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Unformatted text preview: 4 rad/s) sin 27° = 12 V
76. REASONING AND SOLUTION
2 The energy stored in a capacitor is given by 2 Equation 19.11b as CV . The energy stored in an inductor is given by Equation 22.10 as
2 L I 2 . Setting these two equations equal to each other and solving for the current I, we get C
3.0 × 10 −6 F
(35 V ) = 0.86 A
5.0 × 10 −3 H
I= 77. REASONING The magnitude ξ of the average emf induced in the triangle is given by
(see Equation 22.3), which is Faraday’s law. This expression can be used
directly to calculate the magnitude of the average emf. Since the triangle is a single-turn coil,
the number of turns is N = 1. According to Equation 22.2, the magnetic flux Ф is ξ = −N 1236 ELECTROMAGNETIC INDUCTION Φ = BA cos φ = BA cos 0° = BA (1) where B is the magnitude of the field, A is the area of the triangle, and φ = 0° is the angle
between the field and the normal to the plane of the triangle (the magnetic field is
perpendicular to the plane of the triangle). It is the change ∆Φ in the flux that appears in
Faraday’s law, so that we use Equation (1) as follows:
∆Φ = BA − BA0 = BA where A0 = 0 m2 is the initial area of the triangle just as the bar passes point A, and A is the
area after the time interval ∆t has elapsed. The area of a triangle is one-half the base (dAC)
times the height (dCB) of the triangle. Thus, the change in flux is ∆Φ = BA = B ( 1 dACdCB )
2 The base and the height of the triangle are related, according to dCB = dAC tan θ, where
θ = 19º. Furthermore, the base of the triangle becomes longer as the rod moves. Since the
rod moves with a speed v during the time interval ∆t, the base is dAC = v∆t. With these
substitutions the change in flux becomes
∆Φ = B ( 1 d AC dCB ) = B 1 dAC ( d AC tan θ ) = 1 B ( v∆t )2 tan θ
2 (2) SOLUTION Substituting Equation (2) for the change in flux into Faraday’s law, we find that
the magnitude of the induced emf is ∆Φ
ξ = −N
2 B ( v∆t ) tan θ
2 ∆t = N 1 Bv 2 ∆t tan θ
2 = (1) 1 ( 0.38 T )( 0.60 m/s ) ( 6.0 s ) tan19° = 0.14 V
2 Chapter 22 Problems 1237 78. REASONING AND SOLUTION
a. When the magnet is above the ring its
magnetic field points down through the
ring and is increasing as the magnet
The induced magnetic field
attempts to reduce the increasing field
and points up. The induced current in
the ring is as shown in the drawing at the
right, and then the ring looks like a
magnet with its north pole at the top,
repelling the north pole of the falling
magnet and retarding its motion.
When the magnet is below the ring, its magnetic field still points down through the ring but
is decreasing as the magnet falls. The induced magnetic field attempts to bolster the
decreasing field and points down. The induced current in the ring is as shown in the
drawing, and the ring then looks like a magnet with its north pole at the bottom, attracting
the south pole of the falling magnet and retarding its motion.
b. The motion of the magnet is unaffected, since no induced current can flow in the cut ring.
No induced current means that no induced magnetic field can be produced to repel or attract
the falling magnet.
79. SSM REASONING According to Equation 22.3, the average emf ξ induced in a single
loop (N = 1) is ξ = −∆Φ / ∆t . Since the magnitude of the magnetic field is changing, the
area of the loop remains constant, and the direction of the field is parallel to the normal to
the loop, the change in flux through the loop is given by ∆Φ = ( ∆B )A . Thus the magnitude
ξ of the induced emf in the loop is given by ξ = − ( ∆B ) A / ∆ t .
Similarly, when the area of the loop is changed and the field B has a given value, we find
the magnitude of the induced emf to be ξ = − B ( ∆A ) / ∆ t .
a. The magnitude of the induced emf when the field changes in magnitude is ∆B 2
–3 = (0.018 m )(0.20 T/s) = 3.6 × 10 V ∆t ξ = −A b. At a particular value of B (when B is changing), the rate at which the area must change
can be obtained from B∆A
∆t or ∆A ξ 3.6 × 10 –3 V
= 2.0 ×10 m / s
1.8 T 1238 ELECTROMAGNETIC INDUCTION In order for the induced emf to be zero, the magnitude of the magnetic field and the area of
the loop must change in such a way that the flux remains constant. Since the magnitude of
the magnetic field is increasing, the area of the loop must decrease, if the flux is to remain
constant. Therefore, the area of the loop must be shrunk .
80. REASONING AND SOLUTION If the rectangle is made ∆I = If wide, then the top of the
rectangle will intersect the line at LIf. The work is one-half the area of the rectangle, so
2 ( LI f ) I f = 1 LI f
2 2 ____________...
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