Chapter 29--Examples
1
Problem
a)
Derive the equation relating the total charge
Q that flows through a search coil
(Conceptual Example 29.3 in Section 29.2) to
the magnetic field B. The search coil has N
turns, each with area A and the flux through
the coil is decreased from its initial value to
zero in a time Dt. The resistance of the coil is
R and the total charge is Q=I Dt where I is the
average current induced by the change in
flux
2
From Example 29.3


A search coil is a practical way to measure magnetic
field strength. It uses a small, closely wound coil with
N turns. The coil, of area A, is initially held so that its
area vector A is aligned with a magnetic field with
magnitude B. The coil is quickly pulled out of the field
or rotated.
Initially the flux through the field is F=NBA. When it
leaves the field or rotated, F goes to zero. As F
decreases, there is a momentary induced current
which is measured. The amount of current is
proportional to the field strength.
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Solution

EMF=-DF/Dt

DF=F2 –F1
Where F1 =NBA and F2 =0
 DF=-NBA



EMF=NBA/Dt
V=iR where V=EMF
EMF=NBA/Dt=iR
 i=NBA/RDt


Q=iDt=NBA/R
4
Part B

In a credit card reader, the magnetic strip
on the back of the card is “swiped” past a
coil within the reader. Explain using the
ideas of the search coil how the reader
can decode information stored in the
pattern of magnetization in the strip.
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Solution
The card reader contains a search coil.
 The search coil produces high EMF and
low EMF as the card is swiped.
 A high EMF is treated as a binary 1 and
a low EMF is treated as a binary 0
 Ascii information (character codes
between 1 and 64) can be stored there in
a few bits (6-bits or 26).

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Problem
A circular loop of flexible iron wire has an initial
circumference of 165 cm but its
circumference is decreasing at a rate of 12
cm/s due to a tangential pull on the wire. The
loop is in a constant, uniform magnetic field
oriented perpendicular to the plane of the
loop and with a magnitude of 0.5 T
a) Find the EMF induced in the loop at the
instant when 9 s have passed.
b) Find the direction of the current in the loop as
viewed looking in the direction of the
magnetic field.
7
EMF=-DF/Dt=-B*DA/Dt

F=BA


dF/dt=BdA/dt
c=2pr so A=pr2



dA/dt=d/dt(c2/4p)= (2c/4p)dc/dt
dF/dt= B(c/2p)dc/dt





r=c/2p so A=c2/4p
Where B=0.5
dc/dt=0.12 m/s
At 9s, c=1.65-.12*(9s)=0.57 m
dF/dt=5.44 x 10-3 V
Since F is decreasing, the EMF is positive. Since
EMF positive, point thumb along A and look at fingers.

They curl counterclockwise.
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Problem
Suppose the loop in the figure below is
a)
Rotated about the y-axis
b)
Rotated about the x-axis
c)
Rotated about an edge parallel to the z-axis.
What is the maximum induced EMF in each case if A=600 cm2, w=35
rad/s, and B=.45T?
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Case a—Rotating about y-axis
d
d  
d
EMF = - F = - B  A = - BA cos 
dt
dt
dt
d
but  = t - BA cos t = BA  sin t
dt
EMFmax = BA  = 0.45 * 0.06 * 35 = 0.945V
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Case b—rotating about the x-axis
The normal to the surface is in the same
direction as B during this rotation. Thus
BA=constant
 d/dt(constant)=0 so EMF=0

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Case c—Rotating about z-axis
d
d  
d
EMF = - F = - B  A = - BA cos 
dt
dt
dt
d
but  = t - BA cos t = BA  sin t
dt
EMFmax = BA  = 0.45 * 0.06 * 35 = 0.945V
12
Problem
The figure below 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. Consequently, the magnetic field due to
current i in the larger loop is nearly constant throughout the
smaller loop. Suppose that x is increasing at a constant rate, v.
1)
Determine the magnetic flux though area of the smaller loop as
a function of x.
2)
In the smaller loop find
1)
2)
The induced EMF
The direction of the induced current.
Radius=r
Radius=R
x
i
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Magnetic Field of a Circular Loop
B=
 0iR 2
2(x  R
2
2
)
3

2
 0iR 2
2 x3
By the RH Rule, the
field is upward in the
small ring
B=
 0iR 2
and
A = pr 2
2x3
   0iR 2 2
F = B A =
pr
3
2x
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EMF in smaller loop
   0iR 2 2
F = B A =
pr
3
2x
 0iR 2 2 d  1 
d
d  0iR 2 2
EMF = - F = pr = pr
 
3
dt
dt 2 x
2
dt  x 3 
d1
3 dx
dx
but
=v
 3=- 4
dt  x 
x dt
dt
EMF =
 0iR 2
2
pr 2 v
3
x4
15
Direction of current
Assume that normal to smaller loop is
positive upwards
 B, then, is in positive direction
 But F is decreasing or has a negative d
dF /dt
 A negative * negative=positive so EMF is
+
 RH Thumb in up direction, fingers curl
counter clockwise!

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