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Unformatted text preview: change in flux ∆Φ is given by
∆Φ = Β ∆Α cos φ, where ∆Α is the area of the loop that leaves the region of the magnetic
field in a time ∆t. This area is the product of the width of the rectangle (0.080 m) and the
length v ∆t of the side that leaves the magnetic field, ∆A = (0.080 m) v ∆t.
∆Φ = B ∆A cos φ = (2.4 T) (0.080 m)(0.020 m/s)(2.0 s) cos 0.0° = 7.7 × 10 –3 Wb ______________________________________________________________________________ 1200 ELECTROMAGNETIC INDUCTION 17. SSM WWW REASONING The general expression for the magnetic flux through an
area A is given by Equation 22.2: Φ = BA cos φ , where B is the magnitude of the magnetic
field and φ is the angle of inclination of the magnetic field B with respect to the normal to
SOLUTION 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 .
The flux through the 1.2 m by 0.30 m face is
Φ = BA cos φ = ( 0.25 T ) (1.2 m ) ( 0.30 m ) cos 0.0 ° = 0.090 Wb 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 = 53°
0.40 m φ = 90° – tan −1 Therefore, Φ = BA cos φ = ( 0.25 T ) (1.2 m ) ( 0.50 m ) cos 53 ° = 0.090 Wb
______________________________________________________________________________ 18. REASONING An emf is induced during the first and third intervals, because the magnetic
field is changing in time. The time interval is the same (3.0 s) for the two cases. However,
the magnitude of the field changes more during the first interval. Therefore, the magnetic
flux is changing at a greater rate in that interval, which means that the magnitude of the
induced emf is greatest during the first interval.
The induced emf is zero during the second interval, 3.0 – 6.0 s. According to Faraday’s law
of electromagnetic induction, Equation 22.3, an induced emf arises only when the magnetic
flux changes. During this interval, the magnetic field, the area of the loop, and the
orientation of the field relative to the loop are constant. Thus, the magnetic flux does not
change, so there is no induced emf.
During the first interval the magnetic field in increasing with time. During the third
interval, the field is decreasing with time. As a result, the induced emfs will have opposite
polarities during these intervals. If the direction of the induced current is clockwise during
the first interval, it will be counterclockwise during the third interval. SOLUTION
a. The induced emf is given by Equations 22.3 and 22.3: Chapter 22 Problems 1201 0–3.0 s: ξ =−N BA cos φ − B0 A cos φ
t − t0 B − B0
= − NA cos φ t −t
0 0.40 T − 0 T 2 = − ( 50 ) 0.15 m ( cos 0° ) = −1.0 V 3.0 s − 0 s ( ) 3.0–6.0 s: B − B0 0.40 T − 0.40 T 2 = − ( 50 ) 0.15 m ( cos 0° ) = 0V 6.0 s − 3.0 s t − t0 ( ξ = − NA cos φ ) 6.0–9.0 s: B − B0 0.20 T − 0.40 T 2 = − ( 50 ) 0.15 m ( cos 0° ) = +0.50 V t − t0 9.0 s − 6.0 s ( ξ = − NA cos φ ) b. The induced current is given by Equation 20.2 as I = ξ/R.
0–3.0 s: I= 6.0–9.0 s: I= ξ
R = −1.0 V
= −2.0 A
0.50 Ω = +0.50 V
= +1.0 A
0.50 Ω As expected, the currents are in opposite directions.
19. REASONING According to Faraday’s law, as given in Equation 22.3, the magnitude of the
emf is ξ = − (1)
, where we have set N = 1 for a single turn. Since the normal is
parallel to the magnetic field, the angle φ between the normal and the field is φ = 0° when
calculating the flux Φ from Equation 22.2: Φ = BA cos 0° = BA. We will use this
expression for the flux in Faraday’s law. SOLUTION Representing the flux as Φ = BA, we find that the magnitude of the induced
∆ ( BA)
∆t In this result we have used the fact that the field magnitude B is constant. Rearranging this
equation gives 1202 ELECTROMAGNETIC INDUCTION ∆ A ξ 2.6 V
= 1.5 m 2 /s
B 1.7 T
20. REASONING An emf is induced in the body because the magnetic flux is changing in time.
According to Faraday’s law, as given by Equation 22.3, the magnitude of the emf is
. This expression can be used to determine the time interval ∆t during which
ξ = −N
the magnetic field goes from its initial value to zero. The magnetic flux Φ is obtained from
Equation 22.2 as Φ = BA cos φ, where φ = 0° in this problem. SOLUTION The magnitude ξ of the induced emf is BA cos φ − B0 A cos φ ∆t ∆Φ ξ = −N = −N ∆t Solving this relation for ∆t gives
∆t = = − NA cos φ ( B − B0 ) ( ) = 4.8 s
21. ξ = − (1) 0.032 m 2 cos 0° ( 0 − 1.5 T ) SSM REASONING According to Equation 22.3, the average emf induced in a coil of N
loops is ξ = –N ∆Φ / ∆ t .
SOLUTION For the circular coil...
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