Physics 112
Homework7 (Ch20 &21)
1. Three long parallel wires are 3.8 cm from one another. (Looking along them, they are at three
corners of an equilateral triangle.) The current in each wire is 8.00 A, but its direction in wire
M is opposite to that in wires N and P (see figure below). Determine the magnetic force per
unit length on each wire due to the other two.
Solution
a) The forces on wire M due to the other wires are repelling forces, away from FMP
IIl
the other wires. Use equation F  0 1 2 to calculate the force per unit
2r
length on wire M due to each of the other wires, and then add the force vectors
together. The horizontal parts of the forces cancel, and the sum is vertical.
FM net y
lM


FMN
lM
cos 30o 
FMP
lM
FMN
30o 30o
cos 30o
0 I M I N
 I I
1m  cos 30o  0 M P 1m  cos 30o
2 d MN
2 d MP
 4 10
2
7
TmA
2
 8.00 A 
2
 0.038 m 
cos 30o  5.8  10 4 N m , 90 o
b) The forces on wire N due to the other wires are an attractive force towards
wire P and a repelling force away from wire M. Calculate the force per unit
length on wire N due to each of the other wires, and then add the force vectors
together. From the symmetry, we expect the net force to lie exactly between
the two individual force vectors, which is 60o below the horizontal.


2
Fnet FMN FMP  0 I M I P
4  10 7 Tm / A 8.00 A




 3.4  10  4 N / m ;
l
l
l
2d MP
2 0.038m
c) The forces on wire P due to the other wires are an attractive force
towards wire N and a repelling force away from wire M. From symmetry,
this is just a mirror image of the previous solution, and so the net force is as
follows.
4
FP net  3.4  10 N m
  240
o
FNP
90o
FNM
30o
  300 
FPN
90o
FPM
30o
Physics 112
Homework7 (Ch20 &21)
2. The north pole of the magnet in figure below 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.
3. What is the direction of the induced current in the circular loop due to the current shown in
each part of figure below?
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.
Physics 112
Homework7 (Ch20 &21)
4. Part of a single rectangular loop of wire with dimensions shown in figure below 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 equation,
E  Blv .
Because the inward flux is decreasing, the induced flux will be into the page, so the induced
current is clockwise, given by equation
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 
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
5. 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
Epeak  NB A   
f 
Epeak
NBA

120 V
320  0.650 T  0.210 m 
2
 13.08 rad s
 13.08 rad s

 2.08 rev s
2 2 rad rev
6. What is the inductance L of a 0.60-m-long air-filled coil 2.9 cm in diameter containing
10,000 loops?
Solution
L
0 N 2 A
l
 4 10

7


T m A 10000   1.45 102 m
2
0.60 m

2
 0.14 H
Physics 112
Homework7 (Ch20 &21)
7. 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 equation:
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
8. After how many time constants does the current in figure below 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 
I  I max 1  et    et   1 
 t   ln 1 

I max
 I max 

I 
(a) t   ln 1 
   ln 1  0.9   2.3
 I max 

I 
(b) t   ln 1 
   ln 1  0.99   4.6
 I max 

I 
(c) t   ln 1 
   ln 1  0.999   6.9 .
 I max 
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.
Physics 112
Homework7 (Ch20 &21)
9. At what frequency will a 2.40- F capacitor have a reactance of 6.70 k?
Solution
We find the frequency from eqution 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



10. 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 from equation X L  2 fL .
V  IX L  X L 
V
I
X L  2 fL  L 
XL
2 f

V
2 fI

240 V
2  60.0 Hz 12.8 A
 4.97  10 2 H