so the flux into the loop is increasing.
Assess: If the original flux were decreasing then the induced current would be in the
opposite direction from that drawn to produce an induced field into the page to oppose
the decrease. That isn’t the case, so we know the field isn’t decreasing.
Q25.14. Reason: (a) As the north magnetic pole of the magnet approaches the coil, there is an increasing
upward magnetic flux through the coil. The induced current must be in such a direction as to oppose this increase
in magnetic flux. If the induced current is clockwise (as viewed from above), the induced magnetic field will be
downward. As the coil is pushed into the field, the upward magnetic flux increases, the induced clockwise
current increases, the induced magnetic field increases and the induced magnetic flux directed in the downward
direction increases.
(b) When the magnet is being pushed toward the coil, the magnetic situation at the site of the coil is as shown in
the figure below. The magnetic field at the site of the coil may be resolved into a vertical and horizontal
component.
Recall that the current is clockwise as indicated by the tail of an arrow representing the current at the left and the
head of an arrow at the right. Notice that as the induced current interacts with the magnetic field of the
approaching magnet, the vertical component of the magnetic field results in forces that try to collapse the coil
and the horizontal component of the magnetic field results in forces that try to push the coil vertically upward.
Assess: If the magnetic field associated with the magnet is strong enough and magnet is
thrust toward the coil fast enough, the resulting induced forces will support the coil.
P25.1. Prepare: Please refer to Figure P25.1. The magnetic field is perpendicular to the wire. The positive
charges experienced a magnetic force to the left. We will assume the magnetic field to be uniform.
Solve: This is a straightforward use of Equation 25.3. We have
B=
ε
vl
=
0.050 V
= 0.10 T
(5.0 m/s)(0.10 m)
By the right-hand rule the field must be out of the page so that the magnetic force on the positive charges is to
the left.
Assess: This is reasonable. Laboratory fields are typically up to a few teslas in
magnitude.
P25.4. Prepare: The motional emf induced may be determined by ε = Blv and the current is determined
by I = ε /R.
Solve: The motion emf is ε = Blv = (1.4 T)(0.15 m)(3.5 m/s) = 0.74 V. The current induced is I = ε /R =
0.74 V/0.65 Ω = 1.1A.
Assess: This is a reasonable potential difference and current for this situation.
P25.12. Prepare: Please refer to Figure P25.12. If the changing field produces a changing flux in the loop
there will be a corresponding induced emf and current. We will use Equations 25.9 and 25.11 and Lenz’s law.
Solve: (a) The induced emf is ε = |∆Φ/∆t| and the induced current is I = ε /R. The field B is changing, but the
area A is not. Take A to be out of the page and parallel to B, so from Equation 25.9 Φ = AB. Thus,
ε= A
I=
ε
R
∆B
∆B
= π r2
= |π ( 0.0 50 m )2 (0.50 T/s) | = 3.93 × 10 −3 V = 4 mV
∆t
∆t
=
3.9 3 × 10 −3 V
= 3.93 × 10 −2 A = 40 mA
0.1 Ω
The field is increasing out of the page. To prevent the increase, the induced field needs to point into the page.
Thus, the induced current must flow clockwise.
(b) As in part a, ε = A(∆B/∆t) = 4.0 × 10−3 V and I = 40 mA. Here the field is into the page and decreasing. To
prevent the decrease, the induced field needs to point into the page. Thus the induced current must flow
clockwise.
(c) Now A (left or right) is perpendicular to B and so ABcosθ = 0 Wb. That is, the field does not penetrate the
plane of the loop. If Φ = 0 Wb, then ε = |∆Φ/∆t| = 0 V/m and I = 0 A. There is no induced current.
Assess: Note that the induced field opposes the change.
P25.16. Prepare: Please refer to Figure P25.16. Assume the field is uniform across the loop. There is a
current in the loop so there must be an emf that is due to a changing flux. With the loop fixed the area is constant
so the change in flux must be due to a changing field strength.
Solve: The induced emf is ε = |∆Φ/∆t| and the induced current is I = ε IR. The B field is changing, but the area
A is not. Take A as being into the page and parallel to B, so Φ = AB and ∆Φ/∆t = A(B/∆t). We have
ε=
∆Φ
∆B
=A
∆t
∆t
∆B IR (150 × 10−3 A)(0.10 Ω)
=
=
= 2.3 T/s
∆t
A
(0.080 m)2
The original field and flux is into the page. The induced counterclockwise current produces an induced field and
flux that is out of the page. Since the induced field opposes the change, the field must be increasing.
Assess: This range of changing magnetic field is rather large, but still reasonable.
P25.18. Prepare: The induced emf is determined by ε induced = ∆Φ/∆t = A(∆B/∆t). The induced emf is in the
opposite direction of the applied voltage (the 9.0 volts from the battery, so the net emf is ε net = εapplied − ε induced.
The net emf and current are related by ε net = IR. Combining these expressions and solving for the current we
obtain: I = [εapplied − A(∆B/∆t)]/R.
Solve:
The current in the circuit and hence the resistor is
−4
−3
I = [ εapplied − A(∆B/∆t)]/R = [9.0 V − (25× 10 m )(0.50 T/10 × 10 s)]/(20 Ω) = 0.44 A
2
Assess: The current in the circuit is decreased slightly due to the fact that the induced
emf opposes the applied voltage.
P25.54. Prepare: The emf induced between the ends of the wire may be determined by
ε= N
Solve:
∆Φ
∆(BA)
∆A
π r2
=N
= NB
= NB
∆t
∆t
∆t
∆t
The induced emf between the ends of the wire is
ε = NB
π r2
∆t
−5
4
= (10 )(5.0 × 10 T)
π (4.0 × 10−2 m) 2
6.3 × 10−2 s
= 40 mV
Assess: This is a small but acceptable value for the induced emf.
P25.59. Prepare: Please refer to Figure P25.59 and assume that the magnetic field is uniform over the loop.
We are just interested in the emf due to the motion of the eye so we can ignore the details of the time dependence
of the change. The motion of the eye will change the orientation of the loop relative to the fixed field direction,
resulting in an induced emf.
Solve: From Faraday’s law,
ε co i l = N
=
∆Φ
∆t
=N
∆Φ
∆t
= N AB
∆ c os θ
∆t
=
Nπ r 2 B| cos θf − cos θ i |
20 π (3 × 10 −3 m )2 (1.0 T) | cos8 5° − cos 90° |
0.20 s
∆t
= 2.5 × 1 0 −4 V
Assess: This is a reasonable emf to measure, although you might need some
amplification.