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Worksheet 21 - Induction, Faraday's Law

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Worksheet 21 - Induction,
Faraday's Law
Q1: A circular loop of radius 0.1 m is rotating in a uniform magnetic field of 0.2
T. Find the magnetic flux (in mWb) through the loop when the plane of the loop
and the magnetic field vector are at an angle of 30.0°. (mWb = 10 -3 T.m2)
Answer: 3.14
∅ = 𝐡𝐴 cos πœƒ = 0.2(πœ‹(0.1)2 ) cos(90 + 30) = 3.14 π‘šπ‘Šπ‘
Q2: A 11 loop coil of area 0.29 m2 is in a 0.047 T uniform magnetic field
oriented so that the maximum flux goes through the coil. The coil is then rotated
so that the flux through it goes to zero in 0.40 s. The average emf (in volts)
induced in the coil during the 0.40 s is
Answer: 0.37
𝑉=
𝑁 𝐡 𝐴 11(0.047)(0.29)
=
= 0.37 𝑉
𝑇
0.4
Q3: A conducting loop of wire has an area of 180 cm 2 and a resistance of 22 Ω.
There is a magnetic field of 200 T perpendicular to the loop. At what rate must
this field be reduced (in T/s) to induce a current of 0.10 A in the loop?
Answer:122.2
𝐼=
𝑉
𝑁𝐴 𝑑𝐡
=−
( )
𝑅
𝑅 𝑑𝑑
𝑑𝐡
−𝐼 𝑅
0.1(22)
( )=
=−
= −122.2 𝑇/𝑠
𝑑𝑑
𝑁𝐴
180 × 10−4
Q4: A uniform magnetic field is applied perpendicular to the plane of a 90-turn
circular coil with a radius of 6.0 cm and a resistance of 0.8 Ω. If the magnetic
field increases from 0.2 T to 2.1 T in 0.30 s, what is the emf (in V) induced in
that coil?
Answer: 6.45
𝑉=
βˆ†π΅
2.1 − 0.2
𝑁𝐴 =
(πœ‹(0.06)2 )(90) = 6.45 𝑉
βˆ†π‘‡
0.3
Q5: A circular coil of radius r = 5 cm and resistance R = 0.3 Ω is placed in a
uniform magnetic field perpendicular to the plane of the coil. The magnitude of
the field changes with time according to B = 0.6 e -0.1 t T. What is the magnitude
of the current (in mA) induced in the coil at the time t = 3.5 s?
Answer: 1.11
𝐼=
𝑉
𝑁𝐴 𝑑𝐡
=−
( )
𝑅
𝑅 𝑑𝑑
πœ‹(0.05)2
𝐼=−
(−0.1(0.6)𝑒 −0.1(3.5) ) = 1.11 π‘šπ΄
0.3
Q6: You are designing a generator with a maximum emf 9.0 V. If the generator
coil has 200 turns and a cross-sectional area of 0.060 m2, what would be the
frequency (in Hz) of the generator in a magnetic field of 0.035 T?
Answer: 3.41
πΈπ‘šπ‘Žπ‘₯ = 𝑁 𝐴 𝐡 πœ” = 𝑁 𝐴 𝐡 (2πœ‹ 𝑓)
𝑓=
πΈπ‘šπ‘Žπ‘₯
9
=
= 3.41 𝐻𝑧
𝑁 𝐴 𝐡 (2πœ‹) 200 (0.06) (0.035) (2πœ‹)
Q7: A 46-turn circular coil (radius = 6.0 cm, total resistance = 0.12 Ω) is placed
in a uniform magnetic field directed perpendicular to the plane of the coil. The
magnitude of the magnetic field varies with time as given by B = 40 sin(7π t) mT
where t is measured in s. What is the magnitude of the induced current (in A) in
the coil at 0.12 s?
𝐼=
Answer: 3.34
(Use Radians while
solving)
𝐼=−
𝑉
𝑁𝐴 𝑑𝐡
=−
( )
𝑅
𝑅 𝑑𝑑
(46)πœ‹(0.06)2
(7πœ‹(40) cos(7πœ‹(0.12)) × 10−3 ) = 3.34 𝐴
0.12
Q8: In the arrangement shown, a conducting bar of negligible resistance slides
along horizontal, parallel, frictionless conducting rails connected as shown to a
2.0-Ω resistor. A uniform 1.5-T magnetic field is perpendicular to the plane of
the paper. If L = 70 cm, at what rate is thermal energy (in Watt) being generated
in the resistor at the instant the speed of the bar is equal to 4.3 m/s?
𝑃=
𝑉 2 (𝐡𝑙𝑣)2 (1.5(0.7)(4.3))2
=
=
= 10.19 π‘Š
𝑅
𝑅
2
Answer: 10.19
Q9: A circular wire ring is situated above a long straight wire, as shown in the
figure. The straight wire has a current I flowing to the right, and this current is
decreasing at a constant rate. Which of the following statements is true?
a) There is an induced current in the wire ring, directed in a counterclockwise
orientation.
b) There is no induced current in the wire ring.
c) There is an induced current in the wire ring, directed in clockwise orientation.
Q10: In the figure, a bar magnet moves away from the solenoid. The direction of
the induced current through the resistor R is
a) from b to a.
b) The direction of current will keep changing.
c) No current is induced.
d) from a to b.
Q11: A square loop of wire moves with a constant speed v from a field-free
region into a region of uniform B field, as shown. Which of the five graphs
correctly shows the induced current i in the loop as a function of time t?
IV
ο‚· III
II
I
V
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