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PHY 303L
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Storr
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Physics

PHY 303L

Storr

Summer

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rashid (sar3553) – HW9 electromagneticInduction – storr – (37537) 1
This print-out should have 11 questions.
Multiple-choice questions may continue on I
the next column or page – ﬁnd all choices
before answering. R
001 10.0 points
A coil is wrapped with 125 turns of wire on
the perimeter of a circular frame (of radius What is the magnitude of the average emf
25 cm). Each turn has the same area, equal generated in the loop?
to that of the frame. A uniform magnetic
ﬁeld is directed perpendicular to the plane ofCorrect answer: 51941.7 ▯V.
the coil. This ﬁeld changes at a constant rate
Explanation:
from 28 mT to 49 mT in 22 ms. The initial angle between the normal to the
What is the magnitude of the induced E in plane of the circular loop and the magnetic
the coil at the instant the magnetic ﬁeld has
ﬁeld is
a magnitude of 43 mT? ◦ ◦ ◦
θi= 51 − 90 = −39
Correct answer: 23.4281 V. θ = −39 + 180 = 141 . ◦
f
Explanation: dΦ B
From Faraday’s Law E = −N for a
Basic Concepts: dt
solenoid the average emf is
dΦ B
E = −N
dt Eavg= −N ∆Φ B = −N ∆B · A
Z ∆t ∆t
B π r2
ΦB ≡ B · dA = B · A = −N (cosθf− cosθ i
∆t
(74 ▯T)π (0.66 m)2
Solution: = −(29.7 turn)
0.09 s
dΦ B × (cos141 − cos−39 )
E = −N dt
= 51941.7 ▯V
= −N A ∆B
∆t |Eavg| = 51941.7 ▯V .
(B − B )
= −N π r 2 2 1
∆t 003 10.0 points
= −(125)π (25 cm)2 A magnetic ﬁeld of 0.0913 T exists in the re-
(49 mT) − (28 mT) gion enclosed by a solenoid that has 510 turns
× 22 ms and a diameter of 9.41 cm.
Within what period of time must the ﬁeld
= −23.4281 V
|E| = 23.4281 V. be reduced to zero if the average magnitude
of the induced emf within the coil during this
time interval is to be 15.1 kV?
002 10.0 points
Correct answer: 2.14453 × 10 s.
A 29.7 turn circular loop of wire has a diam-
eter of 1.32 m. The plane of the circular loopExplanation:
makes an angle of 51 to the Earth’s magnetic Basic Concept:
ﬁeld. In 0.09 s the coil is ﬂipped at a locationraday’s Law:
where the magnitude of the Earth’s magnetic dΦ B
ﬁeld is 74 ▯T. E = −N
dt rashid (sar3553) – HW9 electromagneticInduction – storr – (37537) 2
From Faraday’s Law, we get a = 7.6 cm,
b = 19 cm, and
|E| = −N dΦ B I = 54 A.
dt 0
dB
= N A dr
dt
∆B
= N A )
∆t δ
B +
= N A t ω t
r ℓ
So, the time needed equals sin(
I
N AB
t = I
E
(510 turns)(0.00695455 m )(0.0913 T) a b
=
(15.1 kV)
= 2.14453 × 10−5 s.
Magnetic ﬁeld near a long wire is
▯0I
004 (part 1 of 2) 10.0 points B = 2π r .
A long, straight wire carries a current and liFaraday’s Law is
in the plane of a rectangular loops of wire, as
shown in the ﬁgure. dΦ
E = − B .
dt
→
i δ 120 loops (turns) The magnitude of the magnetic ﬁeld is
+
t ▯0I
s) B = 2π r .
/
Thus the ﬂux linkage is
29 cm
h (205 rad ΦB = N Φ B1
▯ N I ℓ Z a+b dr
= 0
19 cm 2π a r
. cm ▯ N I ℓ a + b
(54 A)sin = 0 0 ln sin(ω t + δ).
2π a
Determine the maximum emf, |E|, induced Finally, the induced emf E is
in the loop by the magnetic ﬁeld created by
the current in the straight wire. dΦ B
E = −
dt
Correct answer: 96.5219 mV. ▯0N I 0ω a + b
= − 2π ln a cos(ω t + δ),
Explanation:
which is a maximum when the cosine function
Let : N = 120, yields 1.
ω = 205 rad/s, ▯0N a + b
E max= I0ℓω ln
ℓ = 29 cm, 2π a rashid (sar3553) – HW9 electromagneticInduction – storr – (37537) 3
(1.25664 × 106 N/A )(120)
=
2(3.14159)
× (54 A)(29 cm)(205 rad/s)
R l Fapp
× ln (7.6 cm) + (19 cm)
(7.6 cm)
m mV
× 10 −2 103
cm V At what speed should the bar be moved to
produce a current of 0.44 A in the resistor?
= 96.5219 mV .
Correct answer: 0.40357 m/s.
Explanation:
Let : R = 5.96 Ω,
005 (part 2 of 2) 10.0 points
If at t = 0, δ=0, and a positive I denotes ℓ = 1.9 m,
B = 3.42 T, and
an upward moving current in the ﬁgure, then
which statement below is correct at t = 0. I = 0.44 A.
The emf inside the loop can be calculated
1. There is zero current in the loop. using Ohm’s Law
E = I R
2. The current in th

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