Chapter 23
4.
Picture the Problem
: A house has a floor of dimensions 22 m by 18 m. The local magnetic field due
to Earth has a horizontal component 2.6×10
5
T and a downward vertical component 4.2×10
5
T.
Strategy:
The horizontal component of the magnetic field is parallel to the floor, so it does not
contribute to the flux. Use equation 231 to calculate the flux using the vertical component.
Solution:
Calculate the
magnetic flux:
Insight:
The flux through the vertical walls of the house is determined by the horizontal component of
the magnetic field instead of the vertical component.
11.
Picture the Problem
: The figure shows the flux through a
single loop coil as a function of time.
Strategy:
Use equation 233 to calculate the emf at the times
t
=
0.05 s, 0.15 s, and 0.50 s. Use the graph to find the change
in flux.
Solution:
1. (a)
Calculate
the emf at
t
=
0.05 s:
2.
(b
Calculate the emf
at
t
=
0.15 s:
3. (c)
Calculate the emf
at
t
=
0.50 s:
Insight:
When the slope of the flux is constant, the emf is constant. The emf is 0.04 kV from
t
= 0.2 s
to
t
= 0.6 s.
13.
Picture the Problem
: The magnetic flux through a coil
oscillates in time as indicated by the graph at the right.
Strategy:
Use equation 234 to estimate the
magnitude of the emf from the graph.
Solution:
1. (a)
The magnitude of the induced
emf is greater
near
t
= 0.5 s, because
is changing more rapidly
there. (In fact, it is not changing at all at the
instant
t
= 0.4 s.)
2. (b)
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 Spring '08
 Dr.G
 Physics, EMF, Magnetic Field, Energy density, Lenz

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