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Homework 8
Ch21: P 3, 9, 15, 23, 41, 47, 51, 55, 59
3.
(I) The north pole of the magnet in Fig. 21–47 is being inserted into the coil. In
which direction is the induced current flowing through the resistor R?
Solution
As the magnet is pushed into the coil, the magnetic flux increases to the right. To oppose
this increase, flux produced by the induced current must be to the left, so the induced
current in the resistor will be from right to left.
9.
(II) What is the direction of the induced current in the circular loop due to the
current shown in each part of Fig. 21–49?
Solution
(a) The increasing current in the wire will cause an increasing field out of the page
through the loop. To oppose this increase, the induced current in the loop will
produce a flux into the page, so the direction of the induced current will be
clockwise.
(b) The decreasing current in the wire will cause a decreasing field out of the page
through the loop. To oppose this decrease, the induced current in the loop will
produce a flux out of the page, so the direction of the induced current will be
counterclockwise.
(c) The decreasing current in the wire will cause a decreasing field into the page through
the loop. To oppose this decrease, the induced current in the loop will produce a flux
into the page, so the direction of the induced current will be clockwise.
(d) Because the current is constant, there will be no change in flux, so the induced
current will be zero.
15. (II) Part of a single rectangular loop of wire with dimensions shown in Fig. 21–51 is
situated inside a region of uniform magnetic field of 0.550 T. The total resistance of
the loop is 0.230 . Calculate the force required to pull the loop from the field (to
the right) at a constant velocity of 3.40 m s . Neglect gravity.
Solution
As the loop is pulled from the field, the flux through the loop decreases, causing an
induced EMF whose magnitude is given by Eq. 21-3,
E  Blv .
Because the inward flux is decreasing, the induced flux will be into the page, so the
induced current is clockwise, given by
I E R.
Because this current in the left-hand side of the loop is in a downward magnetic field,
there will be a magnetic force to the left. To keep the rod moving, there must be an equal
external force to the right, given by F  IlB .
F  IlB 
*23.
E
R
lB 
Blv
R
lB 
B 2l 2 v
R
 0.550 T   0.350 m   3.40 m s 
2

2
0.230 
 0.548 N
(II) A simple generator has a 320-loop square coil 21.0 cm on a side. How fast must
it turn in a 0.650-T field to produce a 120-V peak output?
Solution
From Eq. 21-5, the peak voltage is Epeak  NB A . Solve this for the rotation speed.
Epeak  NB A   
f 
Epeak
NBA

120 V
320  0.650 T  0.210 m 
 13.08 rad s

 2.08 rev s
2 2 rad rev
2
 13.08 rad s
*41. (I) What is the inductance L of a 0.60-m-long air-filled coil 2.9 cm in diameter
containing 10,000 loops?
Solution
Use the relationship for the inductance of a solenoid, as given in Example 21-14.
L
 4 10

0 N 2 A
l
7


T m A 10000   1.45 102 m
2
0.60 m

2
 0.14 H
*47. (I) The magnetic field inside an air-filled solenoid 36 cm long and 2.0 cm in
diameter is 0.80 T. Approximately how much energy is stored in this field?
Solution
The magnetic energy in the field is derived from Eq. 21-10.
u
Energy stored
Volume
Energy 
1
2
B2
0

1
2
B2
0
 Volume  

1
2
B2
0
r L 
2
1
2
 0.80 T 
 4 10
7
2
Tm A

  0.010 m   0.36 m   29 J
2
*51. (III) After how many time constants does the current in Fig. 21–33 reach within
(a) 10%, (b) 1.0%, and (c) 0.1% of its maximum value?
Solution
“Within (a) 10%, (b) 1.0%, and (c) 0.1% of maximum value” means: (a) I I max  0.9 ,
(b) I I max  0.9 9 , and (c) I I max  0.9 99 . If we assume that the current is increasing
(battery is connected), than for an LR circuit, we have
I  I max 1  e t   .
Solve for t :

I  I max 1  et 

I
(a) t   ln 1 
 I max

I
(b) t   ln 1 
 I max

I
(c) t   ln 1 
 I max

 e t   1 
I
I max

I 

I max 
 t   ln 1 


   ln 1  0.9   2.3


   ln 1  0.99   4.6


   ln 1  0.999   6.9 .

If we assume that the current is decreasing (battery is disconnected), than for an LR
circuit, we should use:
I  I max e  t /  ,
and we will find different numerical results.
*55. (I) At what frequency will a 2.40- F capacitor have a reactance of 6.70 k?
Solution
We find the frequency from Eq. 21-12b for the reactance of a capacitor.
1
1
1
XC 
 f 

 9.90 Hz
3
2 fC
2 X C C 2 6.70  10  2.40  106 F



*59. (II) An inductance coil operates at 240 V and 60.0 Hz. It draws 12.8 A. What is the
coil’s inductance?
Solution
We find the reactance from Ohm’s law, and the inductance by Eq. 21-11b.
V
V  IX L  X L 
I
X
V
240 V
X L  2 fL  L  L 

 4.97  102 H
2 f 2 fI 2  60.0 Hz 12.8 A