8th Ed 【CH30】Inductance
9th Ed 【CH30】Induction and Inductance
8th Ed ：Homework of CH 30
1, 5, 7, 11, 13, 15, 21, 29, 31, 34, 37, 39, 45, 47, 51, 57, 63, 65, 73, 75
（圖30-8）
dB A dt
= 22
(423)(82)2d r t t t dt π++=+
独角仙图片
2
(0.2)[8(10)2] 5.152 5.22
V V +=≈
5.1522 1.58 1.6V V A A −=≈蓝印花布图案
where B Φ is in milliwebers and t is in conds. (a) What is the magnitude of the emf induced in
the loop when 2
t s =? (b) Is the direction of the current through R to the right or left?
圖30-37，線圈的磁通量根據267B t t Φ=+的關係增加，其中B Φ的單位為milliwebers 和t 的單位為秒。（a）當2t s =時，線圈的感應電動勢的大小為何？（b）通過電阻R 的電流方向是向左邊或是向右邊？
（圖30-39）
<；解>：The total induced emf is given by
()20002()()1.5 A (120)(4T m A)(22000/m)0.016m 0.025 s 0.16V.
B d dB d di di N
NA NA ni N nA N n r dt dt dt dt dt εμμμπΦ⎛⎞=−=−=−=−=−⎜⎟⎝⎠netlimiter
⎛⎞
=−×⋅⎜⎟
⎝⎠
=-7p 10p
（圖30-40）
The field (due to the current in the straight wire) is out-of-the-page in the upper half of the
circle and is into the page in the lower half of the circle, producing zero net flux, at any time. There is no induced current in the circle.
因長直導線電流產生的磁場，在長直導線的上方是向外出紙面，在長直導線的下方是A square wire loop with 2.00 m sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field as shown in Fig. 30-44.The loop contains an ideal battery with emf 20V ε=. If the magnitude of the field varies with time according to 0.0420.87B t =−, with B in teslas and t in conds, what are (a) the net emf in the circuit and (b) the direction of the (net) current around the loop?
一個正方形線圈邊長2.00米，與均勻磁場垂直，有一半的面積在磁場內，如圖30-44所示。 線圈含有一個理想的電池，電動勢為20V ε=。如果磁場大小變化跟時間的關係為
0.0420.87B t =−，其中B 的單位為tesla 和時間t 的單位為秒（c），問（a）電路的淨電動勢，及（b）線圈淨電流的方向？
（圖30-44）
be the length of a side of the square circuit. Then the magnetic flux through the
commercial alternating-current generator. (b) What value of Nab gives an emf with 0150V ε= when the loop is rotated at 60.0 rev/s in a uniform magnetic field of 0.500 T?
一個N 札的矩形線圈，長度a 和寬度b，在均勻磁場B G
中，以頻率f 旋轉，如圖30-46所示。
维稳ௐ6ࢱĂВ28ࢱ
線圈連接到同部轉動的圓柱，而不是與滑動的金屬刷連接。（a）證明線圈上的感應電動勢
為02sin(2)sin(2)fNabB ft ft εππεπ==。這是商業交流發電機的原理。
（b）問當線圈以60rev/s 速率，在0.5T 磁場中轉動時，怎樣的Nab 大小，可以得到0150V ε=的電動勢。
圖30-48，堅硬的鋼絲彎曲成一個半圓，半徑2a cm =，在均勻20mT 的磁場中，以固定角速率40rev/s 旋轉，問線圈感應電動勢的（a）頻率和（b）振幅？
游戏搞笑名字
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8th
Ed 【Problem 30-21】
：9th
Ed 【Problem 30-23】 Figure 30-50 shows two parallel loops of wire having a common axis. The smaller loop (radius r )
is above the larger loop (radius R ) by a distance x R >>. Conquently, the magnetic field due to the counterclockwi current i in the larger loop is nearly uniform throughout the smaller loop.
Suppo that x is increasing at the constant rate
心之所属
dx
v dt
=. (a) Find an expression for the magnetic flux through the area of the smaller loop as a function of x. (Hint: See Eq. 29-27.) In the smaller
loop, find (b) an expression for the induced emf and (c) the direction of the induced current.
（圖30-50）
(a) In the region of the smaller loop the magnetic field produced by the larger loop may be
taken to be uniform and equal to its value at the center of the smaller loop, on the axis.
ε=−=−
F H
G I K J F
H G
I K
J =−F H G I
K J −F H G I K J =dt
dt x x dt x元宵节古诗词
B 03
4
4
222. (c) As the smaller loop moves upward, the flux through it decreas, and we have a situation
like that shown in Fig. 30-5(b). The induced current will be directed so as to produce a magnetic field that is upward through the smaller loop, in the same direction as the field
of the larger loop. It will be counterclockwi as viewed from above, in the same direction as the current in the larger loop.
8th Ed 【Problem 30-29】：9th Ed 【Problem 30-31】
If 50.0 cm of copper wire (diameter 1mm =) is formed into a circular loop and placed perpendicular
8th Ed 【Problem 30-31】：9th Ed 【Problem 30-29】
In Fig. 30-56, a metal rod is forced to move with constant velocity v G
along two parallel metal rails,
connected with a strip of metal at one end. A magnetic field of magnitude 0.35B T = points out of the page. (a) If the rails are parated by 25L cm = and the speed of the rod is 55.0 cm/s, what emf is generated? (b) If the rod has a resistance of 18Ω and the rails and connector have negligible范文学
resistance, what is the current in the rod? (c) At what rate is energy being transferred to thermal energy?
（圖30-56）
=BLv (.0350T)(0.250m)(0.55(b) By Ohm’s law, the induced current is 0.04810.0026718V
i =
=Ω
current is clockwi in Fig. 30-56. 20.000129i R W =
【Problem 30-34】
E dt r ==×=×−−T /s m
V /m.c
h b g
b g
8th
Ed 【Problem 30-39】：9th
Ed 【Problem 30-39】
The magnetic field of a cylindrical magnet that has a pole-face diameter of 3.3 cm can be varied
