Download Homework Hints Assignment (1-87)

Homework Hints
Assignment (1-87)
Read (22.3 - 22.4), Do PROBLEMS # (2, 15, 23) Ch. 22 + AP 1995 #7 (handout)
2.
Near San Francisco, where the vertically downward component of the earth’s magnetic field is
4.8 x 10-5 T, a car is traveling forward at 25 m/s. An emf of 2.4 x 10-3 volts is induced between
the sides of the car.
REASONING AND SOLUTION
a.
Which side of the car is positive ?
The right-hand rule shows that positive charge would be forced to the driver's side .
b.
What is the width of the car ?
According to the equation emf = BLv , the width of the car is
L=
2.4 × 10 –3 V
Emf
=
= 2.0 m
vB
( 25 m/s ) 4.8 × 10 –5 T
(
)
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15.
A five-sided object, whose dimensions are shown, is placed in a uniform magnetic field. The magnetic
field of magnitude 0.25 T points along the positive y-direction.
Determine the magnetic flux through each of the five sides.
1.20 m
.50 m
B
REASONING
The general expression for the magnetic flux through an
area A is given by the equation Φ = BA cosφ ,
where B is the magnitude of the magnetic field
and φ is the angle of inclination of the magnetic
field with respect to the NORMAL to the surface.
.30 m
.40 m
SOLUTION
<i> Since the magnetic field B is parallel to the surface for the triangular ends and the bottom
surface, the flux through each of these three surfaces is 0 Wb .
<ii> The flux through the 1.2 m by 0.30 m face is
Φ = (0.25 T)(1.2 m)(0.30 m) cos 0.0° = 0.090 Wb
<ii> For the 1.2 m by 0.50 m side, the area makes an angle φ with the magnetic field B,
where
 0.30 m 
φ = 90° – tan −1
comp
??  = 53°
 0.40
m
Therefore,
Φ = (0.25 T)(1.2 m)(0.50 m) cos 53° = 0.090 Wb
23.
A 1.5 m aluminum rod is rotating about an axis that is perpendicular to one end. A 0.16 T magnetic
field is directed parallel to the axis. The rod rotates through ¼ of a circle in 0.66 sec. What is the
magnitude of the average emf generated between the ends of the rod during this time ?
REASONING AND SOLUTION
The rod may be viewed as sweeping out an area whose initial value is zero (0) and final value is π r2/4.
The flux change due to this change in area is:
2
∆Φ = Bπ r /4
Faraday's law gives the magnitude of the emf to be
Emf = ∆Φ/∆t = (1/4)(0.16 T)(π)(1.5 m)2/(0.66 s) = 0.43 V